Dietmar Gross · Wolfgang Ehlers Peter Wriggers · Jörg Schröder Ralf Müller
Dynamics – Formulas and Problems Engineering Mechanics 3
1 3
Dynamics – Formulas and Problems
Dietmar Gross Wolfgang Ehlers Peter Wriggers Jörg Schr öder Ralf Müller •
•
Dynamics – Formulas and Problems Engineering Mechanics 3
1 3
Dietmar Gross Division of Solid Mechanics TU Darmstadt Darmstadt Germany
Jörg Schröder Institute of Mechanics Universität Duisburg-Essen Essen Germany
Wolfgang Ehlers Institute of Applied Mechanics Universität Stuttgart Stuttgart Germany
Ralf Müller Engineering Mechanics TU Kaiserslautern Kaiserslautern Germany
Peter Wriggers Institute of Continuum Mechanics Leibniz Universität Hannover Hannover Germany
ISBN 978-3-662-53436-6 DOI 10.1007/978-3-662-53437-3
ISBN 978-3-3-662-53437-3
(eBook)
Library of Congress Control Number: 2016951667 © Springer-Verlag
Berlin Heidelberg 2017 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, speci�cally the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on micro�lms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a speci �c statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and info inform rmat atio ion n in this this book book are are beli believ eved ed to be true true and and accu accura rate te at the the date date of publication. Neither the publisher nor the authors or the editors give a warranty, express or implied, with respect to the material contained herein or for any errors or omissions that may have been made. Printed on acid-free paper This Springer imprint is published by Springer Nature The registered company is Springer-Verlag GmbH Germany The registered company address is: Heidelberger Platz 3, 14197 Berlin, Germany
Preface This 3rd volume of the Formulas and Problems concludes the series to the basic courses in Engineering Mechanics. Experience shows that the field of Dynamics is particularly difficult for students, because besides the concept of force now additional kinematic quantities occur, which must be brought into relation with each other and with the forces. Therefore, with numerous purely kinematical problems, we tried to deepen the understanding of the relevant geometric quantities and their description in different coordinate systems. Likewise, only by exercises, i.e. by an independent treatment of problems, one can gain experience, which basic principle leads to the solution in the simplest way. Often there are several approaches possible. Therefore we demonstrate this frequently so that the reader can realize the advantages and disadvantages of the alternatives. As in the 1st and 2nd volume, we deliberately placed the emphasis on the principal way how to to apply the theory and not in numerical results. The correct formulation of the relevant basic equations and their solution is in the beginning much more important than numerical calculations without a deeper understanding of the background. Experience also shows that it is an illusion to believe that simply reading and trying to comprehend the presented solutions leads to an understanding of the theory. Neither does it improve the problem solving skills. Therefore, we strongly recommend that the reader first tries to solve the problems independently, possibly by using other approaches. Let us emphasize that a collection of formulas and examples is only an additional aid when studying mechanics and it cannot replace a textbook. When the reader is not familiar with one or the other formula or concept, it is necessary to brush up the theory with the help of a textbook; a number of titles can be found in the list of references. Darmstadt, Stuttgart, Hannover, Essen and Kaiserslautern, Summer 2016
D. Gross W. Ehlers P. Wriggers J. Schr ¨ oder R. M uller ¨
Table of Contents Literature, Notation ...............................................
IX
1
Kinematics of a Point..............................................
1
2
Kinetics of a Point Mass ..........................................
29
3
Dynamics of a System of Point Masses .......................
59
4
Kinematics of Rigid Bodies ......................................
77
5
Kinetics of a Rigid Body ..........................................
99
6
Impact ................................................................ 147
7
Vibrations ............................................................
169
8
Non-Inertial Reference Frames .................................
195
9
Principles of Mechanics ...........................................
211
10
Hydrodynamics .....................................................
227
IX
Literature Textbooks
Gross, D., Hauger, W., Schr¨ oder, J., Wall, W., Govindjee, S., Engineering Mechanics 3, Dynamics, 2nd edition, Springer 2014 Gross, D., Hauger, W., Wriggers, P., Technische Mechanik, vol 4: Hydromechanics, Elements of Avanced Mechanics, Numerical Methods (in German), 9th edition, Springer 2014 Beatty, M.F., Principles of Engineering Mechanics, vol 2: Dynamics, Springer 2005 Beer, F., Johnston, E.R., Cornwell, P., Vector Mechanics for Engineers: Dynamics, 10th edition, McGraw-Hill Education 2012 Hibbeler, R.C., Engineering Mechanics: Dynamics, 14th edition. Pearson 2016 Meriam, J.L., Kraige, L.G., Bolton, J.N., Engineering Mechanics: Dynamics, 8th edition, Wiley 2016 Plesha, M., Costanzo, F., Gray, G., Engineering Mechanics: Dynamics, 2nd edition, McGraw-Hill 2012 Pytel, A., Kiusalaas, J., Engineering Mechanics: Dynamics, 4th edition, Cengage Learning 2016 Shames, I.H., Engineering Mechanics: Dynamics, 4th edition, Pearson 1996 Collection of Problems
Beer, F., Johnston, E.R., Cornwell, P., Vector Mechanics for Engineers: Dynamics, 10th edition, Solution Manual, McGraw-Hill 2012 Nelson, E.W., et al. Engineering Mechanics - Dynamics, 765 fully solved problems, Schaum’s Outlines, McGraw-Hill Education 2010 Gray, G.L., Costanzo, F., Plesha, M.E., Solutions Manual, Engineering Mechanics: Dynamics, 1st edition, McGraw-Hill 2009 Hibbeler, R.C., Practice Problems Workbook, Engineering Mechanics: Dynamics, Pearson 2015
Notation In the problem solutions the following symbols are used:
↑:
A:
abbreviation for equation of motion (impulse law) in direction of arrow. abbreviation for angular momentum theorem relative to point A with given positive rotation direction. abbreviation for it follows.
Chapter 1
Kinematics of a Point
1
2
Kinematics
The position of a point P in space is described by the position vector ds dr
r (t) .
As P moves, its path is given by r(t). From the displacement dr of point P in a neighboring position during time dt follows its velocity v
dr = = r˙ . dt
P
r
z
y x
s
path
The velocity is always tangent to the path (trajectory). With the arc– length s and dr = ds the speed of P is given by
| |
v=
ds = s˙ . dt
The change of the velocity vector dv(t) during time dt is called acceleration a =
dv ¨. = v˙ = r dt
The acceleration generally is not directed tangent to the path (trajectory)! The vectors r , v and tems as follows:
a can
be represented in different coordinate sys-
a) Cartesian Coordinates with the unit vectors r v
ex , ey , ez :
= x ex + y ey + z ez ,
z
= x˙ ex + y˙ ey + z ˙ ez ,
a =
x ¨ ex + y¨ ey + z¨ ez .
(¨ r
− rϕ˙ ) e
r
y
y er , eϕ , ez :
z
r
z
= r˙ er + rϕ˙ eϕ + z ˙ ez ,
a =
x
P
= r er + z ez ,
2
z
ey ex
b) Cylindrical Coordinates with the unit vectors
v
r
ez
x
r
P
path
eϕ
ez
ϕ
+ (rϕ ¨ + 2r˙ ϕ) ˙ eϕ + z¨ ez . x
y
er
r
path
x
y
3
of a point
c) Serret-Frenet Frame with the unit vectors et , en , eb in tangential, principal normal and binormal direction. C v
= v et , ρ
2
a =
v˙ et +
v ρ
en
.
ds dt dv at = v˙ = dt 2 v an = ρ v = s = ˙
path
en
s
Here are: ρ
ρ
P
et
=
radius of curvature (distance between P and center of curvature C ),
=
speed,
=
tangential acceleration,
=
normal acceleration (centripetal acceleration).
two acceleration components at , an are located in the so-called osculating plane . The acceleration vector points always to the ’interior’ of the path. Remarks: The
Rectilinear motion Position Velocity Acceleration
x(t) ,
P
0
dx v= = x˙ , dt dv a = = v˙ = x ¨. dt
x
x(t)
Circular motion (r = const) Position Velocity
s = rϕ(t) , ˙ v = r ϕ = rω ,
Tangential acceleration
at = r ϕ = r ¨ ω˙ ,
Centripetal acceleration
v2 an = = rω 2 r
with ω = ϕ = ˙ angular velocity.
P r
ϕ(t)
s(t)
4
Kinematic Basic Problems
Planar motion in polar coordinates From the relations for cylindrical coordinates follow for z = 0, ϕ = ω ˙ v
= v r er + vϕ eϕ ,
a
= a r er + aϕ eϕ
with path
radial velocity
vr = r˙ ,
circular velocity
vϕ = rω ,
radial acceleration
ar = r¨
circular acceleration
aϕ = r ω + ˙ 2rω ˙ .
y
− rω
2
P r
er
,
ϕ
eϕ
x
case of a central motion the circular acceleration vanishes. From aϕ = r ω˙ + 2rω ˙ = (r 2 ω) /r = 0 then follows the ’Law of Equal Areas’ (Kepler’s 2nd Law) r2 ω = const . Remark: In
·
Kinematic basic problems for a rectilinear motion At initial time t0 the initial position x0 and initial velocity v0 are assumed to be given. Given
Sought
a = 0
v = v 0 = const , x = x 0 + v0 t uniform motion
a = a 0 = const
v = v 0 + a0 t , x = x 0 + v0 t + 21 a0 t2 uniform acceleration
t
v = v 0 +
a = a(t)
x = x 0 +
t0
v
t = t 0 +
a = a(v)
v0
t
a(t¯)dt¯,
v(t¯)dt¯
t0
d¯ v = f (v) , a(¯ v)
t
x = x 0 +
F (t¯)dt¯
t0
with the inverse function v = F (t) a = a(x)
v
2
= v 02
x
a(¯ x)d¯ x,
+2
x0
x
t = t 0 +
x0
d¯ x = g(x) v(¯ x)
the inverse function of t = g(x) gives x = G(t) Remarks:
• •
The relations above can also be used for a general motion by replacing x through s and a through the tangential acceleration at . The normal acceleration then follows from an = v 2 /ρ. If the velocity is given as a function of the position, the acceleration is found from dv d v2 a = v = ( ) . dx dx 2
Rectilinear Motion
5
Problem 1.1
The minimum distance b between two vehicles shall be as big as the distance which the rear vehicle covers within ts = 2 s at its constant velocity. a) Determine the minimum distance xp required for passing. b) Determine the minimum time tp a car (length l1 = 5 m, constant speed v1 = 120 km/h) needs staying on the fast lane for passing a truck (length l2 = 15 m, speed v2 = 80 km/h) correctly? Disregard the time for changing the lanes.
Solution
l1
b1
l2
b2
x2
l1
x p
a) For uniform motion the minimum distances follow with 1 km/h = 1000 m/3600 s as b1 = v 1 ts =
120 200 2= m , 3.6 3
b2 = v 2 ts =
·
80 400 2= m. 3.6 9
·
Thus, the required distance for passing is given by xp = b 1 + l2 + x2 + b2 + l1 . Furthermore, the relations x2 = v 2 tp ,
xp = v 1 tp
hold. Elimination of tp yields b1 + b2 + l1 + l2 xp = = 1 vv21
−
b)
200 + 400 + 5 + 15 1180 3 9 = = 393, 33 m . 80 3 1 120
−
Thus, the minimum time for passing is tp =
xp 1180 3, 6 = = 11, 8 s . v1 3 120
·
·
P1.1
6 P1.2
Rectilinear
Problem 1.2
To simulate absence of gravity, vacuum drop-shafts are used. Given is a shaft with a depth of l = 200 m. Determine the maximum available test time t1 and test distance x1 during free fall, when the sample after passing the test distance is decelerated with aII = 50 g to v = 0?
−
Solution Because
the sample is released from rest (x0 = v0 = 0), during free fall with aI = const = g, the velocity and position are g vI = gt , xI = t2 . 2
sample t = 0 x
(aI = g) x1 l
During the deceleration phase with aII = 50 g, velocity and position are given by
−
vII = v II
− 50 gt ,
t = t 1 (aII = −50 g) t = t 2
0
xII = x II + vII t 0
0
2
− 50 gt /2 .
It shall be noted that the integration constants xII and vII have no direct physical meaning. 0
0
For t = t 2 the following conditions must hold: vII (t2 ) = 0
vII = 50 gt 2 ,
xII (t2 ) = l
xII = l
0
0
−v
50 2 gt 2 = l 2
II 0 t 2 +
− 25 gt
2 2
.
From the transition conditions vI (t1 ) = v II (t1 )
xI (t1 ) = x II (t1 )
gt 1 = 50 g(t2 g 2 t1 = l 2
− t ) , 1
− 502 gt + 50 gt t − 502 gt 2 2
1 2
the test time t1 and subsequently x1 can be determined: t1 =
100 l = 51 g
x1 = x I (t1 ) =
100 200 = 6.32 s , 51 9.81
·
·
g 2 g 100 l 50 t1 = = l = 196 m . 2 2 51 g 51
2 1
Motion
7
Problem 1.3
Between 2 stations an underground covers a distance of 3 km. Given are the starting acceleration aa = 0.2 m/s2 , the braking deceleration ad = 0.6 m/s2 and the maximum speed v ∗ = 90 km/h.
−
Determine the acceleration distance, the deceleration distance, the distance during uniform motion and the travel time. Solution
From the constant acceleration aa within the starting phase the velocity follows as va = a a t . With the given maximum speed we obtain the starting time v∗ 90 1000 ta = = = 125 s aa 3600 0.2
·
·
and the acceleration distance sa =
1 1 aa t2a = 0.2 1252 = 1563 m . 2 2
·
·
During braking with constant deceleration ad the velocity is given by v = v ∗ + a t . d
d
Thus, the time td until stop (vd = 0) is td =
−
v∗ = ad
90 · 1000 − 3600 · (−0.6) = 41.67 s ,
and for the associated braking distance follows 1 90 1000 ad t2d = 41, 67 2 3600 = 1041.75 520.92 = 521 m .
·
sd = v ∗ td +
·
− 21 · 0.6 · 41.67
2
−
For the phase with constant velocity v ∗ remains a distance of s∗ = 3000 s s = 916 m
− − a
d
and an associated time s∗ 916 3600 ∗ t = ∗ = = 36.64 s . v 90 1000
· ·
Thus, the total travel time is T = t a + t∗ + td = 203.31 s = 3.39 min .
P1.3
8 P1.4
Rectilinear
Problem 1.4
A car driver approaches a traffic light with the speed of v0 = 50 km/h. At a distance of l = 100 m the lights turn to ’Red’. The ’Red’ and ’Yellow’ phase takes t∗ = 10 s. The driver wants passing the traffic lights just when the lights turn back to ’Green’. a) Determine the necessary constant deceleration a0 , when the driver is braking along the entire distance? b) Determine the velocity v1 of the car when arriving at the lights? c) Draw the diagrams a(t), v(t) and x(t).
Solution
For constant acceleration a0 we have with x(t = 0) = 0 v0
v = v 0 + a0 t , t2 x = v 0 t + a0 . 2
l x
a) The 2nd equation leads with the condition x(t∗ ) = l to 2 a0 = ∗2 (l t
−
2 v0 t ∗ ) = 2 10
50 1000 10 3600
· 100 −
·
=
−0.78 sm . 2
The negative sign indicates that the car decelerates.
b) With the now known deceleration during braking, the 1st equation yields v1 = v(t∗ ) = 50 = 6.09
· 1000 − 0.78 · 10 3600
m km = 21.9 . s h
a [m/s2 ]
t [s]
-0,78 50
c) Integration of the constant acceleration yields a linear velocity plot, a second integration a parabolic path-time diagram.
10
v [km/h] 21,9
x [m]
10
t [s] 100
10
t [s]
Motion Problem 1.5
A vehicle moves according to the given speed-time diagram.
P1.5
v [m/s] 20
Determine the occuring accelerations, the covered distance and draw the diagrams x(t), a(t), v(x) and a(x). Solution
9
300 360
100
t [s]
It is advantageous to devide the motion into 3 time sections: t1 0
t2
t3 300
100
t [s]
360
1. Section 0 t1 100 s (starting with constant acceleration a1 ): From v1 = a 1 t1 follows v (100) 20 = 1 m/s2 , a1 = 1100 = 100 x1 = 21 a 1 t21 , 5
≤ ≤
s1 = x 1 (100) = 2. Section 0 results
1 2
·
2 1 (100) 5
= 1000 m ,
√ 2a x
v1 (x1 ) =
1 1
.
≤ t ≤ 200 s (uniform motion): From v = 20 m/s = const 2
a2 = 0 ,
2
x2 = v 2 t2 ,
s2 = x 2 (200) = 20 200 = 4000 m .
·
3. Section 0 t3 60 s (braking with constant deceleration a3 ): With v3 = 20 m/s + a3 t3 we obtain 2 2 1 1 a3 = 20 = m/s , x 3 = 20 t3 + 2 a3 t3 , 3 60
≤ ≤
−
−
s3 = x 3 (60) = 20 60
1 2
· − ·
2 1 3 (60)
v3
= 600 m ,
√ = 400 + 2a x
3 3
In total, the vehicle covers the distance s = s 1 + s2 + s3 = 1000 + 4000 + 600 = 5600 m . a [m/s2 ] a1
a1
300 360 100
5600 5000
a [m/s2 ]
a3
t [s]
5000 5600
1000
x [m]
a3
x [m]
v [m/s] 20
1000
x [m] 100
300 360 t [s]
1000
5000 5600
.
10 P1.6
Rectilinear
Problem 1.6
Taking air resistance into account, the acceleration of a free falling body can be described approximately by a(v) = g αv2 . Here g is the gravity acceleration and α a constant.
−
Determine the velocity v(t) of the body that is released from rest. Solution
According to the table on page 4 we have for a given a(v)
v
t = t 0 +
v0
d¯ v . g α¯ v2
−
If the motion starts at t0 = 0 we obtain with the initial condition v(t0 ) = v0 = 0
− − − − √ − v
t =
0
d¯ v 1 = g α¯ v2 α
v
0
d¯ v
(
g α
v¯)(
g α + v¯)
and after partial fraction decomposition 1 t = α
=
v
1
2
g α
1 2 gα
1 g α
0
ln(
g α
1 d¯ v g α + v¯
+
v¯
v
g + v¯) α
v¯) + ln(
=
0
1 ln √ 2 gα
Solving for v yields
√ 2 gα t e =
−
g +v α g α v
v=
−
g +v α . g α v
√ g e 1 √ . α e2 gα t + 1 2 gα t
−
eϕ e−ϕ e2ϕ 1 With the hyperbolic function tanhϕ = ϕ = 2ϕ the result e + e−ϕ e +1 also can be written as v g v= tanh gα t . g α
−
√
−
α
From this last representation the limit value lim v(t) = t→∞
t
g/α ,
can be recognized, i.e. after sufficiently large time the body practically falls with constant speed (a = 0 v = g/α).
Motion
11
Problem 1.7 On
account of the gas expansion, a piston (diameter d) moves V 0 in a cylinder. Here, the acceleration d p0 a of the piston is proportional to the current gas pressure p, i.e. a = c0 p, where l0 for the gas pressure Boyle-Mariotte’s gas law pV = const is valid. The initial state is given by the pressure p0 , the piston location l0 and the initial velocity v0 = 0. Determine the velocity v of the piston as a function of its position. Solution
For a piston displacement x we obtain, according to the gas law p0 V 0 = pV , the relation πd 2 πd2 (l0 + x) , p0 l0 = p 4 4
x
l0 V
where the pressure is given by
p
l0 p = p 0 . l0 + x This leads to the acceleration a(x) = c 0 p = c 0 p0
1 1 = a 0 . 1 + x/l0 1 + x/l0
Here a0 = c0 p0 is the initial acceleration at x = 0. Using the table on page 4 for a given a(x), the velocity is determined by
x
v 2 = v 02 + 2
x
a(¯ x)d¯ x = 2
x0
v=
0
a0 x d¯ x = 2l0 a0 ln 1 + 1+x ¯/l0 l0
2 l0 a0 ln 1 +
a
x l0
.
v
a0
2
4
6
8
x/l0
2
4
6
8
x/l0
Remark: Because the acceleration decreases with increasing x, the velocity increase drops continuously.
P1.7
12 P1.8
Rectilinear Motion
Problem 1.8 The
acceleration of a point P Z P , moving along a straight line, is directed to the point Z and its magnitude is x inverse proportional to the distance x. For t = 0 the point P has the distance x 0 = 2 m, the velocity v0 = 4 m/s and the acceleration a0 = 3 m/s2 .
−
a) Determine the velocity v1 for the distance x1 = 3 m. b) At what distance x2 the velocity is zero?
Solution
a=
−
According to the problem description, the acceleration is c/x, where c can be determined from the given initial conditions:
c =
−a x
0 0
=
−(−3) · 2 = 6 (m/s)
2
.
Knowing a(x), the velocity is obtained from (see table on page 4)
x
v 2 = v 02 + 2
x
a(¯ x)d¯ x = v 02 + 2
x0
v(x) =
a)
± − v02
−
b)
− 2c ln xx
x0
0
2c ln x/x0 .
− 16
12 ln
3 m = 3, 34 . 2 s
The velocity is zero for: v=0
v02 2c ln
−
x =0 x0
Remarks:
•
c d¯ x = v 02 x ¯
Hence, the velocity for x1 = 3 m results as v1 = (+)
•
−
The velocity-position diagram is symmetric with respect to the x-axis. The motion can also take place in the domain of negative x. Because of the discontinuity at x = 0, the equations then must be formulated new, considering the direction change of a.
2 0
x2 = x 0 ev
/2c
= 2 e4/3 = 7, 59 m .
v v0 1
x x0
1 2
ev0 /2c
Vertical Throw
13
Problem 1.9
A ball is thrown vertically upwards with an initial velocity v01 = 20 m/s. Two seconds later, a second ball likewise is thrown vertically upwards with v02 = 18 m/s. Determine the height H where the two balls meet.
Solution
We start counting time t when launching the 1st ball. Considering the given initial values, the direction of gravity g and the time difference ∆t = 2 s, we have z 1 = v 01 t
②
z 1
g 2 t , 2
−
z 2 = v 02 (t
①
z 2
− ∆t) − 2g (t − ∆t)
2
.
From the condition z 1 = z 2 for meeting, the meeting time t∗ is obtained: v01 t∗
− 2g t∗
2
= v 02 (t∗
− ∆t) − 2g (t∗ − 2t∗ ∆t + ∆t ) 2
2
1 g∆t ∆t v02 + 2 t∗ = = 3.16 s . v02 v01 + g∆t Introducing t∗ into the equation for z 1 (or z 2 ) yields the height:
−
H = z 1 (t∗ ) = 20 3.16
·
2
− 4.9 · 3.16
= 14.27 m .
The solution can be illustrated by the position-time diagram:
z [m] H 1max = 20.38
z 2 (t)
H = 14.27 10
v01 t
z 1 (t)
v02 t
2
t = 3.16 ∗
t [s]
P1.9
14 P1.10
Vertical Throw
Problem 1.10
From the rim of a cli cliff, ff, 50 m abo above ve the sea level, level, a bal balll is thrown vertically upwards with an initial velocity of 10 m/s. a) Determine the maximum height the ball reaches above sea level. b) When the ball impinges on sea surface? c) Determine the velocity of the ball when impinging on sea surface.
Solution
With a = velovity v0 we have v = z ˙ = v 0 z = v 0 t
−
−g and the initial H
− gt ,
z
g 2 t . 2
hmax
a) The The rise rise time T follows T follows from the condition v(T T )) = 0: v0
− gT = 0
T =
h = 50 m
v0 . g
Therefore, the throw the throw height H H is is given by v02 H = z (T T )) = g
−
v02 v02 = . 2g 2g
With v0 = 10 m/s the maximum height hmax is obtained as hmax
v02 102 = h + H = h + = 50 + = 55 55..1 m . 2g 2 9.81
·
b) The time ti until the ball impinges on sea surface is obtained from the condition z (ti ) = h :
−
−h = = v v t − 0 i
g 2 t 2 i
1 ti = g
− v0 ( + )
v02 + 2gh 2gh
= 4.37 s .
Note: The formally possible minus sign in front of the root is inapplicable. It would lead to a negative time ti .
c) With t = = t t i the impact velocity follows as v (ti ) = v 0
− gt
i
= 10
m − 9.81 · 4.37 = −32 32..87 . s
The minus sign indicates that the velocity is downwards directed, i.e. opposite to the chosen coordinate z .
Vertical Throw
15
Problem 1.11
The cr The crew ew of a ba ball lloo oon, n, th that at is mo movi ving ng up upw war ards ds in a cloud with constant speed v0 , wants to determine the current hight h0 above ground. For this purpose a gauging member is released from the gondola that falls down and explodes when hitting the ground. After time t1 the crew hears the detonation. Determine Determ ine the hei heigh ghtt h0 for the fo foll llo owi wing ng da data: ta: v0 = 5 m/s, g = 2 9.81 m/s , t1 = 10 s, c = 330 m/s (speed of sound). Solution
Introducing x downwards from the position where the member is launched (t (t = 0), the falling time t time t m until hitting the ground is obtained from 1 x(tm ) = g t2m 2 as tm =
v0 g
−v t
0 m
1( + )
1+
−
t = t1
= h 0
2gh 0 v02
v0 t1
. t=0
Only the positive root is meaningful since tm must be positive. The sound covers the distance h0 + v 0 t1 because during time t1 the baloon is rising the distance v0 t1 . Th Ther eref efor oree the so sound und re requi quire ress the time ts =
x h0
sound distance h0 + v0 t1 = . sound speed c
The total time is given by the sum of falling time and the time of sound: t1 = t m + ts =
v0 g
1+
1+
2gh 0 v02
+
h0 + v0 t1 . c
After rearranging and squaring we obtain h0 = =
c
− − − v0
g
325 9, 81
gt 1 + c 1( + )
98,, 1 + 330 1 98
1+2
gt 1 c v0
1+2
−
98, 1 98, 325
= 338 m . m .
The solution with the positive root leads to a mechanically unreasonable result.
P1.11
16 P1.12
Circular Motion
Problem 1.12
The tw two poin p oints ts P A and P B start start the their ir mo moti tion on al along ong a ci circ rcul ular ar path at the same time t = 0 at A and B , respectively.. Poin respectively Pointt P A move movess wi with th the given giv en con consta stant nt spee speed d v0 and poi point nt P B starts from rest with a constant angular accele acc elerat ration ion ω˙ 0 .
P B
B r C
a) Deter Determi mine ne ω˙ 0 such that both points arrive at C C at at the same time. b) Calculate the angular velocity of P B when passing A?
A
P A
c) Determine the normal acceleration of both points at C .
Solution
For or P P A follows from v from vA = s˙ A = const = v = v 0 and by considering the initial condition sA (0) = 0, sA = v 0 t . For P B we we ob obta tain in from from ω˙ B = ϕ ¨B = const = ω˙ 0 with ϕB (0) = 0, ϕ˙ B (0 (0)) = 0 ϕB =
1 ω˙ 0 t2 , 2
ϕ˙ B = ω B = ω˙ 0 t
sB =
1 ω˙ 0 r t2 , 2
vB = ω˙ 0 r t .
a) Because both points shall arrive at the same time tC at at C C ,, it follows with the different distances sA (tC ) = π πr/ r/2 2 and sB (tC ) = 3πr πr/ /2 1 πrr = v 0 tC , , π 2
3 1 πr = ω˙ 0 r t2C 2 2
πr tC = , 2v0
12v02 12v ω˙ 0 = . πr 2
b) With With the an angl glee π between B and A and and the know known ω˙ 0 we we ca can n calculate the time tA when P B arrives at A and the respective angular velocity: t2A
2π π 2 r2 = = ω˙ 0 6v02
tA =
πr 6v0
√
√ v
ωB (tA ) = ω˙ 0 tA = 2 6
0
r
.
c) The normal accelerations at C are are found by using the relation an = 2 2 rω = v /r /r:: anA
v02 = , r
2 anB = ω B (tC ) r
=
ω˙ 02 t2C r
v02 = 36 . r
Circular Motion Problem 1.13
At the fixed poin pointt A a point P P starts moving along a circular path with radius r. Its motion is described by the relation s = = c c t2 .
17 P1.13
y B
P r
ϕ
s
Determine: A x a) the ve veloc locit ity y co compo mponen nents ts vx (t) and vy (t), b) the velocity at point B , c) the tangential acceleration at (s) and the normal acceleration an (s). Solution
From s = c t2 the speed follows as
v = s˙ = 2c t .
a) Because the velocity is always tangential to the path, its cartesian components at an arbitrary point are vx =
−v sin ϕ ,
vy = v cos ϕ .
Hence, with s c t2 ϕ = = r r
−
vy ϕ
vx s ϕ
we obtain vx =
v
c t2 2c t sin , r
c t2 vy = 2c t cos . r
b) In point B we have πr s(tB ) = = c t2B 2 v (tB ) = 2c tB = 2c
πr = 2c
tB =
√
r
πr , 2c
2πrc .
c) From at = v˙ ,
v2 4c2 t2 an = = , r r
with v˙ = 2c and c t2 = s s,, follow the results 4cs at = 2c , an = . r Remark: While the tangential acceleration remains constant, the normal Remark: While acceleration increases linearly with s.
18 P1.14
Circular Motion
Problem 1.14
A point M M moves along a half circle. The proj ojeection of its mot otio ion n on th thee di diam amet eter er AB is is an uniform motion with the speed c.
v M A
r ϕ
B
Determine a) the speed v (ϕ) and the magnitude a(ϕ) of the acceleration, b) the angle between the acceleration vector and the diameter AB AB.. Solution
a)
From the condition condition v sin ϕ = = c c follows v
c v (ϕ) = . sin ϕ
ϕ
M
The acceleraton components at and an are determined with rϕ˙ = = v v as
c ϕ
dv dv c v at = = ϕ˙ = cos ϕ = dt dϕ r sin2 ϕ v2 c2 an = = . 2 r r sin ϕ
−
−
c2 cos ϕ , r sin3 ϕ
This finally leads to a =
|a| =
a2t
c2 = r sin3 ϕ
b)
2
+
a2n
c = r
cos2 ϕ 1 + sin6 ϕ sin4 ϕ
2 c cos2 ϕ + sin ϕ = . r sin3 ϕ 2
From the the figure figure can be seen: seen: at
2
tan ψ = tan(ϕ tan(ϕ =
−
at α) = = an
cos ϕ − cr sin ϕ 3
2
c r sin2 ϕ
− cot ϕ = + ta tan( n(ϕ ϕ − π/ π/2) 2)
M
α ϕ
π α = , 2
ψ
a
an
i.e. the acceleration vector is perpendicular to AB AB.. Remark: the latter result can also be found without any calculation: if a component of v is constant then there is no acceleration in this direction, i.e. a is perpendicular to the direction of this component.
Circular Motion
19
Problem 1.15
A bik bikee B starts moving at A from rest along a circular path (radius r). It accelerates according to the law at (v ) = a0 (1 κv κv)) until the speed vC is re reac ached hed at C . Here a0 is the in initi itial al acceleration and κ is a constant. a) Determine the acceleration time tC and b) the acceleration distance sC . c) Calculate the normal acceleration an at at t t = = t t C /2 for κ = 1/(2 (2vvC ).
P1.15
−
vC
C B s
r
A
a) From the given given acceleratio acceleration n law at (v ) = v˙ = a 0 (1 follows with the initial velocity v0 = 0 and t0 = 0 (see page 4) Solution
1 t(v ) = a0
v
0
d¯ v = 1 κv¯
−
− κv κv))
− a1κ ln(1 − κv κv)) . 0
The acceleration time is determined from the condition v (tC ) = v C : tC = t t((vC ) =
− a1κ ln(1 − κv
C ) .
0
b)
The inver inverse se function function of t of t(v ) is given by e−a
0
κt
=1
− κv
−
1 v (t) = 1 κ
e−a
0
κt
.
This leads by integration and by considering the initial condition s0 = s(0) = 0 to
t
s(t) =
0
1 v (t¯)dt¯ = a0 κt + e−a 2 a0 κ
0
κt
and thus to the acceleration distance 1 sC = s s((tC ) = a0 κ2
−
ln(1
− κv
C ) +
1
− κv
C
.
c) The normal acceleration within the time span 0
− − −
v2 1 an = = 2 1 r rκ
e−a
0
κt
2
≤t≤t
a is
given by
.
Inserting t = = t t C /2 and κ = 1/2vC , we obtain 2 4vC an = 1 r
1
vC 2vC
1/2 2
− √
2 4vC 3 = r 2
2 vC 2 = 0.343 . r
20 P1.16
Circular Motion
Problem 1.16
A po poiint mass P is rele re leas ased ed fr from om re rest st at A on a circular path in a vertical plane. Due to gravity it experiences a tangential acceleration g cos ϕ.
Determine the velocity and the magnitude of acceleration in dependence on the angle ϕ.
Solution
A
ϕ R
s P
g
at B
From the tangential acceleration
at = g cos ϕ = v˙ = a t (ϕ) follows with s = = R R ϕ and the initial velocity v0 = 0 (see page 4) ϕ
v2 = 2
g cos ϕ¯ R dϕ ¯ = 2gR sin ϕ
0
v=
2gR sin ϕ .
Hence, the acceleration components are at = g cos ϕ ,
v2 an = = 2g sin ϕ . R
Therewith, the magnitude of acceleration is given by a =
|a| =
a2t + a2n = g
a = = g g
cos2 ϕ + 4 si sin n2 ϕ
1 + 3 si sin n2 ϕ .
Remarks:
•
At A the acceleration components are at = g and an = 0. (Point P has solely a tangential acceleration.) P has
•
At B we have at = 0 and an = 2g . (Pure normal acceleration upwards directed.)
•
With the height height difference difference h h = = R R sin ϕ the velocity can be represented by v = 2gh gh..
√
Circular Motion
21
Problem 1.17
A satel satellite lite S is is moving along a circular path around the earth if its normal acceleration just equals the gravitational acceleration g R2 /r2 (gravity on earth surface g = 9.81 m/s2 , earth radius R = 6370 km).
P1.17
S r
a) In whi hicch dis dista tanc ncee H H above above ea eart rth h su surf rfac acee a satellite is circling that has the spe peeed of 25000 km/h?
R
b) Determine the required speed of a satellite whose orbit lies 1000 km above earth surface. c) What time requires a satellite for 1 circle (orbital period) in a height of H = = 400 km? d) Determine the height of a geostationary satellite. Solution
a) From the necessary normal acceleration
an = v 2 /r = g R2 /r 2 follows with the given speed R2 r = g 2 v
H = r
−
R R = = R R g 2 v
−
1 = 1884 km . km .
b) From the same equation with a given distance yields v = R
g = R r
g = 26457 km/h 26457 km/h . R + H
c) With the given speed v speed v = R g/r and the arc length L length L = = 2πr of the orbit, the orbital period is obtained: L r3/2 T = = 2π = 5547 5547 s = 1. 1.54 h . v R g
√
d) A geostationary satellite has the same angular velocity as the earth: ωE = 2π/ π/(24 (24 h) . h) . Thus, we have 2 an = r rω ωE
R = g r
2
or numerically evaluted
R2 r= g 2 ωE 2
(6370) 2 r = 9.81 10−3 (3600)2 24 (2π (2 π )2 H = r R 36000 km .
·
·
− ≈
1/3
1/3
= 4.22 104 km ,
·
22 P1.18
Planar Motion
Problem 1.18
A point is moving with constant speed v0 in a plane along the given path r(ϕ) = b e ϕ (logarithmic spiral). For t = 0 the angle is ϕ = 0. Determine the angular velocity ϕ in ˙ dependence on ϕ and t as well as the radial velocity r. ˙ Solution
v=
In polar coordinates the speed is represented by
r˙ 2 + r2 ϕ˙ 2 .
Introducing v = v 0 and r˙ =
dr dϕ dr = ϕ˙ = ϕb ˙ eϕ dϕ dt dϕ
we obtain v0 =
ϕ˙ 2 b2 e2ϕ + b2 e2ϕ ϕ˙ 2 =
√
2 b e ϕ ϕ˙ .
Solving for ϕ yields ˙ ϕ = ˙
√ v2 b e− 0
ϕ
and thus
√
ϕ
r˙ = ϕb ˙ e = v 0 / 2 = const . To determine the dependence on time we find from ϕ = ˙
dϕ v0 −ϕ = e dt 2b
√
by separation of variables eϕ dϕ =
√ v2 b dt 0
and integration, taking into account the initial condition ϕ(t = 0) = 0, eϕ
− 1 = √ v2 b t 0
eϕ = 1 +
Introduction into ϕ(ϕ) ˙ finally yields ϕ = ˙
√ v2 b 0
1 . 1 + v0 t 2b
√
√ v2 b t . 0
in Polar Coordinates
23
Problem 1.19
A bar rotates about A according to the law ϕ = κt. Along the bar a knuckle K slides with prescibed speed r˙ = v0 at and initial conditions r(0) = 0, ϕ(0) = 0.
P1.19
−
l r
a) Calculate κ and a such that rmax = l is reached at ϕ∗ = 2π. b) Determine the path r(ϕ) of K and c) the magnitude of velocity and acceleration at ϕ∗∗ = π.
K ϕ(t)
A
Solution
From the given functions and initial conditions we first determine the components of position, velocity and acceleration in polar coordinates: r˙ = v 0 ϕ = κt
− at
r = v 0 t
˙ ϕ = κ ,
2
− at /2 ,
r¨ =
−a ,
¨ 0 . ϕ =
a) rmax is reached when r˙ = 0, i.e. at time t∗ = v0 /a. Thus, the prescribed conditions lead to r(t∗ ) = l
ϕ(t∗ ) = ϕ ∗ = 2π
v02 v02 = l a 2a v0 κ = 2π a
−
v02 a = , 2l v0 κ = π . l
b) By eliminating t from r(t) and introducing a and κ we find ϕ r(ϕ) = v 0 κ
−
−
ϕ2 ϕ a 2 = l 2κ π
ϕ2 . 4π 2
c) The magnitude of velocity and acceleration is obtained from the respective components at ϕ = ϕ ∗∗ = π, i.e. at time t∗∗ = π/κ = l/v0 : vr = r˙ = v0 v=
vr2
− +
v02 l v0 = , 2l v0 2 2 vϕ
=
− −
l vϕ = r ϕ = ˙ v0 v0
1 v02 l2 v0 3π π = v0 , 2 2l v02 l 4
1/4 + 9π 2 /16 v0 = 2.41 v0 ,
v02 3l 2 v02 v0 1 3 ar = r¨ r ϕ˙ = π 2 = + π2 = 2l 4 l l 2 4 2 v0 v0 v aϕ = r ϕ ¨ + 2r˙ ϕ = ˙ 2 π = π 0 , 2 l l
−
a =
2
− −
a2r + a2ϕ =
−
√ 7.90
2
+ π 2 v02 /l = 8.50 v02 /l .
v02 7.90 , l
24 P1.20
Planar Motion
Problem 1.20
From the planar motion of a point we know the radial velocity vr = c0 = const and the radial acceleration ar = a0 = const.
−
Determine for the initial conditions r(t=0) = 0 and ϕ(t=0) = 0: a) the angular velocity ω(t), b) the path (trajectory) r(ϕ), c) the circular acceleration aϕ (t).
Solution
a) From vr = r˙ = c 0 follows with r(t = 0) = 0
r¨ = 0
r = c 0 t .
and
Therewith, from ar = r¨
−a
0
=
−c tω 0
2
− rω
2
we obtain
a0 . c0 t
ω=
b) By integrating ω = ϕ we ˙ find with ϕ(t=0) = 0
√ t
ϕ =
ω dt¯ =
0
a0 c0
t
0
dt¯ = ¯ t
√
a0 2 t c0
ϕ2 c0 t = . 4 a0
Introduction into r = c 0 t yields the equation for the path trajectory c20 2 r(ϕ) = ϕ . 4a0
c) The circular acceleration can be determined from aϕ = r ϕ ¨ + 2r˙ ϕ˙ with
− −
ϕ = ω ˙ = as
aϕ = c 0 t
a0 and c0 t 1 2
a0 c0 t 3
ϕ = ¨
+ 2c0
1 2
a0 3 = c0 t 2
path v
a0 c0 t3
a 0 c0 . t
Remark: The circular velocity is given by vϕ
vr
vϕ r ϕ
√ = rω = a c t = c ϕ/2 . 0 0
0
in Polar Coordinates
s
Problem 1.21
An obeserver watches the flight of an low flyer P , flying with constant speed v0 in a flight altitude h.
P1.21
P v0 h
Calcuate the angular acceleration ϕ(ϕ) ¨ of his head and the radial acceleration r¨. Sketch both diagrams.
Solution
25
r ϕ
O
The position of P is given by
s = h tan ϕ
s ϕ = arctan h
r=
h2 + s2 .
Differentiation, considering s = v ˙ = v 0 = const and s¨ = 0, leads to s˙ v0 h v0 v0 2 = 2 = = cos ϕ, 2 h h + s2 h(1 + tan ϕ) h
1
ϕ = ˙ 1+
s h
2
−
v0 2 v0 ϕ = ¨ 2 cos ϕ( sin ϕ)ϕ = ˙ 2 sin ϕ cos3 ϕ , h h 2ss˙ v0 tan ϕ r˙ = = = v 0 sin ϕ , 2 2 2 1 h +s 1 + tan ϕ v02 r¨ = v 0 cos ϕ ϕ = ˙ cos3 ϕ . h
−
√
ϕ¨
−
π 2
π 2
ϕ
π 2
ϕ
r¨ v02 h −
π 2
Remark: The maximum angular acceleration occurs at ϕ = 30o ; its 3 v0 2 magnitude is ϕ ¨max = 3 . The maximum radial acceleration 8 h occurs at ϕ = 0. Note: the total acceleration of P is zero!
|
|
√
±
26 P1.22
Planar Motion
Problem 1.22
A fox F and a rabbit R spot each other and start running at the same time. The rabbit runs with constant speed vR straight to the save warren W , and the fox, to catch the rabbit, with constant speed vF along the curve r = 4 2 aϕ/π.
√
W vF F
a
Hint:
1 1 + x2 dx = x 2
R
r ϕ
Determine the necessary speed vR of the rabbit such that it will not be caught by the fox and at what time tR it should arrive at W ?
vR
a
1 + x2 + ln x +
1 + x2
Solution
We first determine the time tF the fox needs to reach W . From the path r(ϕ) follow the velocity components
√ 4 2a
dr dϕ = dϕ dt
vr = r˙ =
π
ϕ˙ ,
vϕ = r ϕ = ˙
√ 4 2a π
ϕ ˙ ϕ.
Since vF is constant, we obtain vF =
vr2
+
2 vϕ
=
√ 4 2a π
ϕ˙
1+
ϕ2
=
√ 4 2 a dϕ π
dt
1 + ϕ2 .
Now we separate the variables, integrate over the whole path from start (t = 0, ϕ = 0) until W (t = t F , ϕ = π/4) and find in this way:
tF
vF
0
√ 4 2a
√
dt¯ =
π
4 2a π vF tF = 2π 4 a tF = 1.548 . vF
π/4
1+ ϕ ¯2 dϕ ¯
0
π2 π 1+ + ln + 16 4
π2 1+ 16
= 1.548 a
The time t R the rabbit needs to arrive at W is calculated from its speed and the distance: tR = a/vR . To be not caught by the fox, the rabbit must be earlier at W than the fox, i.e. tR < tF . This condition leads to 1 vF = 0.646 vF , 1.548 a tR < 1.548 . vF vR >
Planer Motion
27
Problem 1.23
The point A moves in a plane with constant speed v0 along the x-axis. In constant distance a the point A is followed by a point B, moving such that its velocity vector always points to A. At time t = 0 point A is located at the origin of the coordinate system and B is on the y-axis.
P1.23
y B a v0
Solution
1 tan2 α2 cos α = . 1 + tan2 α2
2tan α2 sin α = , 1 + tan2 α2
dα α = ln tan , sin α 2
x
A
a) Determine the path of B and b) the speed vB (xA )? Hint:
−
a) With the known motion of point A
x˙ A = v 0 ,
xA = v 0 t
y
and by using the angle ϕ(t) it firstly follow the position coordinates and the velocity components of B: xB = x A a cos ϕ = v 0 t a cos ϕ ,
yB = a sin ϕ ,
x˙ B = v 0 + a ˙ ϕ sin ϕ ,
y˙ B = a ˙ ϕ cos ϕ .
− −
B
yB
y˙ B
−
x˙ B
a
vB ϕ
xB
A v0 xA
x
The condition that the velocity vector of B always points to A is expressed by
−y˙
B
x˙ B
=
yB xA
−x
−a ˙ϕ cos ϕ
v0 + a ˙ ϕ sin ϕ
B
=
a sin ϕ . a cos ϕ
Solving for ϕ yields ˙ step by step
− va sin ϕ . Separation of variables and integration lead for π/2 ≥ ϕ ≥ 0 to (see a ˙ ϕ(sin2 ϕ + cos2 ϕ) =
−v
0
sin ϕ
dϕ = dt
0
hint)
dϕ = sin ϕ
−
v0 a
dt
ln tan
ϕ 2
=
− va
0
t+C.
28
Planar Motion
The integration constant C is determined from the initial condition: ϕ(0) =
π : 2
ln 1 = 0 + C
C = 0 .
Therewith, we obtain the dependence of the angle ϕ on time t and on position xA , respectively (remind: xA = v 0 t): tan
ϕ = e−v 2
0
t/a
or
tan
ϕ = e−xA /a . 2
By the latter equation, the path of B is uniquely described through x A and the accompanied angle ϕ. The parameter representation of the path in cartesian coordinates is obtained by using the formulas of the hint: 1 tan2 ϕ2 1 e−2xA /a xB = xA a cos ϕ = x A a = x A a , − 2xA /a 1 + tan2 ϕ2 1+e 2tan2 ϕ2 2e−2xA /a yB = a sin ϕ = a = a . 2xA /a − 1 + tan2 ϕ2 1+ e
−
−
−
−
−
b) For the velocity we first have with ϕ = ˙
− va
0
sin ϕ
2 2 vB = x˙ 2B + y˙ B = v 02 + 2v0 a ˙ ϕ sin ϕ + a2 ϕ˙ 2 sin2 ϕ + a2 ϕ˙ 2 cos2 ϕ = v 02 + 2v0 a ˙ ϕ sin ϕ + a2 ϕ˙ 2 = v 02 2v02 sin 2 ϕ + v02 sin2 ϕ = v 02 (1 sin2 ϕ) = v 02 cos 2 ϕ
−
−
vB = v 0 cos ϕ .
Introducing the result of a) we finally obtain 1 tan2 vB (xA ) = v 0 1 + tan2
−
ϕ 2 ϕ 2
= v 0
1
2xA /a
− e−
. − 2xA /a 1+ e
Remarks:
• • •
For the limit case xA /a we obtain ϕ 0, yB 0 and xB xA a. The velocity of B then is given by vB v0 . The representation with the angle ϕ is shorter and more practical than that by cartesian coordinates. The results can also be represented by using hyperbolic functions. For example, recasting leads to vB (xA ) = v 0 tanh xA /a.
→
−
→∞
→
→
→
Chapter 2
Kinetics of a Point Mass
2
30
Newton’s Law
NEWTON’s 2nd Law (law of motion): The motion of a point mass under the action of forces is described by d(m v ) = p˙ = F dt with F =
F i and the momentum
p = m v .
Since the mass is constant, Newton’s law can also be expressed as m a = F
mass
× acceleration
= force .
As an example, this leads for cartesian coordinates to max =
F x ,
may =
F y ,
maz =
F z .
Remarks: Newton’s law is valid in this form only in an inertial reference frame (= reference system that is absolutely at rest or in uniform, rectilinear motion, see also chapter 8), Bodies with finite dimensions can be regarded as point masses if their dimensions have no influence on the motion.
• •
Impulse Law: Time integration of the law of motion leads to
t
mv
− mv
0
=
t0
F dt¯
bzw.
p
− p
0
= F
t where F = t F dt¯ is the linear impulse . When no forces are acting (F = 0), the linear momentum is conserved: 0
p = m v = const .
Angular Momentum Theorem: The vector product of Newton’s law with the position vector r yields dL(0) = M (0) , dt where L(0) = r M (0)
× p = angular momentum with respect to the fixed point 0, = r × F = moment with respect to the fixed point 0.
31
Work and Energy Theorem
If the moment vanishes (M (0) = 0), the angular momentum is conserved: L
(0)
= r
× m v = const .
path
dr
In this case, with 1 dA = r 2
× dr
dA
dA 1 = r dt 2
×v
0
m r
the law of areas (Kepler’s 2nd law) is obtained (see page 4): A˙ = const .
Work–Energy Theorem: Path integration of the law of motion yields mv12 2
−
r mv02 = F dr 2 r 1
0
U =
Work of Force F :
− T
0
= U ,
1 m v2 , 2
T =
Kinetic Energy :
T 1
or
·
· dU =
F dr ,
Remarks:
• •
α F
dU = F dr = F dr cos α.
·
dr
| || |
m
Forces orthogonal to the path (α = π/2), do not execute work. For a rotation holds dU = M dϕ.
·
Conservation-of-Energy Law: If the forces according to F =
− grad V =
−
∂V ∂V ∂V ex + ey + ez ∂x ∂y ∂z
can be derived from a potential V (= conservative forces ), the work is path independent, i.e. given by the potential difference:
r1
U =
r0
F dr = V 0
·
− V
1
.
From the Work-Energy Theorem then follows T 1 + V 1 = T 0 + V 0 = const . In words: When the applied forces possess a potential, then the sum of potential energy V and kinetic energy T remains constant during the motion.
32
Potential, Projectile Motion
Several Potentials Gravitational Potential (near earth’s surface)
V = mg z
Gravitational Potential (general)
V =
mg
−G Mrm
M
·
V =
Potential of a spring
11
−
G = 6, 673 10
Gravitational constant
z
m
r
m3 /kg s2 x
1 kx2 2
F
k
Power dU = F v = Power of a force, dt dϕ P = M = M ω = Power of a moment. dt P =
·
·
·
Projectile Motion Parabolic trajectory of motion: z = z 0
−
g 2
−
2
x x0 v0 cos α
+ (x
− x )tan α , 0
Maximum height: h =
z
1 (v0 sin α)2 , 2g
z 0
v0
α
h
Flight time: td =
v0 sin α 1+ g
1+
2gz 0 v02 sin 2 α
,
d
Flight distance: d = v 02
sin α cos α 1+ g
1+
2gz 0 v02 sin 2 α
Special case z 0 = 0 : td =
2 v0 sin α , g
d =
xd
x0
1 2 v0 sin 2α . g
.
x
Rectilinear Motion Problem 2.1
A box of weight W is pushed downwards a rough inclined plane (kinetic friction coefficient µ) with an initial velocity v0 . a) Determine the velocity in dependence on the distance. b) At what distance xE the box comes to rest? Under what circumstances is this possible?
33
v0 W
µ α
a) The law of motion yields in x- and in y-direction
Solution
y
m¨ x = G sin α
: :
−R ,
x
− G cos α .
0 = N
R
In conjunction with the friction law R = µN , the acceleration follows as x ¨ = g(sin α
− µ cos α) = a
0
N
α
W = mg
.
Twice integration, taking into account the initial conditions x(0) = 0, v(0) = v 0 , leads to: 1 v(t) = x = v ˙ x(t) = v 0 t + a0 t2 . 0 + a0 t , 2 Therewith, by eleminating the time, we obtain t =
v
−v
0
a0
v(x) =
x = v 0
v
−v
0
a0
a0 v 2 + 2
− 2vv + v 0
a20
2 0
v2 v02 = 2a0
−
v02 + 2a0 x .
b) From the condition v(xE ) = 0 (rest), the covered distance xE is determined: v02 2 0 = v 0 + 2a0 xE xE = . 2a0
−
From the condition xE > 0 follows a0 < 0, i.e. µ > tan α. The same results can be found easier by applying the work-energy theorem T 1 T 0 = U . It leads with 1 1 U = (mg sin α)x Rx, T 0 = mv02 , T 1 = mv 2 , R = µmg cos α 2 2
−
−
and by solving for v directly to v(x) =
v02 + 2g(sin α
− µ cos α)x .
P2.1
34 P2.2
Rectilinear
Problem 2.2
Two cars, one with and one without ABS, are stopping from the speed v0 = 100 km/h by full breaking. The first without ABS by blocking the wheels, i.e. sliding (’kinetic friction’), the second with ABS with still rolling wheels (ideal ’limiting static friction’ assumed). Determine for both cars the time tB and the distance sB for stopping if the coefficients of static and kinetic friction between pavement and tire are µ0 = 0.7 and µ = 0.45, respectively. Solution
→:
From the equation of motion of the first car (sliding)
mv˙ = m¨ s =
−R ,
↑:
− mg
0 = N
and the friction law
v s
mg
R = µN , R
it follows v˙ =
N
−µg .
Integration yields with v(t =0) = v 0 and s(t = 0) = 0 1 v(t) = v 0 µgt , s(t) = v 0 t µgt2 . 2
−
−
The stopping time and distance are calculated from the condition v = 0: v0 tB = , µg
v02 sB = s(tB ) = µg
−
v02 v02 = . 2µg 2µg
With the given coefficient of kinetic friction, we obtain 100 = 6.3 s , tB = 3.6 0.45 9.81
·
·
sB =
3.62
1002 = 87 m . 2 0.45 9.81
· ·
·
For the second car the wheels are still rolling under the limit condition of static friction (ABS), i.e. the friction force is now given by H = H 0 = µ 0 N . This means that in the calculation above only R must replaced by H 0 and µ by µ0 , respectively. Thus, for the car with ABS, we obtain tB = 6.3
· 0.45 = 4.05 s , 0.7
sB = 87
· 0.45 = 56 m . 0.7
Remark: Note that stopping time and distance are inverse proportional to the friction coefficient. Note also that, because of the neglected reaction time, the numbers for tB and sB in reality might be higher!
Motion
35
Problem 2.3
A parachutist (weight W including parachute) has the initial velocity v0 imediately after the parachute opens. a) Determine the velocity v in dependence on t if the air drag is assumed to obey the law F d = kv 2 . b) What limit speed vl reaches the parachutist? Solution
The law of motion yields
↓:
ma = m x ¨ = mg
x ¨ =
dv = g dt
− kv
F d = kv 2
2
or
−k v 1
2
k1 =
with
k . m
x˙
a) Separation of variables and integration leads to v t d¯ v = dt¯ , 2 k1 v¯ v g 0
−
0
W = mg
where the time t is counted from the opening of the parachute. With the basic integral
√ − √ dz = A Bz 2
1 artanh( AB
B/A z )
v
we obtain
1 artanh gk 1
vl
v
k1 /g v¯
= t
v0
v0
or by solving for v(t) v(t) =
b) For t vl =
g tanh k1
gk 1 t + artanh
k1 v0 g
t
.
→ ∞, it follows (tanh z → 1 for z → ∞)
g = k1
W . k
The same result can be found from the consideration that in the limit case, the acceleration is zero: a = g
2 1 l
−k v
=0
vl =
g = k1
W . k
P2.3
36 P2.4
Rectilinear
Problem 2.4 A
computer (weight W =100 N) in a packing case is protected against impact by foam plastics (spring stiffness k = 100 N/cm).
W
From what height h the case may impinge h a hard surface, if the acceleration of the computer shall not be bigger than four times the gravity?
k
Solution
During free fall, the case experiences the acceleration g. After impinging the surface, the foam plastics (= linear spring) will be compressed and the computer will be accelerated upwards. Then the motion is described by
↑:
m a =
mg
δ
−mg + k δ .
k
From the condition amax = 4g follows the maximum spring compression δ max =
5mg = 5 cm . k
Knowing this limit compression, the allowable height of fall can be determined from the conservation of energy law T 1 + V 1 = T 2 + V 2 as follows: state 1
state 2
l h
T 1 = 0 , V 1 = mg(l + h) , T 2 = 0 , V 2 = mg(l Introducing these quantities yields h =
δ max
l
1 k 2 δ max 2 mg
− δ
max
=
15 mg = 7.5 cm . 2 k
− δ
max ) +
1 2 kδ max . 2
Motion
37
Problem 2.5
On a rough inclined plane (inclination angle α, kinetic friction constant µ), a block (mass m1 ) is moving which is connected by a rope with a body of mass m 2 . Pulley and rope are regarded as massless. a) Determine the accelerations when m1 slides upwards and downwards, respectively. b) What force acts in the rope?
P2.5
m1 m2 µ α
a) We cut the rope and formulate for the 3 parts the basic equations, where we first assume upward sliding: Solution
W 1 = m1 g x1
S 2
A
S 1
S 1 S 2
N
R
W 2 = m2 g
: m a = S − R − W sin α, A : S : N = W cos α, R = µN . 1 1
x2
1
1
1
= S 2 ,
↓: m a
2 2
= W 2
− S , 2
1
With the kinematic condition (unextensible rope) v1 = v2 and consequently a1 = a 2 = a , we obtain a(u) = a = g
m2
− m (sin α + µ cos α) . 1
m1 + m2
For upward sliding, the acceleration must be positive, a > 0, and therefore m2 > m1 (sin α + µ cos α)! For downward sliding, only the direction of R must be changed. Then it follows m1 (sin α µ cos α) m2 a(d) = a = g . m1 + m2
−
−
−
This case occurs for a < 0, i.e. for m1 (sin α
− µ cos α) > m . 2
b) Independent on the sliding direction, the force in the rope is S = S 2 = S 1 = W 2
−m a
2 2
= m 2 (g
− a) .
Introducing the respective accelerations yields S (u) =
m1 m2 g(1 + sin α + µ cos α) , m1 + m2
S (d) =
m1 m2 g(1 + sin α µ cos α) . m1 + m2
−
38 P2.6
Rectilinear Motion
Problem 2.6
A chain (mass m, length l) lying on an inclined frictionless support starts sliding from the sketched position.
g
m
a
α a) Determine the position and velocity as functions of time. b) Determine the velocity as function of the position by applying the conservation-of-energy law.
a) Each chain link experiences the same velocity and acceleration. We therefore consider the chain as a single body of mass m which is driven by a force which depends on the length x of the overhanging part. Thus, the equation of motion reads Solution
x g sin α x ¨ κ2 x = 0 . l where κ2 = g sin α/l. This differential equation has the solution mx ¨ = m
−
x m g sin α l
x α
x(t) = A cosh κt + B sinh κt x = Aκ ˙ sinh κt + Bκ cosh κt . The integration constants follow from the initial conditions x(0) ˙ =0
B = 0,
x(0) = a
A = a
what finally leads to x(t) = a cosh κt ,
x(t) ˙ = aκ sinh κt ,
κ2 = g sin α/l.
Note that this solution is only valid for a x l. When the complete chain is on the inclined plane, the driving force remains constant!
≤ ≤
b) If we use as reference position for zero potential energy the upper horizontal plane, the energy terms in the initial and in the displaced position are given by
− a2 sin α m al g , x x = − sin α m g , 2 l
x/2
V 0 =
T 0 = 0 ,
V 1
1 T 1 = ˙x2 m . 2
x sin α/2
Introducing into V 0 + T 0 = V 1 + T 1 and solving for x leads ˙ to x(x) ˙ =
− (x2
a2 )g sin α .
Again, this solution is only valid for a
≤ x ≤ l.
Projectile Motion
39
Problem 2.7 Determine
the geometric locus of all points P 1 that is given by the position of all point masses at time t = t1 , that are thrown at time t = 0 with the same initial velocity v0 from a point P under different angles α with respect to the horizontal. Assume that all trajectories are located in the same vertical plane and that there is no air drag.
P2.7
t = t1 v0
v0 α
t=0 P
Solution
For convenience, the origin of the coordinates is chosen at P . Then it follows from
↑:
m¨ z =
→:
−mg ,
m¨ x = 0
z
with the initial conditions
P 1 v0
x(0) = z (0) = 0 , P
mg
α x
x(0) ˙ = v 0 cos α , z ˙ (0) = v 0 sin α by integration x = v 0 t cos α ,
z =
− 12 gt
2
+ v0 t sin α .
Since the solution is sought at time t1 for arbitrary angles, the angle α must be eliminated. Squaring and adding yields x21
2
2
= (v0 t1 ) cos α ,
g z 1 + t21 2 x21 +
2
= (v0 t1 )2 sin2 α
g z 1 + t21 2
2
2
= (v0 t1 ) .
Accordingly, all points P 1 are located on a circle with the radius r = v 0 t1 and the center C at z =
− g2 t
2 1
.
z t1
t1
α3 α1 r
P α2 C
t1
x
40 P2.8
Rectilinear Motion
Problem 2.8
From the top of a tower, two point masses are thrown with the same initial velocity v 0 under two different angles α1 and α2 . It is recognized that both masses impinge the surface at the same location.
v0 α2 α1
Determine the height of the tower.
Solution
z
− z
0
The parabolic trajectory of motion is given by (see page 20) =
− − g 2
x x0 v0 cos α
2
+ (x
− x )tan α . 0
We chose the origin of the coordinates at the top of the tower. Then we have x0 = z 0 = 0, and the point where the masses impinge the surface has the unknown coordinates x = l, z = h. For both throws holds:
z x
h
−
−h = −
g l2 + l tan α1 , 2 v02 cos 2 α1
−h = −
g l2 + l tan α2 . 2 v02 cos 2 α2
l
Equating both expressions leads for the horizontal distance l to 2v02 1 l= . g tan α1 + tan α2 Herewith, from the 1st equation, the height is determined as
g 2v02 h = + 2 2v0 g
2
1 1 cos2 α1 tan α1 + tan α2
2v02 1 = . g (tan α1 + tan α2 )tan(α1 + α2 )
− 2
2v02 tan α1 g tan α1 + tan α2
Remark: For the solution, the following formulas are used: 1 cos2 α2
− cos1 α 2
1
= tan2 α2 = (tan α2
2
− tan α − tan α )(tan α + tan α ) ,
tan α1 + tan α2 = tan(α1 + α2 ) . 1 tan α1 tan α2
−
1
1
2
1
Projectile Motion
41
Problem 2.9
To rescue shipwrecked persons, a rescue package of mass m is dropped from an airplane, flying with speed v0 = 200 km/h in a height of h = 150 m. a) Determine the distance s B from launching the package until it impacts on sea surface. b) Calculate the impact velocity vB Assuming a high horizontal velocity component vh , the air drag shall be taken into account by a horizontal drag force D = κm vh2 with κ = 0.003 m 1 .
A v0
−
h B
sB
a) We introduce an appropriate coordinate system and sketch the free body diagram with the acting drag force D and the weight mg. With vh = x, ˙ the equations of motion in x- and in z direction read Solution
m¨ x =
←:
−mκ x˙
2
,
↑:
m¨ z =
−mg . D
Integration yields with the initial conditions x(0) = 0, z (0) = h, x(0) ˙ = v 0 , z ˙ (0) = 0
− x˙
v0
t
dx ¯˙ = x ¯˙ 2
κ
t
d¯ x =
z ˙ =
−gt
x = ˙
0
x
0
dt¯
1
0
1 + κt¯ v0
,
z=
dt¯
− g2 t
2
1
,
The impact time tB follows from z B = 0 as tB =
2h/g ,
and thus, we obtain for the distance 1 sB = x(tB ) = ln 1 + κv0 2h/g κ
= 218 m .
b) From the velocity components at impact, v0 x(t ˙ B) = = 104 km/h , 1 + κv0 tB z ˙ (tB ) = gt B = 54.2 m/s = 195 km/h ,
−
−
−
results the velocity as vB =
x˙ 2 (tB ) + z ˙ 2 (tB ) = 221 km/h .
m x˙
x
1 B v0 + κt 1 x = ln(1 + κv0 t) c0
+h.
z
mg
A
P2.9
42 P2.10
Conservation of Energy
Problem 2.10
A rocket without an own propulsion is catapulted vertically upwards from earth’s surface with an initial velocity v0 . a) Determine the maximum flight hight H by considering the change of gravitation and neglecting drag forces. b) What magnitude of v0 is required when the rocket shall escape from the gravitation field of earth? (Earth’s radius R = 6370 km)
a) Since only conservative forces are acting, the conservationof-energy law Solution
T 1 + V 1 = T 0 + V 0 is appropriate as solution method. The gravitational potential V = GMm/r according to (force on earth’s surface = weight mg)
−
mg =
− dV dr
= G r=R
Mm R2
GM = gR2
can be written as V =
−
R2 mg . r
Thus, the different energies on earth’s surface (r = R) and final flight height (r = R + H ) are 1 T 0 = mv02 , V 0 = mgR , 2 R2 T 1 = 0 , V 1 = mg . R + H Introduction into the energy conservation law yields
− −
−
R2 1 mg = mv02 R + H 2
− mgR
v02 H = R . 2gR v02
−
∗
b) The ‘escape velocity’ v0 is found from H
→ ∞
∗
v0 =
2gR = 11180
m s
≈ 40000 km . h
Remarks: Note that a rocket in real cases is not launched from earth’s surface without an own propulsion! The required kinetic energy to reach the ‘escape velocity’ is T 0 = mgR.
• •
Circular Motion
43
Problem 2.11
Which minimum initial velocity v0 in position ① is necessary, such that the ② ② body with mass m reaches position ② if g a) b) a) it slides along a frictionless circular path (radius l), l m l m b) it is fixed at a rigid 60 60 massless rod (length l) ? ① ① ◦
◦
Solution
In both cases the initial velocity v0 is connected with v2 at position ② by the energy-conservation law T 2 + V 2 = T 1 + V 1 . Choosing zero potential energy at position ①, we obtain with V 1 = 0 m 2 m 2 v2 + mg(l + l cos 60 ) = v0 2 2 ◦
v02 = v 22 + 3gl .
a) The necessary velocity v2 is obtained from the condition that the normal force N between the path and the body is non-negative (otherwise the body looses contact with the path). From the law of motion
:
man = N
− mg cos ϕ
2
with an = v /l we obtain at position (ϕ = π) for the limit case N = 0: mv22 = mg l
v22
an
②
ϕ
N mg
= gl .
Hence, it follows v02
= gl + 3gl
v0 = 2
gl .
b) The initial velocity for the mass fixed at the rod will take a minimum if it comes to rest in position ②. For v2 = 0, the energy-conservation law directly yields v0 =
√
3
gl .
Remark: In case b) the force S in the rod may get negative. For example, in ② (v2 =0) the force is S = mg.
−
P2.11
44 P2.12
Simple Pendulum
Problem 2.12
Determine the velocity v(ϕ) of mass m of a simple pendulum in dependence on the maximum amplidude ϕ0 .
l ϕ
Discuss the result for characteristic angles ϕ.
g m
Solution
Since the velocity shall be determined in dependence on the position, the energy-conservation law is the first choice as solution method. As reference position for the potential energy, we choose the horizontal position ϕ = π/2 and find from
l cos ϕ
ϕ0 ϕ
T (ϕ) + V (ϕ) = T 0 + V 0 with v(ϕ0 ) = 0, i.e. T 0 = 0: m 2 v mgl cos ϕ = 0 mgl cos ϕ0 2
−
−
v=
±
2gl(cos ϕ
− cos ϕ ) . 0
The same result can be found by integrating the law of motion in tangential direction. From
:
¨ mat = ml ϕ =
−mg sin ϕ
ϕ
with ϕdϕ = ¨ ϕd ˙ ϕ and ˙ the initial condition ϕ(ϕ ˙ 0 ) = 0, we obtain
0
˙¯ ϕ d ¯˙ ϕ =
ϕ˙
−
ϕ0
g l
ϕ
at
an 2
sin ϕ d ¯ ϕ ¯
S
− ϕ˙2
W = mg
g = (cos ϕ0 l
− cos ϕ)
or with v = l ˙ ϕ again v 2 = 2gl(cos ϕ
− cos ϕ ) . 0
The maximum speed occurs for cos ϕ = 1, i.e. at ϕ = 0: vmax =
2gl(1
− cos ϕ ) = 0
ϕ0 2gl2sin =2 2 2
gl sin
ϕ0 . 2
For a small maximum amplitude ϕ0 also the angle ϕ remains small, and we obtain by truncated series approximation cos ϕ
2
≈ 1 − ϕ /2 , sin(ϕ /2) ≈ ϕ /2 v = gl(ϕ − ϕ ) , v = gl ϕ 2
0
2 0
2
0
max
0
.
Circular Motion
45
Problem 2.13
In a clamped frictionless pipe elbow (radius R) glides a sphere (weight W = mg) with zero initial velocity downwards from the top.
P2.13
ϕ R
Determine the support reactions at the clamping in dependence on the position ϕ of the sphere. At which ϕ the reactions take extreme values? Solution Newton’s
W g
law yields in com-
ϕ
ponents:
: :
an N
mat = mg cos ϕ , at
man = N
− mg sin ϕ .
mg
With a t = R ϕ, a ¨ n = R ϕ˙ 2 and ϕdϕ = ¨ ϕd ˙ ϕ, ˙ it follows from the 1st equation by integration
ϕ˙
ϕ
˙¯ ¯˙ ϕ = ϕ d
0
g cos ϕ d ¯ ϕ ¯ R
ϕ˙ 2 g = sin ϕ . 2 R
0
Therewith, we obtain the normal force from the 2nd equation as N (ϕ) = mg sin ϕ + mR ˙ ϕ2 = 3W sin ϕ . The equilibrium conditions for the elbow lead to the support reactions:
↑: ←:
AV = N sin ϕ = 3W sin2 ϕ , ϕ
AH = N cos ϕ = 3W sin ϕ cos ϕ =
A : M A =
− 32 W sin2ϕ ,
N M A AV
−NR cos ϕ = − 32 W R sin 2ϕ .
AH
R cos ϕ
A
The clamping moment M A and the horizontal force A H take their maximum, when the sphere is at ϕ = π/4. The vertical force AV is maximal for ϕ = π/2: M A max =
− 32 W R ,
AH max =
3 W , 2
AV max = 3W .
46 P2.14
Circular Motion
Problem 2.14
A hockey puck (mass m) in an ideally smooth ice field (no friction) is shot with speed v0 into the half-circular part of the boards and slides along the boards. At the end of the curved part, the speed is measured to be 0.6 v0 .
v0 r µ
0.6 v0
Determine the friction coefficient µ of the boards.
Solution
m
With aid of the free body diagram, we obtain the equations
of motion
:
mat =
−R ,
:
man = N .
R
Introducing the kinematic relations at = v, ˙ 2 an = v /r and the friction law R = µN yields
N
2
mv˙ =
−µm vr
an at
.
Separation of variables and integration leads to the velocity v(t):
v
v0
d¯ v = v¯2
−
t
0
µ ¯ dt r
v(t) =
v0 . 0 1 + µv t r
From repeated integration, the path s(t) is determined: s
t
d¯ s = v 0
0
0
dt¯ 1 + µ vr0 t¯
s(t) =
r µv0 ln (1 + t) . µ r
The time t1 , until the speed is reduced to 0.6 v0 , is calculated as 0.6 v0 =
v0 µv 1 + r 0 t1
t1 =
2 r . 3 µv0
Thus, we obtain for the corresponding path length at the end of the curved boards s1 = s(t1 ) =
r µv0 r 5 ln (1 + t1 ) = ln . µ r µ 3
Equalizing it with the length of the half-circle yields s1 = rπ
µ =
ln(5/3) = 0.16 . π
Motion on a Curved Path
47
Problem 2.15
A car (weight W = mg) passes a bend (static friction coefficient µ0 ), whose curvature 1/ρ increases proportional to the covered distance s, i.e. s = A 2 /ρ (clothoid). At time t0 = 0, the car is at s0 = 0 and has an initial speed v0 . At which speed v, where and when the car ‘skids off the bend’ if a) it moves with constant speed, b) it brakes with constant deceleration a0 ? Given.: A = 35 m, µ0 = 0.6, a0 = g/4, v0 = 72 km/h. Solution
tion force
a) For at = 0, Newton’s law yields with an = v02 /ρ the fricH n
m
v02 : H = H n = ma n = m . ρ
an
The car leaves its path when the limit friction force is attained:
v
ρ
v02 v02 H = µ 0 mg m = µ 0 mg ρ1 = . ρ1 µ0 g This leads with s = A 2 /ρ = v 0 t to (v1 = v 0 )
µ0 gA 2 s1 = = 18 m , ρ1
t1 =
s1 = 0.9 s . v0
b) When the motion is decelerated, an additional force acts in tangential direction. With at = a0 follows
: :
mat =
−ma 2
man = m
0
− = −H , t
an
v = H n . ρ
H t ρ
Static limit friction is attained for H =
H n2 +
H t2
= µ 0 mg
v2 = ρ2
2 v 0 s2 = 4a ( + ) 0
− v02
µ20 g 2
mv 2 ρ2
−a
2 0
2a0 s = v0
v02 4a0
2
A2 2a0
2a0 s2 = 61.2 km/h ,
2
v
+ (ma0 )2 = µ 0 mg
.
− − − −
Thus, with v = v02 s = A 2 /ρ, we obtain
v2 =
H n
m
a0 t (constant deceleration) and
µ20 g 2
−a
2 0
= 22.7 m ,
t2 = v0 a0 v2 = 1.22 s .
−
P2.15
48 P2.16
Planar Motion
Problem 2.16
The potential of a free movable point mass m in a horizontal plane is given by V (x, y) = k(x2 + y 2 )/2.
y x
a) Determine the acting forces and formulate the equations of motion. b) Determine the path of the point mass in parameter and implicit representation for the initial conditions x(0) = a, x(0) ˙ = 0, y(0) = 0, y(0) ˙ = v 0 . Solution
F x =
m v0
y
x a
a) The forces follow from the derivatives of the potential as
− ∂V = −kx , ∂x
Fy =
− ∂V = −ky , ∂y
which leads to the equations of motion or
→:
m¨ x = F x =
−kx ,
x ¨ + ω 2 x = 0 ,
↑:
m¨ y = F y =
y¨ + ω 2 y = 0 ,
−ky
where
ω 2 = k/m .
b) Both differential equations describe free undamped vibrations, whose solutions are given by (cf. chapter 7) x = A cos ωt + B sin ωt ,
y = C cos ωt + D sin ωt ,
where A, B, C, D are constants. They are determined by using the initial conditions: x(0) = a
A = a ,
y(0) = 0
x(0) ˙ =0
B = 0 ,
y(0) ˙ = v 0
C = 0 , v0 D= . ω
Thus, in parameter representation, the path is described by y(t) =
x(t) = a cos ωt ,
v0 sin ωt . ω
The implicit representation is found by eliminating t through squaring and adding, resulting in
x a
2
+
y v0 /ω
2
= 1 .
Accordingly, the point mass moves counterclockwise along an ellipse with half-axes a and v0 /ω.
on a Curved Path
49
Problem 2.17
A space vessel shall be lifted from a circular path with radius r1 around the earth E (earth’s radius R) to a more distant circular orbit of radius r2 . The transition is carried out by changing in A and in B suddenly the magnitude of the velocity of the vessel.
P2.17
B E r2
r1 A
Determine the necessary velocity change in A.
Solution
According to Kepler’s 1st law, the vessel moves between A and B along an ellipse, whose one focus coincides with the earth. Since in A and B r2 only the magnitude of velocity and not its B a b direction shall be changed, the transition ellipse in A and in B must be tangential to E e the circles. From this condition the ellipse r1 parameters follow as A a=
r1 + r2 , 2
e = a
−r
1
=
r2
−r
1
2
,
b2 = a 2
−e
2
= r 1 r2 ,
and the curvature radius at A (vertex of the ellipse) yields b2 2r1 r2 ρ = = . a r1 + r2 In A, the gravitational force has the magnitue F = mg(R/r1 )2 . At this location, the law of motion normal to the circular path (before velocity change) leads to v12 m = mg r1
R r1
2
−
v1 = R
g r1
and normal to the elliptic orbit (after velocity change) to 2 vA m = mg ρ
R r1
2
vA = R
2r2 . r1 + r2
g r1
Thus, the necessary velocity change is given by ∆vA = v A
−v
1
= R
g r1
2r2 r1 + r2
1
.
50 P2.18
Work
Problem 2.18
A barge K is towed in a channel by a haul engine L. In the towing rope acts a force S = 9 kN, which is inclined by an angle α = 28 with respect to the rail track.
L
◦
Determine a) the work for a covered distance s = 3 km, b) the power for a towing speed v = 9 km/h.
K
a) For the work
Solution
U =
α
· | |
F cos α dr ,
F dr =
| |
◦
we obtain with F = const = S , cos α = const = cos 28 and dr = ds:
| |
| |
U = S cos α s = 9 0.883 3000 = 23800 kNm = 23800 kJ = 23.8 MJ .
·
·
b) The power is given by P = F v = S cos α v = 9 0.883
·
P2.19
·
9 · 3.6 = 19.9 kJ/s = 19.9 kW .
Problem 2.19 Determine
the necessary work for lifting a body of weight W = 1 N from erth’s surface (earth’s radius R) into the distance r0 of the moon (r0 = 60 R). Solution
According to the gravitational law, the gravitational force varies inverse to the squared distance from the earth’s surface. Thus, the ‘weight’ in distance r is F = W
R r
2
.
Therewith, the work follows as 60R
U = W
R r
2
dr =
59 WR . 60
R
With R = 6370 km and W = 1 N, we obtain U =
59 6370 Nkm = 6264 kJ = 6.3 MJ . 60
·
and Power
51
Problem 2.20
A motor winch M tows a body of weight W = mg with constant speed v0 upwards a rough inclined plane (coefficient of kinetic friction µ). Determine the necessary electric power P A of the winch if its efficiency η is known.
P2.20
M W µ α
Solution
For uniform motion (v˙ = 0), the force in the rope S follows from the equilibrium conditions
:
S = W sin α+R ,
:
v
R
N = W cos α
and the friction law R = µN as
S
W
N
S = W (sin α + µ cos α) . Thus, the power generated by the winch is P = S v0 = W (sin α + µ cos α)v0 . The power absorbed by the winch having an effienciency η is given by P A =
P W = (sin α + µ cos α)v0 . η η
Problem 2.21
A big container vessel with a drive power of 80000 kW covers in 7 days 4000 nautical miles. Determine the average drag force F d Solution
Using the conversion 1 nautical mile = 1.852 km and 1 kW = 1 kNm/s, we obtain from P = F d v with the drag force F d =
v=
4000 1852 m = 12.25 7 24 3600 s
· · ·
P 80000 kNm/s = = 6531 kN = 6.53 MN . v 12.25 m/s
P2.21
52 P2.22
Work and Power
Problem 2.22
In a centrifuge of radius r, rotating with constant angular velocity ω0 , a body (point mass m) is accelerated by dynamic friction (friction coefficient µ) from its initial angular velocity ω(0) = ω0 /2 to the final angular velocity ω0 .
ω0
Determine the required acceleration time tr , the drive torque M (t), the power P (t) and the work U done by the centrifuge.
m
r
Solution
During acceleration, the point mass rotates with angular velocity ω(t). With the accelerations at = r ω, ˙ an = rω 2 , the equations of motion are given by mr ω˙ = R ,
↑:
←:
mrω 2 = N .
at
ω0
R
an m
N
N
R
Introducing the friction law R = µN , eliminating N and using ω(0) = ω 0 /2 leads to ω˙ = µω 2
dω = µ ω2
dt
2 ω0
− ω1 = µt .
The acceleration time tr is obtained from the condition ω(tr ) = ω 0 : 2 ω0
−
1 = µt r ω0
tr =
1 . µω0
Since the centrifuge is not accelerated, the driving torque is given by the moment of the friction force R: µmr 2 ω02 M (t) = rR = mr ω˙ = µmr ω = . (2 µω0 t)2 2
2
2
−
Because M and ω0 are coaxial, the required power is given by µmr2 ω03 P (t) = M ω 0 = M ω0 = . (2 µω0 t)2
·
−
The total work U done by the centrifuge (strictly speaking, the friction force) is calculated easiest from the difference of kinetic energies: U =
1 mv 2 (tr ) 2
− 12 mv (0) = 21 m(rω ) − 12 m(rω /2) 2
0
2
0
2
3 = mr 2 ω02 . 8
Impulse Law
53
Problem 2.23
A soccer player kicks the ball from a distance e with a kick-off angle α = 45 against a vertical wall. The impact at the wall is assumed to be idealelastic. ◦
What initial velocity v0 of the ball is necessary if a) it shall bounce back exactly to the foot of the player, b) it shall bounce back to the head (height H ) of the player?
v0 H
α e
Solution
Since no energy gets lost when the impact against the wall is ideal-elastic, the magnitudes of impact velocity v1 and rebound velocity v2 must be equal. Then, from the impulse law follows (reflection law)
↑:
mv2 cos β 2 mv1 cos β 1 = 0
−
β 2 v2 v1
β 1 = β 2 .
β 1
Hence, we can replace the problem of reflection at the wall by a mirroring problem, where we imagine the trajectory being continued through the wall. z
H α e
x
e ◦
a) The ‘flight distance’ d = 2e follows with α = 45 and z 0 = 0 as (see page 32) d = 2e = v 02
sin2α g
v0 =
2ge = sin2α
2ge .
b) We introduce the coordinates of the kick-off point x0 = z 0 = 0 and end point x = 2e, z = H into the parabolic trajectory of motion (page 32) and obtain H =
− g 2
2e v0 cos 45
◦
2 ◦
+ 2e tan 45
v0 = 2e
2e
g
− H .
P2.23
54 P2.24
Impulse Law
Problem 2.24
A body (mass m) is driven along a rough horizontal path (kinetic friction coefficient µ) by a periodically acting force, such that v(t0 ) = v0 and v1 = v(t1 ) = v0 /2. During the ∆t T F driving phase ∆t, the force g F 0 profile F (t) is triangular. m
F
a) Determine the period T and the required peak force F 0 for a given ∆t.
µ
t
t1
t0
b) Calculate the work U done by F during a period T .
a) We consider one period and start counting time at the end of the driving phase. First we apply F the impulse law over the full period mg F 0 T . With R = µN = µmg = const F and the given F (t) profile, the linear impulses of the friction force and R the driving force are given by Solution
R = µmgT ,
F =
N
0
F (t)dt = F 0 ∆t/2 ,
t1
t0
t
T
∆t
Thus, with v(0) = v(t0 ) = v 0 , the impulse law leads to
→:
m v0
− mv
0
=0=
−µmgT + 21 F ∆t . 0
In the same way, we obtain with v(t0 ) = v 0 and v(t1 ) = v0 /2 from the impulse law applied over the driving phase ∆t
→:
m v0
− m v /2 = −µmg∆t + 21 F ∆t . 0
0
From these two equations for the two unknowns T and F 0 , it follow
v0 T = ∆t + , 2µg
v0 F 0 = 2µmg 1 + . 2µmg∆t
b) The work U done by F (t) during time T is determined by using the work-energy theorem, i.e. from the difference of kinetic energies at t0 and t1 : U = T (t0)
− T (t ) = 21 mv − 21 m(v /2) 1
2 0
0
2
=
3 mv02 . 8
Remark: The work of F and of R during T are equal but have opposite signs. This easily allows calculating the covered distance: l = 3v02 /(8µg).
Angular Momentum
55
Problem 2.25
A point mass, fixed at a massless thread, rotates along a horizontal circular path. At time t = 0, the radius is r0 and the angular velocity is ω0 .
P2.25 ez
ω0 m
a) Determine r(t) and ω(t), when the thread is pulled with constant speed u0 downwards through the sketched vertical pipe. b) At what time t1 the angular velocity has doubled and how big is the associated radius r1 ?
r0 u0
c) Determine for this case the change of kinetic energy ∆T of the point mass.
a) Because there acts no external moment on the point mass with respect to the center of its path, the angular momentum remains conserved: Solution
× m v = const . With r × v = rv e and v L = r
ϕ
2
z
ϕ
= m r02 ω0
L = m r ω
= rω, it follows r02 ω = ω 0 2 . r
The dependence of r(t) on time is given by the constant thread speed r˙ = u0 :
−
r(t) = r 0
−u t. 0
Inserting into ω leads to ω0 r02 . ω(t) = (r0 u0 t)2
−
b) From the condition ω(t1 ) = 2ω0 , it follows r0 t1 = u0
− √ 1 2 2
1
and
r1 = r 0
−u t
0 1
=
√ 2 2
c) The energy change is calculated as m 2 (vϕ + u20 ) 2 m 2 = r0 2ω0 2 2
∆T =
1
− m2 (v + u ) − m2 (r ω ) = 21 mr ω
√
2 ϕ0
2 0
2
The kinetic energy has doubled.
0
0
2
2 0
2 0
.
r0 .
56 P2.26
Angular Momentum
Problem 2.26
In an upright standing frictionless hollow cylinder (radius R), a little sphere (point mass m) is inserted at point A with a horizontal initial speed v 0 .
v0
A
g R h
a) What angle α to the horizontal plane has the velocity vB at point B lying in height distance h below A? b) What speed v0 is necessary such that the sphere impinges on ground at C with an angle of 45 and what magnitude has vC in this case?
B α
H
vB
C
◦
a) The speed of the sphere in point B follows from the energy conservation law T A + V A = T B + V B : Solution
1 1 2 mv02 + mgh = mvB 2 2
vB =
v02 + 2gh .
Since there acts no external moment with respect to the cylinder axis, the angular momentum (moment of momentum) with respect to this axis is conserved: L = const
LA = L B .
With LA = R(mv0 ) and LB = R(mvB cos α), the angle α follows as cos α =
v0 = vB
v0 . 2 v0 + 2gh
◦
b) With α = 45 and h = H , we obtain at C 1 cos 45 = 2 ◦
√
2=
v0 v02 + 2gH
or after squaring and solving for v0 v0 =
2gH .
Thus, after falling the height distance H , the velocity is vC =
v02
+ 2gH =
2gH + 2gH = 2
gH .
Remark: Because on the sphere acts only the weight (vertical) and the normal force from the cylinder wall (normal to the wall) the horizontal velocity component remains unchanged v0 .
Reaction Forces and Bending Moment
57
Problem 2.27 A
cosine-shaped arch of a roller coaster is in A and in B pin-supported. The arch is passed without friction by a car (mass m) that has in point A the initial y g velocity v0 = ga/10. m v0 Determine the support reactiy = a cos2πx/l a B A ons and the bending moment x a at C when the car just passes point C . The weight of the C arch shall be disregarded. l/2 l/2
Solution
The velocity of the car at point C follows from the energy conservation law T A + V A = T C + V C : 1 1 2 mv02 + mga = mvC 2 2
− mga
2 vC = v 02 + 4ga =
41 ga . 10
The derivatives y = (2πa/l)sin2πx/l and y = (4π 2 a/l2 )cos2πx/l yield y (l/2) = 0 and y (l/2) = 4π 2 a/l2 . Hence, the curvature radius ρ of the path at C is given by
−
1 4π 2 a y (l/2) = = 2 . [1 + y 2 (l/2)]3/2 l ρC
−
ρC
With the normal acceleration an = v2 /ρ, the law of motion allows to determine the force F at C , which acts from the arch onto the car:
↑:
2 vC = F m ρC
− mg
mg vC F
2 vC 164 π 2 a2 F = mg + m = mg 1 + . 10 l2 ρC
Knowing F , the support reactions and the bending moment in C can be calculated: F B
A
mg 164 π 2 a2 A = B = 1+ , 2 10 l2
M C
A
l mgl 164 π 2 a2 M C = A = 1+ . 2 4 10 l2
Remark: When the results are evaluated for the data m = 500 kg, 2a = 10m, l = 50 m, we obtain v0 = 7, 97 km/h, vC = 51, 05 km/h, ρC = 12, 66m, F = 12, 84 kN, A = B = 6, 42 kN, M C = 160, 55kNm .
P2.27
Chapter 3
Dynamics of a System of Point Masses
3
60
Dynamics of a
m j
System of Point Masses: The mass mi is subject to the external force F i and the internal forces F ij = F ji . If the distances rij between the masses remain constant, the system of point masses is rigid .
C
−
F ji
r ij F ij
r C
z mi
ri
0
F i
y
x
Law of Motion for the Center of Mass: The center of mass C moves such as a point mass with the same total mass subject to the resultant of all external forces acting on the system:
− r c = p˙ c = m¨
m =
mi
F i
=
pc = m r˙ c =
total mass,
mi r˙ i =
pi
=
total linear momentum.
Impulse Law: Time integration of the law of motion yields mv c
mv c0 = F ,
with the linear impulse t
F i d τ .
F =
t0
Conservation of Linear Momentum: In case that the resultant external force is zero, the linear momentum is conserved: mv c =
mi v i = const .
Angular Momentum Theorem: The time rate of change of the total angular momentum with respect to a fixed point 0 is equal to the resultant moment of all external forces about the same point: dL(0) = M (0) dt
System of Point Masses (0)
=
L
(0)
M
× × ri
=
(0)
mi v i =
ri
F i
Li
=
=
61
total angular momentum,
total external moment.
Special Case: For the rotation of a rigid system of point masses about a fixed axis a-a follows Θa ϕ = M ¨ a,
ri2 mi = mass moment of inertia relative to axis a-a,
Θa = ri
=
orthogonal distance between mass mi and axis a-a.
Work-Energy Theorem: The change of kinetic energy is equal to the sum of the work U (e) of all external and U (i) of all internal forces: T
− T
0
T =
= U (e) + U (i)
· · mi vi2 /2
=
U (e) =
F i dri
U (i) =
F ij dr ji
kinetic energy,
=
work of external forces,
=
work of internal forces.
For rigid constraints (drji = 0) holds W (i) = 0.
Conservation of Energy Law: If the external forces can be derived from a potential V (e) and the internal forces from a potential V (i) , the work-energy theorem results in the conservation of energy law (i)
T + V (i) + V (e) = T 0 + V 0
(e)
+ V 0
= const .
Bodies with variable mass: The motion of a body with variable mass (e.g. a rocket) is described by m(t) a = F + T where
F = external force,
m(t) = time dependent mass, T = µ w = thrust, where µ = m = ˙ rate of mass change (mass flow), w = mass flow velocity relative to the body.
− −
62 P3.1
Equations of Motion
Problem 3.1
On a frictionless horizontal plane, two wedges of masses m1 and m2 are placed on top of each other. The wedges can slide frictionless against each other. m2 m1 Determine the accelerations of both wedges. α Check the result by considering the limit cases m1 and α π/2.
→ ∞
→
Solution
We separate the two bodies and formulate the equations of motion in x- and in y-direction: ①
②
←: ↓: ←: ↓:
x ②
N 1
m1 x ¨1 = N 1 sin α , m1 y¨1 = m 1 g m2 x ¨2 =
− N + N cos α 2
1
①
,
m1 g
−N sin α , m y¨ = m g − N cos α . 1
2 2
2
m2 g
y
N 2
1
Since wedge ① moves horizontally and at its top side wedge downwards, the kinematic relations read
② slides
x1
y¨1 = 0 ,
α
y2 = (x1
− x )tan α
2
y¨2 = (¨ x1
− x¨ )tan α .
y2 x2
2
Thus, we have six equations for the six unknowns (¨ x1 , y¨1 , x ¨2 , y¨2 , N 1 , N 2 ). By eliminating N 1 and N 2 it follows m2 g tan α m1 x ¨1 = , m 2 1 + 1 + m21 tan α ¨2 = x
−
y¨1 = 0 ,
g tan α , 2 m 2 1 + 1 + m1 tan α
m2 g tan2 α 1+ m 1 y¨2 = . 2 m 2 1 + 1 + m1 tan α
For the two limit cases we obtain: a) m 1
→ ∞ : π b) α → : 2
x ¨1
and y¨2 /¨ x2
→ 0
|
x ¨1 = x ¨2
| → tan α (inclined plane),
→ 0 u n d y¨ → g 2
(free fall).
Remark: Adding the equations of motion in x-direction and time-integration confirms conservation of linear momentum: m1 x ¨1 + m2 x ¨2 = 0 m1 ˙x1 + m2 ˙x2 = C , i.e. the total linear momentum is constant!
System of Point Masses
63
Problem 3.2
The two massless pulleys are connected by a massless inextensible rope. The system is subject to the weights W 1 = m1 g and W 2 = m 2 g.
P3.2
g
Determine the force in the rope, the acceleration of the mass m1 and its velocity in dependence of the covered distance.
m2 m1
Solution
We first separate the system, draw the free body diagrams including all acting forces and introduce the coordinates x1 and x2 (counted from an arbitrary initial position). From the equilibrium of moments about the centers of the pulleys follows
S 1 S 2
S 1
② ①
m2 g
The equations of motion yield
↑:
m1 x ¨1 = 2S
x2
x1
S 1 = S 2 = S 3 = S .
①
S 3
m1 g
−m g ,
②
1
m2 x ¨2 = m 2 g
↓:
−S .
When the body ② moves downwards by x2 the body ① is lifted by x1 = x2 /2. Therefore, with x ¨1 = x ¨2 /2 we otain by solving for the unknows 2m2 m1 x ¨1 = g , m1 + 4m2
−
x2
3m1 m2 g S = . m1 + 4m2
x2
x1
The relationship between velocity and covered distance is most easily determined from the energy conservation law T + V = T 0 + V 0 . With T 0 = 0, V 0 = 0 follows 1 1 m1 ˙x21 + m2 ˙x22 + m1 gx 1 2 2
− m gx 2
2
=0
or 1 1 m1 ˙x21 + m2 (2x˙ 1 )2 = 2m2 gx 1 m1 gx 1 2 2 2m2 m1 v1 (x1 ) = x˙ 1 = 2gx 1 . m1 + 4m2
−
−
Remark: In the special case 2m2 = m 1 , we obtain x ¨1 In this case, the system is in equilibrium!
≡ 0 and S = W . 2
64 P3.3
Equations of Motion
Problem 3.3 A
hinged pendulum is modeled as a massless rigid rod with two attached point masses of weights W 1 = m1 g and W 2 = m 2 g.
ϕ
a a
Formulate the equation of motion.
Solution
g
m1 m2
We will solve the problem by two different approaches.
1st approach: Because the distance of the masses is constant, the pendulum represents a rigid system of point masses. Therefore, the angular momentum theorem with respect to the fixed hinge A ΘA ¨ ϕ = M A can be applied. With 2
2
A
2
ΘA = a m1 + (2a) m2 = a (m1 + 4m2 ) the equation of motion follows as (notice the positive sense of rotation!)
2 ¨ A : a (m1 + 4m2 )ϕ =
ϕ ¨+
ϕ
a
m1 g
−m ga sin ϕ − m g 2a sin ϕ 1
2
g m1 + 2m2 sin ϕ = 0 . a m1 + 4m2
a m2 g
2a sin ϕ
2nd approach: We start from the conservation of energy law T + V = const, where the zero level for the potential energy is chosen at ϕ = π/2 (hinge A): 1 1 m1 v12 + m2 v22 2 2
− m ga cos ϕ − m g 2a cos ϕ = const . 1
2
With v1 = a ˙ ϕ and v2 = 2a ˙ ϕ follows 1 (m1 + 4m2 )a2 ϕ˙ 2 2
− ag cos ϕ(m + 2m ) = const 1
2
and differentiation with respect to time leads to (m1 + 4m2 )a2 ϕ ˙ ¨ϕ + ag sin ϕ(m1 + 2m2 )ϕ = ˙ 0. Since ϕ is ˙ nonzero for all times t, it remains ¨+ ϕ
g m1 + 2m2 sin ϕ = 0 . a m1 + 4m2
System of Point Masses
65
Problem 3.4 The
constant acceleration a0 of a truck (mass M ) is such high that the box (weight W = mg) starts to slide on the loading area (coefficients of static and kinetic friction µ0 , µ). l
a) Determine the minimum acceleration a0 for the onset of sliding and the corresponding horizontal force F excerted by the truck to the pavement. b) Calculate the time T , the box requires to bounce against the rear wall.
W µ0 , µ
∗
a0
∗
a) As long as the box is not sliding, the equations of motion for truck and box are given by Solution
mg
① ②
→: →:
M a0 = F
− H ,
ma0 = H ,
H
where the static friction force H is limited by H 0 = µ 0 N = µ 0 mg. Introducing H = H 0 leads to the limit values of acceleration and force: ∗
a0
②
N N
H
a0
①
F
∗
a0 = µ 0 g ,
F = (M + m)µ0 g
b) During sliding, the friction force is R = µN = µmg, which for the box leads to the equation of motion mg
②
→:
ma2 = R
s2
a2 = µg .
②
R
Integration of the two constant accelerations a1 = a0 and a2 of truck and box, considering the initial conditions s1 (0) = s 2 (0) = s0 , v1 (0) = v 2 (0) = v 0 , yields
a1
s1
N a0 ①
v1 = a 0 t + v0 , v2 = µgt + v0 , t2 t2 s1 = a 0 + v0 t + s0 , s2 = µg + v0 t + s0 . 2 2 The box bounces at time T against the rear wall, when ∆s = s 1
−s
2
= l
(a0
2
− µg)T
= 2l
T =
a0
2l
− µg .
P3.4
66 P3.5
Equations of Motion
Problem 3.5
A big block (mass M ), initially resting on a frictionless ground, starts to move due to a little block of mass m, which is shot with speed v0 to its rough upper surface. The kinetic friction coefficient v0 µ between the two bodies is µ. m a) Determine the time T , that is required, until both bodies move with the common speed v . b) Determine the covered distance of M , until the common speed is reached.
g
M
∗
Solution
We separate the two bodies, sketch the free body diagrams and introduce position coordinates. Then, the equations of motion read ① ②
→: →:
mg
x2
②
R N
M ¨ x1 = R , m¨ x2 = R .
R
−
x1
Introducing the friction law R = µN = µmg leads to
①
Mg
m µg , x ¨2 = µg . M By integration, considering the initial conditions x˙ 1 (0) = 0, x1 (0) = 0, x˙ 2 (0) = v 0 , x2 (0) = 0, we obtain x ¨1 =
x˙ 1 =
−
m µgt , M
x˙ 2 = v 0
1m µgt 2 , 2 M 1 x2 = v 0 t µgt 2 . 2
x1 =
− µgt ,
−
∗
a) The condition x˙ 1 (T ) = x˙ 2 (T ) = v leads to m µgT = v 0 M and ∗
− µgT
v = x˙ 1 (T ) =
T =
v0 µg(1 + m/M )
m 1 µgT = v 0 . M 1 + M/m
b) The covered distance of M is given by 1m 1 v02 m/M 2 x1 (T ) = µgT = . 2 M 2 µg(1 + m/M )2
System of Point Masses Problem 3.6 In
an experiment, a particle of mass m is shot with speed v0 against a second resting particle of mass 2m After impact, three particles are observed, where the sketched directions and the following masses and velocities are detected during measurements: m1 = m, v1 = 2v0 , v2 = v 0 /2.
v2
m2 ◦
60
v0
P3.6
after impact
before impact
2m
67
v1 y
m1 45◦
x
45◦
m3 v3
m
Determine m2 , m3 and v3 .
Solution
The momentum of the particle system is unchanged after impact. Thus, with cos 45 = sin 45 = 2/2, cos 60 = 1/2, sin 60 = 3/2 we obtain in components ◦
√
√
◦
◦
◦
√ √ √ ↑ : mv = m v 2/2 + m v 3/2 − m v 2/2 , → : 0 = m v √ 2/2 − m v 1/2 − m v √ 2/2 . 0
1 1
2 2
1 1
2 2
3 3
3 3
In conjunction with m1 + m2 + m3 = 3m , we now have three equations for the three unknows m2 , m3 , v3 . To solve them, it is advantageous to introduce the given quantities:
√ √ √ mv = mv 2 + m v 3/4 − m v 2/2 , √ √ 0 = mv 2 − m v 1/4 − m v 2/2 , 0
0
2 0
0
3 3
2 0
3 3
m2 + m3 = 2m .
Subtracting the 2nd from the 1st equation provides directly
√
1+ 3 mv0 = m 2 v0 4
m2 = m
4 = 1.464 m . 1+ 3
√
Thus, from the 3rd equation follows m3 = 2m
−m
2
− √
= m 2
4 1+ 3
√ − √
2 3 2 = m = 0.536 m . 1+ 3
Finally, introducing m2 , m3 into the 2nd equation yields
− √
m v3 = v 0 2 m3
m2 2 2 m 1
= 2.765 v0 .
Remark: Note, that the center of mass stays on the y-axis!
68 P3.7
Impulse Law
Problem 3.7
On a frictionless plane, a body with mass m1 hits with velocity v a second body at rest (mass m2 ) and connects to it. Subsequently, the composite body hits via a spring (spring constant k) a third body at rest (mass m3 ). m1
m2
k
v
m3
g
a) Determine the speed v of the mass m 1 such that m3 remains at rest, if the plane solely at the location of m3 is rough (coefficient of static friction µ0 ). b) Determine the speed of m3 after collision if the plane at the location of m3 is also frictionless.
a) When m 1 and m 2 are connecting, the momentum remains conserved. Thus, the velocity v¯12 of the composite is given by Solution
→:
m1 v = (m1 + m2 )¯ v12
v¯12 =
m1 v. m1 + m2
If m3 is at rest, the maximum compression x of the spring is reached, when during collision, the velocity of the composite body has come to zero. Then, energy conservation yields 1 1 2 (m1 + m2 )¯ v12 = k x2 2 2
x2 =
m1 + m2 2 v¯12 , k
and the horizontal maximum force acting on m3 follows as F k = k x. In order that m3 remains at rest, the equilibrium condition H = F c and the condition fo static friction H < H 0 = µ 0 m3 g must be fulfilled. This leads to µ0 m3 g m1 + m2 k x < µ 0 m3 g v< m1 k
b) The velocity v¯3 of m3 after collision with the composite body can be determined from conservation of linear momentum and energy conservation:
→ : (m + m )¯v 1
2
12
= (m1 + m2 )v¯12 + m3 v¯3 ,
1 1 1 2 2 (m1 + m2 )¯ = (m1 + m2 )v¯12 + m3 v¯32 . v12 2 2 2
Hence, introducing the already know velocity v¯12 of the composite body, we obtain v¯3 =
2m1 v. m1 + m2 + m3
Impulse Law
69
Problem 3.8
Two swimmers (mass m1 = 75kg, m2 = 60 kg) jump into the water from an initially resting rowboat (mass mB = 150 kg), which can ideally glide without any drag. The first swimmer jumps horizontally from the stern with speed v1 = 2 m/s relative to the boat. Subsequently, the second swimmer jumps in the same direction with speed v2 = 3 m/s relative to the boat. a) Determine the velocity of the boat after the jumps. b) What speed has the boat, after both swimmers jumped simultaneously with the speed v3 = 2.5 m/s?
a) We introduce as positive direction the jump direction. Since the momentum of the system initially is zero, the sum of momenta after the first jump must vB1 + v1 vB1 stay zero. Thus, with the absolute vex m2 locities vB and vB + v1 of the boat m1 and the first jumper, we obtain Solution
1
→:
1
mB
(mB + m2 )vB + m1 (vB + v1 ) = 0 1
vB = 1
−m
1
m1 v1 = B + m1 + m2
−0.52 m/s .
The sign indicates that boat and jumper move in opposite directions. vB2 + v2 Before the second jump, the boat inv B2 cluding second jumper has the mom2 mentum (mB +m2 )vB , which is conserved after the jump: mB x 1
→:
(mB + m2 )vB = m B vB + m2 (vB + v2 ) 1
vB = 2
2
2
(mB + m2 )vB m2 v2 = mB + m2
−
1
−1.38 m/s .
b) If both swimmers jump simultaneously, the situation is analogeous to the first jump, but with changed masses and velocities:
→:
mB vB + (m1 + m2 )(vB + v3 ) = 0 3
3
vB = 3
−m
m1 + m2 v3 = + m + m B 1 2
−1.18 m/s .
P3.8
70 P3.9
Impulse Law
Problem 3.9
A rope without any bending stiffness (mass m, length l), moves freely on a frictionless plane, such that its bend K changes its position continuously (similar as it does at a horsewhip). In the initial state, K is in the middle of the rope; the lower part ① has the velocity v10 = v0 and the upper part ② the velocity v20 = 0.
initial state
K
m
②
v0
①
l/2 x K
u v2
②
v1
①
a) Determine the velocities v2 and v1 in dependence on the distance u of the ends of the rope. b) What are the velocities of the parts ① and ② at the instant when the bend K passes the end of the rope? Sketch the functions v1 (u) and v2 (u).
a) Because no external forces are acting, the linear momentum and the total energy at initial state must be conserved. Thus, with the lengths (l + u)/2 and (l u)/2 of the parts ① and ② of the rope, it follows
−
m m l + u m l u v0 = v1 + v2 , 2 2 l 2 l 1 m 2 1 m l + u 2 1 m l u 2 v0 = v1 + v2 2 2 2 2 l 2 2 l
−
momentum conservation:
−
energy conservation: respectively with y = u/l v0 = (1 + y)v1 + (1
− y)v , + (1 − y)v . 2
v02 = (1 + y)v12
2 2
(a)
Solving for v2 leads to the quadratic equation 2y(1
2 2
− y)v − 2(1 − y)v
0
v2
−yv
2 0
=0
with the solution
± − − ∓ − −
v0 v2 = 1 2y
2y 2 1+ . 1 y
Introducing this result into (a), yields v0 v1 = 3y 2y(1 + y)
1
(1
y)
2y 2 1+ . 1 y
Conservation of Energy Law
71
To find the correct sign in front of the root, we consider the initial state v1 = v0 , v2 = 0. By determining the limit values for y 0 it can be seen that the ’lower’ sign leads to the correct initial state. Thus, the solution reads
→
− − − − −
v0 v1 = 3y 2y(1 + y)
1 + (1
2y 2 1+ , 1 y
y)
2y 2 1+ . 1 y
v0 v2 = 1 2y
b) The bend K reaches the end of the rope for u l or for y respectively. Determining the limit values leads to the velocities
→
lim v1 (y) =
y →1
v0 , 2
lim v2 (y) =
y →1
→ 1,
−∞ .
c) In order to sketch v1 (y) and v2 (y), it is advisable first to approximate the functions near y = 0 and y = 1. By series expansion and neglecting higher order terms, we find y (1
1 :
− y) 1 :
v1 (y)
≈ v (1 − y/2), √ 1−y v v (y) ≈ 1 + √ 1
In conjunction with the results from a), b), we obtain the displayed courses of the functions:
v2 (y)
0
0
2
2
≈ −v
0
y/2 ,
√ 2 v √ v (y) ≈ − . 2 1−y
,
2
0
vi /v0 1
v1
1/2 1/2
−1/2 −1
v2
1
y = u/l
Remark: The infinite final velocity of the rope end ② (supersonic speed) may explain the so-called ’crack of the whipe’.
72 P3.10
Motion of Bodies
Problem 3.10
A rocket of initial mass m0 (including propellant mass mT ) is vertically launched at time t = 0. Assume that the mass flow µ and ejection speed w are constant and that at thrust cut-off the propellant is fully consumed. a) Determine the velocity v(t) of the rocket, assuming that there is no air drag and the gravity g is constant . b) What is the velocity at thrust cut-off for mT = 0.8 m0 , thrust duration tT = 2 min and w = 2000 m/s? c) What are the accelerations at lift-off and just before thrust cut-off? Solution
↑: With
a) The rocket motion is described by
m(t)
dv = dt
mg
−m(t)g + T .
T = µw ,
m(t) = m 0
v
− µt
follows the acceleration
T = µw
dv µw a = = dt m0 µt
− − g .
w
Integration and considering the initial condition v(0) = 0 leads to the velocity 1
v(t) = w ln
1
−
−gt .
µ t m0
b) The mass at thrust cut-off is m(tT ) = m 0 µ =
m0
− m(t
T )
tT
−m
T ,
resulting in
mT , tT
=
Thus, the velocity at thrust cut-off is given by v (tT ) = w ln
1
−
1 mT m0
−gt
T =
2000 ln
1 1 0.8
m km − . 9.81 · 120 = 2042 = 7350 − s h
c) The accelerations at lift-off (t =0) and just before thrust-off (t = t T ) follow as a(0)= a(tT )=
µw m0
m −g = m
T
0
µw m0 mT
−
w tT
· 2000 − 9.81 = 3.52 m − g = 0, 8120 s 2
− g = 0.2mm w t − g = 56.84 sm . T
0 T
2
with Variable Mass Problem 3.11
At point A of a drum, a rope (length l0 > h, total mass m0 ) is attached, the lower part of which initially rests on ground C . At time t = 0 the drum starts to rotate such that the rope is lifted with constant acceleration a0 and reeled. For t = 0 point A is at s = 0.
73 P3.11
A s B
h
Determine the force H in the rope at the drum inlet B. Assume that the rope hangs vertical at all times.
g
C Solution
We first determine the reeled rope length s from the known acceleration and the initial conditions s(0) = 0, s(0) ˙ = 0: s¨ = a 0 ,
s = v ˙ = a 0 t ,
s =
1 2 a0 t . 2
In what follows, two cases must be distinguished.
Case a: One part of the rope still rests at the ground (s + h In this case, the hanging part of the rope is regarded as a body with variable mass, in which at the s ground C mass flows in and at B mass flows out. Then, the mass m of the hanging part is m0 m = l0 h
m = m 0
h l0
and the mass flow follows as µ =
T = µ w =
0
H a B m
dm (ds/l0 )m0 m0 = = v. dt dt l0
At the lower end C of the hanging part, the ’ejection velocity’ w (there is no mass outflow but a mass inflow!) has the magnitude w = v, at the upper end at B it is zero (velocity of the outflowing mass = velocity of the hanging part). Accordingly, there exists only at C a thrust of magnitude
≤ l ).
h
mg
C T
m0 2 m0 2 2 v = a t , l0 l0 0
which is downwards directed. Hence, the equation of motion reads
↑:
m a0 = H a
− mg − T .
74
Motion of Bodies
It leads, after inserting T and with v 2 = 2a0 s, to the force
h s H a = m 0 a0 + g + 2a0 . l0 h From this result, for s = 0 and for the limit case s = l 0 the end of the rope lifts-off from ground), we obtain h H a (0) = m 0 (a0 + g), l0
H a (l0
h h) = m 0 l0
−
−
− h (just before
l0 a0 + g + 2a0 . h
Case b: The rope no longer touches the ground (s + h In this case, the mass of the hanging part is m0 m = l0 l0 s
−
m = m 0
l0
m a0 = H b
0
−s .
s
l0
Since there is no mass flow at the lower end and the ’ejection velocity’ at B is zero, there is no thrust anywhere. The equation of motion now reads
↑:
≥ l ).
H b
− mg .
B
m
Thus, the force in the rope follows as H b = m 0
l0
l0
− s (a + g) . 0
l0
−s
mg
For the limit case s = l 0 h (just after lifting the rope from ground) we obtain
−
H b (l0
− h) = m
h (a0 + g) . 0 l0
Remark: When comparing the limit cases of a and b for s = l0 h, it can be recognized that the force in the rope at lift-off experiences a jump of magnitude
−
∆H = H a (l0
− h) − H (l − h) = 2 l l− h m a b
0
0
0
0 0
.
with Variable Mass
75
Problem 3.12 A
whipcord (mass m, length l) is pulled at one end with the constant speed v0 . Given are the initial conditions x1 (0) = l/2, x2 (0) = l/2 and x˙ 1 (0) = v0 , x2 u x˙ 2 (0) = 0. F
Determine the required force F in dependence on the distance u of the cord ends.
m, l
v0 x1
Solution
Considering x˙ 1 = v0 = const and x ¨1 = 0, the kinematic relationships between u and x1 , x2 are given by u = x 1
−x
2
,
u = ˙ x˙ 1
− x˙
2
= v 0
− x˙
2
,
u ¨ =
−x¨
2
and the initial conditions are u(0) = 0 and u(0) ˙ = v 0 . In what follows we consider part ① and part ② of the whipcord as bodies with variable mass. Hence, the equations of motion read ①
②
− T ,
:
0 = F m
:
l
− u x¨ 2l
2
x1
(a)
x2 T
= T .
(b)
T
m
②
l
−u
2l F l + u v 0 m 2l
①
During the time increment dt the distance between the cord ends increases by du where part ② shortens by du/2 and part ① elongates by du/2. Accordingly, for part ② the mass flow and the relative ejection velocity are µ = um/2l ˙ and w = x˙ 1 x˙ 2 = u, ˙ respectively. Thus, the ’thrust’ is given by
−
T =
−µ w = − m u˙ 2l
2
(c) .
Introducing T into (b) and considering x ¨2 = (l
−u¨ yields
2
− u)¨u = u˙
and with u ¨ = ud ˙ u/du, ˙ separation of variables and integration du˙ du = u˙ l u
−
ln
u˙ l u(0) = ln u(0) ˙ l u
− −
This leads with (c) and (a) to m v02 F = T = 2l
− l
l
u
2
.
u = v ˙ 0
l
l
−u .
P3.12
Chapter 4
Kinematics of Rigid Bodies
4
78
Kinematics
The Motion of a Rigid Body can be composed of a Translation and a Rotation .
Spatial Motion The relations between the positions, the velocities and the accelerations of points A and P of a rigid body are given by
P ω
r P = r A + r AP , v P = v A + ω aP = a A
×r + ω˙ × r
AP
A
, z
AP
r AP
+ω
× (ω × r
rA
AP )
y
x
where ω = angular velocity vector.
Remarks:
• •
ω has the direction of the current rotation axis.
From the equations above follow v P and a P for any arbitrary point P of the body if v A , aA , ω and ω˙ are known.
Planar Motion With ω = ω ez = ϕ˙ ez , r AP = r er follows v P = v A + v AP , t n aP = a A + aAP + aAP ,
y
where
P
r
v AP = ω
AP
ϕ
AP )
t aAP
AP
ϕ
AP )
n aAP
er
eϕ
× r = ωre (⊥ to r = ω˙ × r = ωr ˙ e (⊥ to r = ω × (ω × r ) = −ω re ( to r AP
2
r
r AP
A ϕ x
AP )
’The velocity (acceleration) of point P is equal to the velocity (acceleration) of point A plus the velocity (acceleration) caused by the rotation of P about A’.
79
of Rigid Bodies
Velocity and Acceleration Diagram The graphical representation of the velocities and accelerations of plane kinematic problems and their graphical solution is done by using a velocity and a acceleration diagram. The directions of the velocity and acceleration components are taken from the layout diagram .: layout diagram
velocity diagram
acceleration diagram
ω, ω˙ P A
r aA
vP A = rω
vP vA vA
aA
aP
anP A = rω 2
atP A = r ω˙
Instantaneous Center of Rotation The plane motion of a rigid body having an instantaneous angular velocity ω can be considered at each instant as a pure rotation about the instantaneous center Π (instantaneous center of rotation, instantaneous center of zero velocity): vA = ρ A ω ,
vA
A vB
ρA ω Π
B ρB
vB = ρ B ω . The trajectory passed by the instantaneous center of rotation is called centrode .
Rolling Wheel
A
The locus of the center of rotation is given by: v0 = rω ,
r 0 v0 ω
vA = 2rω . Π
vA
80 P4.1
Velocity and Acceleration
Problem 4.1
The end point A of a rigid bar moves in a horizontal channel with speed vA and acceleration aA .
B l ϕ
Determine the velocity and acceleration of B as well as the angular velocity ω = ϕ˙ and the angular acceleration ω of ˙ the bar.
vA , aA A
Solution
The velocity and acceleration of B are vertically directed. With that information, the velocity diagram can be plotted and from the graph we find by inspection: lω =
vA cos ϕ
ω = ϕ = ˙
vA ϕ
vA , l cos ϕ
vB
vBA = lω
vB = v A tan ϕ . From the acceleration diagram follow in the same way aA + ω 2 l tan ϕ cos ϕ 2 aA vA sin ϕ ω˙ = + 2 , l cos ϕ l cos3 ϕ
aA
lω˙ =
ϕ
aB
lω 2 aB = a A tan ϕ + cos ϕ 2 vA = a A tan ϕ + . l cos3 ϕ
The problem can also be solved purely analytically. We introduce an appropriate coordinate system and from the coordinates xA = l sin ϕ ,
n 2 ϕ aAB = lω
atAB = l ω˙
y B ϕ l
yB = l cos ϕ A
x
Rigid Body
81
one obtains with x˙ A = vA and x ¨A = v˙ A = a A x˙ A = l ˙ ϕ cos ϕ
˙ ω = ϕ =
vA , l cos ϕ
2 v˙ A l cos ϕ + vA l ˙ ϕ sin ϕ vA sin ϕ aA ω˙ = = + , l2 cos2 ϕ l cos ϕ l2 cos3 ϕ
y˙ B = v B =
y¨B = a B =
−l ˙ϕ sin ϕ = −v −v˙
A tan ϕ
−
A tan ϕ
,
ϕ˙ vA = cos2 ϕ
−a
A tan ϕ
−
2 vA . l cos3 ϕ
Note that since y is positive upwards directed, the quantities y˙B and y¨B have a negative sign. The velocity vB and the angular velocity ω can also be determined by using the center of instantaneous rotation Π. This point Π is given by the intersection of the perpendiculars to vA and vB , the directions of which are known. In this way we obtain
B
vA = ωl cos ϕ
ω=
vA , l cos ϕ
l sin ϕ
vB
Π l cos ϕ
l
ϕ vA
vB = ωl sin ϕ = v A tan ϕ . A
Remark: From the derived relations it can be seen that ω˙ and aB are nonzero even when point A moves at constant speed (aA = 0).
82 P4.2
Velocity and Acceleration
Problem 4.2
At a crank mechanism, the wheel rotates with constant angular velocity ω and point P moves along a vertical straight guide rail.
P l
Determine analytically the velocity and acceleration of P .
a
r ω
Solution
We introduce a coordinate sytem and the two auxiliary angles ϕ and ψ, where ϕ = ωt and ϕ˙ = ω = const. Then the position, velocity and acceleration of P are given by l
y˙ P = y¨P
2
ϕ = ωt
r 2
ω
cos ψ .
ψ a
yP = r cos ϕ + l cos ψ ,
−rω sin ϕ − lψ˙ sin ψ , = −rω cos ϕ − lψ¨ sin ψ − lψ˙
P
y
x
The still unknown quantities sin ψ, cos ψ, ψ˙ and ψ¨ follow from the condition that P moves along a vertical line: xP = a = x˙ P x ¨P
−r sin ϕ + l sin ψ , = 0 = −rω cos ϕ + lψ˙ cos ψ , = 0 = rω sin ϕ + lψ¨ cos ψ − lψ˙ 2
2
sin ψ .
Solving for the unknowns yields a r sin ψ = + sin ϕ , l l r cos ϕ ψ˙ = ω , l cos ψ
cos ψ = ψ¨ =
−ω
2
− 1
a r + sin ϕ l l
r sin ϕ sin ψ + ψ˙ 2 . l cos ψ cos ψ
Introduction into y˙P and y¨P finally leads to y˙ P =
y¨P =
−rω −rω
sin ψ sin ϕ + cos ϕ cos ψ
2
cos ϕ
−
,
sin ψ r cos2 ϕ sin ϕ + cos ψ l cos3 ψ
.
2
,
Rigid Body
83
Problem 4.3
The boom with a pulley and a carrying rope is reeled in with speed v by a second horizontal cable. There is no sliding between pulley and carrying rope.
P4.3
v
r l α
Determine the angular velocities of the boom and the pulley for α = 45 ?
W
◦
Solution
Because the motion of the boom is a pure rotation about A, the velocity vP is perpendicular to l, and thus we have
vP
l ˙α = v P
45
◦
P
and vP =
v
√
l
2 v .
Hence, the angular velocity of the boom is v α = ˙ 2 . l
α = 45
◦
A
√
To determine the angular velocity ω of the pulley we use its instantaneous center of rotation Π. Its location is given by the intersection of the perpendiculars to vP and vB . Notice: Due to the constraint by the rope, point B can only move in vertical direction.
vB Π B
From
vP
√
2 rω = v P
we obtain vP v ω= = . r 2 r
√
√ 2 r
P
r
84 P4.4
Velocity and Acceleration
Problem 4.4 A
bar of length l moves in plane. In the displayed position, the velocity v A of point A and the angular velocity ω are known. a) Determine the magnitude and direction of the velocity y of point B. b) Where on the bar is point C located, whose instantaneous velocity has the x-direction? c) Identify the location of the instantaneous center of rotation Π.
ω
A
B
C
α b
vA a
l
x
◦
Given: vA = 3 m/s, α = 60 , ω = 6/s, l = 1/2m, a = b = 1/4 m.
a) From the relation v B = v A + ω
Solution
−v = −v
A cos α ex
vA = vB
−v
A sin α ey ,
A cos α ex +
(ω l
×r
AB
ω = ω ez ,
−v
A sin α) ey
and
r AB = l ex ,
vBx = 1.5 m/s , vBy = +0.402 m/s
−
we obtain
2 2 + vBx = 1.55 m/s , vBx
vB =
cos β =
vBx vB
◦
β = 165.4 .
b) The condition vCy = 0 leads with v C = v A + ω
=
−v
y
AC
vB
c
A cos α ex
+(ω c to
×r
A ω
−v
A sin α) ey
α rA
vA sin α c = = 0.433 m . ω
β
C B
vA
r AΠ
b
Π
yΠ a
xΠ
x
c) From the condition for the instantaneous center of rotation Π v Π = v A + ω
×r
AΠ
=0
ω
×v
A
=
−ω × ( ω × r
AΠ )
= ω 2 rAΠ
follows with r Π = r A + rAΠ the location of Π: 1 vA ω v = (a + sin α) ex + (b A ω2 ω vA xΠ = a + sin α = 0.683 m , yΠ = b ω
r Π = r A +
×
− vω − vω
A
cos α) ey ,
A
cos α = 0 .
Rigid Body
85
Problem 4.5 In
a gear, the wheel ① rolls along the fixed circular bearing ②. It is driven by a constant angular velocity Ω. Determine analytically the magnitudes of velocity and acceleration for a point P of the wheel.
P4.5
Ω
R P r
① ②
ϕ
Solution
We introduce a coordinate system and consider an initial and a displaced position of the wheel. With the angles α and β , where α + β = ϕ, it follows for the covered arc length of P y α
(R + r)α = r(β + α) Rα = rβ
R
Differentiation leads with α = ˙ Ω and ˙ = ω to β
α
P β
r P
x
R Ω = r ω
ω = Ω R/r .
The velocity and acceleration components are derived from the position of P : xP = R cos α + r cos β ,
yP = R sin α
x˙ P =
y˙ P
x ¨P
−RΩsin α − rω sin β , = −RΩ cos α − rω cos β , 2
2
y¨P
− r sin β , = RΩcos α − rω cos β , = −RΩ sin α + rω sin β . 2
2
Thus, with R Ω = r ω and α + β = ϕ we obtain 2 2 vP = x˙ 2P + y˙ P = R 2 Ω2 + r 2 ω 2 + 2RΩrω(sin α sin β
= 2R2 Ω2
2
2
cos(α + β ) = 2R2 Ω2 (1 ϕ vP = 2RΩsin , 2
− 2R Ω
− cos α cos β )
2 a2P = x ¨2P + y¨P = R 2 Ω4 + r 2 ω 4 + 2RΩ2 rω 2 (cos α cos β 2 2 R2 Ω2 2 2 2 2 R Ω = (RΩ ) + 2RΩ cos(α + β ) r r
−
aP = RΩ2
2
2
− cos ϕ) = 4R Ω
1 + (R/r)2 + 2(R/r)cos ϕ .
sin2
− sin α sin β )
ϕ 2
86 P4.6
Velocity and Acceleration
Problem 4.6
An idealized fair ride consists of a base plate and attached gondola crosses, both rotating with constant angular velocities ω1 and ω2 = 2ω1 , respectively.
l G ω1
ω2
Determine the magnitudes of velocity and acceleration of a gondola G.
2l A
Solution
We first solve the problem by sketching the velocity and acceleration diagrams. This can be done with the aid of the displayed layout diagram. Using the cosine rule, it can be seen 2 vG = (2lω1 )2 + (lω2 )2
l
y
A
2l
ϕ1 x
0
− 2 · 2lω lω cos(π − ϕ + ϕ ) = 8l ω [1 + cos(ϕ − ϕ )] v ϕ −ϕ = 16l ω cos , 2
2 1
2
1
2
2 1
2
2
2
1
1
2
= lω2
GA
ϕ2
1
2
vG = 4lω1 cos
ϕ2
−ϕ
1
2
a2G = (2lω12 )2 + (lω22 )2
2 1
−ϕ
1
vA = 2lω1
vG
G ϕ2
,
2 2
− 2 · 2lω lω cos(π − ϕ − ϕ ) = 4l ω [5 + 4 cos(ϕ − ϕ )] , 2
4 1
aG =
2
2lω12
2
1
1
5 + 4 cos(ϕ2
− ϕ ) . 1
ϕ2
−ϕ
anGA = lω 22
1
aA = 2lω12 aG
Remark: Because the cosine varies between +1 and 1, the maximum and minimum acceleration are given by aGmax = 6l ω12 and aGmin = 2l ω12 , respectively.
−
Rigid Body
87
The problem can also be solved analytically by considering the components of the position vector and by subsequent differentiation with respect to time. With ϕ˙ 1 = ω 1 and ϕ˙ 2 = ω 2 = 2ω1 we obtain xG = 2l cos ϕ1 + l cos ϕ2 , x˙ G =
−2l ϕ˙ sin ϕ − l ϕ˙ sin ϕ = −2l ω sin ϕ − l 2 ω sin ϕ = −2l ω cos ϕ − l 4 ω cos ϕ , 1
1
x ¨G
2 1
1
2
2
1
2 1
1
1
2
,
2
yG = 2l sin ϕ1 + l sin ϕ2 , y˙ G = 2l ω1 cos ϕ1 + l 2 ω1 cos ϕ2 , x ¨G =
2 1
2 1
−2lω sin ϕ − l 4 ω sin ϕ 1
2
.
Using the addition theorem it follows 2 2 vG = x˙ 2G + y˙ G = (2l ω1 )2 + (l 2 ω1 )2 +
+8l2 ω12 (sin ϕ1 sin ϕ2 + cos ϕ1 cos ϕ2 )
= 8l2 ω12 [1 + cos(ϕ2 ϕ2 ϕ1 vG = 4lω1 cos , 2
−
− ϕ )] 1
2 a2G = x ¨2G + y¨G = (2lω12 )2 + (4lω12 )+
+16l2 ω14 (cos ϕ1 cos ϕ2 + sin ϕ1 sin ϕ2 ) = 4l2 ω12 [5 + 4 cos(ϕ2
aG =
2lω12
5 + 4 cos(ϕ2
− ϕ )] − ϕ ) . 1
1
Remark: The instantaneous center of rotation Π of the gondola cross is located on the perpendicular to vA . Because A rotates about 0 its velocity is vA = 2lω1 . Due to the rotation of the gondola cross AG with the angular velocity ω2 = 2ω1 about Π at the same time the relation vA = ρA ω2 is valid. Thus, it follows ρA = l and therefore vG = ω 2 ρG = ω 2 2l cos
ϕ2
−ϕ
1
2
.
vG G 1 (ϕ2 2
−ϕ ) 1
ω2
Π 0
ρG v A ρA = l
A
l ϕ2
−ϕ
1
88 P4.7
Velocity
Problem 4.7
The crank DC of a crank-rocker mechanism rotates with constant angular velocity ω.
a
Determine the vertical velocity of point B and the angular velocity ψ˙ of the bar AB.
A
Solution
y
From the sketch we obtain the geometric relation
tan ψ =
B
r D ω
ψ
l
r sin ϕ = (l + r cos ϕ)tan ψ
C
a
C r ϕ
ψ A
D
l
r sin ϕ . l + r cos ϕ
B r sin ϕ x
r cos ϕ
This leads to tan ψ
yB = a sin ψ = a
1 + tan2 ψ
= ar
sin ϕ l2 + r2 + 2rl cos ϕ
and for the vertical velocity, using ϕ = ω, ˙ to ω cos ϕ y˙ B = ar
= arω
l2 + r2 + 2rl cos ϕ
sin ϕ) − 2 sinl ϕ+(−r 2rlω + 2rl cos ϕ
2
2
(l2 + r 2 + 2rl cos ϕ)
(r + l cos ϕ)(l + r cos ϕ) . (l2 + r 2 + 2rl cos ϕ)3/2
˙ obtained by differentiating tan ψ with respect The angular velocity ψ is to time: 1 ˙ = r cos ϕ (l + r cos ϕ) r sin ϕ ( r sin ϕ) ω ψ cos2 ψ (l + r cos ϕ)2
−
−
r(r + l cos ϕ) 1 r(r + l cos ϕ) = ψ˙ = ω ω. (l + r cos ϕ)2 1 + tan2 ψ l2 + r2 + 2rl cos ϕ
Angular Velocity
89
Problem 4.8
In a part of a freight lift, three wheels are connected by unrolling vertical cables.
P4.8
R r
Determine the velocities and angular velocities of the wheels ② and ③ when wheel ① rotates with a given angular velocity ω1 .
①
ω1 r2
② ③
r3
Solution Wheel ①
rotates about the fixed point Π1 . Therefore, the velocities of points A and B are given by vA = Rω1 ,
vB = rω1 . vC
Since the cables unroll, vA and vB will be transA ferred unchanged to the wheel ②. Thus, from vA = v 2 + r2 ω2
and
follows with r2 = (R v2 = ω2 =
vB = v 2
−r ω 2
2
− r)/2
vA
vA
r2
Π2
= ω 1 .
In the same way we obtain for wheel v2 = v 3 + r3 ω3 ,
vC = v 3
①
ω2 ②
2
vB
vA
1 1 (vA + vB ) = (R + r)ω1 , 2 2
−v
B Π1 C
−r ω 3
vB
v2 ω3
③ from
Π3 ③
3
v2
vc
v3
with r3 = 2r + r2 = (R + 3r)/4 and vC = rω 1 the results
−
v3 = ω3 =
1 1 (v2 + vC ) = (R 2 4 v2
−v
3
r3
− r)ω
1
,
= ω 1 .
Remark: The instantaneous centers of rotation Π1 , Π2 and Π3 are located on one and the same straight line.
90 P4.9
Plane Motion
Problem 4.9
Point A of a bar of length l is guided horizonally with the constant velocity v0 and slides at B across a post.
P B a
Determine the magnitudes of velocity and acceleration of point P of the bar. Solution
cot ϕ =
l A v0
The angular velocity ϕ of ˙ the bar is determined from x a
P
by time differentiation with x = ˙
− sinϕ˙ ϕ = xa˙
ϕ = ˙
2
−v
0
l
B
as
a
ϕ
v0 sin2 ϕ . a
A
x
Second differentiation leads to the angular acceleration
− − −
v0 v0 ϕ = ¨ 2 sin ϕ cos ϕ ϕ = ˙ 2 a a
2
sin3 ϕ cos ϕ .
Then, by using the cosine rule, we obtain from the velocity diagram 2 vP = v 02 + (lϕ) ˙ 2
vP = v 0
1+
π 2
2v0 l ˙ ϕ cos
l a
ϕ
2
sin4 ϕ
2
l a
lϕ ˙
sin3 ϕ .
vP π 2
− ϕ v0
The acceleration diagram yields aP =
aP
l2 ϕ˙ 4 + l2 ϕ ¨2
v0 = l a
2
v0 = = l a
2
sin3 ϕ
Remark: For ϕ = π/2 vP = v 0 (l/a
lϕ¨
sin8 ϕ + 4 sin6 ϕ cos2 ϕ lϕ ˙ 2
sin2 ϕ + 4 cos2 ϕ .
(point A arrives at the post) follows
− 1) = (l − a)ϕ˙ |
π/2
.
Velocity and Acceleration Diagram
91
Problem 4.10
The wheel of a crank drive rotates with constant angular velocity ω.
P4.10
Determine graphically the velocity and acceleration of the horizontally moving point B for the angle α = 30 .
A
r C α
ω
ϕ
l = 3r
r/2 B
◦
Solution
From the layout diagram the directions of the different velocity and acceleration terms are taken: vA to r, vBA to l, vB horizontal, aA in direction of r, an BA in direction of l, t aBA to l, aB horizontal. With this knowledge, the velocity and acceleration diagrams can be constructed.
⊥
r
A 30
ϕ
◦
C
l
r/2
B
⊥
⊥
velocity diagram
rω
scale:
vA = rω vBA = l ˙ϕ
ϕ vBA = 0.92 rω = l ˙
ϕ = ˙
0.92 r ω = 0.31 ω , l
30
ϕ
◦
vB
vB = 0.81 rω .
acceleration diagram
rω 2
scale:
aB
an BA
2
2
2
= l ˙ ϕ = 3r(0.31 ω) = 0.29 rω ,
aB = 0.99 rω 2 .
30
atBA ϕ
anBA
◦
aA = rω 2
92 P4.11
Instantaneous Center
Problem 4.11
Point A of the mechanism has in the displayed position the velocity vA .
C 2
A
3
B
Determine vA a) for the 4 bars the instantaneous centers of rotation, b) the velocities of points B and C as well as the angular velocities of all 4 bars. Given: a = 30 cm, b = 40 cm, v A = 1 m/s.
a
1
D 4
E
a
a
b
a) The centers Π3 (bar 3) and Π4 (bar 4) are the hinged supports D and E , respectively. The centers Π1 (bar 1) and Π2 (bar 2) are given by the interΠ1 sections of the perpendiculars to v A and v B (vB ω1 vC EB) and to vB and vC C (vC DC ), respectively (see figure). Solution
⊥
⊥
2
3
ω3
vB vA
1
D
B
Π3
A
b) For bar 1 follows: vA = Π1 A ω1 = b ω1 vA 1m ω1 = = = 2.5 s b 0.4 m s
4
vB = Π1 B ω1 =
a2
+
b2
1
−
ω2 = ω 4
, E
ω1 = 1.25 m/s .
For bar 4 and for bar 2 we obtain vB = Π2 B ω4 = a2 + b2 ω4 , ω2 = ω 4 = ω 1 = 2.5 s 1 ,
ω4 = ω 2
−
vC = Π2 C ω2 = (a + b)ω2 = 1.75 m/s . Finally, the rotation of bar 3 about Π3 leads to vC = Π3 C ω3 = a ω3
ω3 =
vC = 5.83 s a
1
−
,
Π2 = Π4
of Rotation
93
Problem 4.12
A toggle lever is driven with the angular velocity ω.
P4.12
A
Determine graphically the velocities of the points B, C and D for a certain angle β by using the centers of instaneous rotation.
l C l
◦
ω = 4s
1
−
l
r β E
ω
Given.: l = 30 cm, a = 40 cm, r = 15 cm, B b = 50 cm, β = 70 ,
D
a
b
.
Solution
The centers of rotation Π1 (bar CD) and Π2 (bar BC ) are given by the intersections of the perpendiculars to v D = rω (vD ED) and vC (vC AC ) and to vC and vB , respectively. This leads in the layout diagram to the depicted representation.
⊥
⊥
vD
scales:
r
A
D
vC
vD
C E α2
vB α2 B
Π2 α1
α1 Π1
With the chosen scale, we read from the figure
vC = 0.9 vD ,
vB = 1.0 vD
and obtain with vD = rω = 0.6 m/s the velocities
vC = 0.5 m/s ,
vB = 0.6 m/s.
94 P4.13
Velocity
Problem 4.13
A mechanism consists of three pin connected bars which are moved by a rope that is hauled in with constant speed vs .
l rope vS ϕ l/ 2 C
A
√
Determine graphically the velocities of the points B and C for ϕ = 60 .
l
D
√ l/ 2
◦
B Solution
We first solve the problem in the layout diagram by using the instantaneous center of rotation Π of bar C B. Π is given by the intersection of the perpendiculars to vB (vB AB) and vC (vC CD, magnitude of vC determinable from vs ).
⊥
A
vC 60
D
◦
Π
vS α
⊥
scale:
vS
C
α
vB
B
From the figure we obtain by inspection:
vC = 1.4 vs ,
vB = 5.4 vs .
The same result can be obtained by using the velocity diagram. Here, the directions of the velocities (e.g. vBC CB) are taken from the layout diagram:
⊥
vC
vC = 1.4 vs , vB =5.4 vs .
vS
vBC vB
and Acceleration
95
Problem 4.14
At the rolling wheel, a bar is connected by a hinge. The bar slides at B along an inclined plane.
P4.14
2.5 r
v0
r
A
2.5 r
Determine graphically the velocities and accelerations of B and C for the displayed position and the given constant speed v0 of the wheel.
B 30
2r
C
◦
Solution
From the layout diagram , using the centers of instantaneous rotation Π1 (wheel) and Π2 (bar), the directions of the velocities and accelerations can be obtained. With that information the velocity diagram is drawn (vA Π1 A, vBA BA, vB Π2 B) and it follows by inspection
⊥
⊥
Π2 A
0
Π1
C
⊥
B
vA
vB = 3.6 v0 ,
vBA = 3.5 v0 vCA = 1.7 v0
vC = 2.1 v0 ,
an BA =
an CA
v02 aC = 4.7 . r
vA0
vCA vC
2 vBA
5r
= 2.45
v02 r
,
vBA vB
2 vCA v02 = = 1.2 . 2.5r r
The acceleration diagram leads with aA = v 02 /r and atCA = a tBA /2 to v02 aB = 8.4 , r
v0
anBA aA
anCA
atCA aC
atBA
aB
v0
96 P4.15
Velocity
Problem 4.15
From the tire of a car (radius r = 30cm, speed v0 = 108 km/h) separates a stone at point P (α = 30 ). Determine ◦
a) the velocity components of the stone at the instant of separation, b) the maximum flight height and the flying distance of the stone, c) the minimum distance of a following car with same speed so that it will not be hit by the stone. The height of the separation point above street level, the air drag and the length of the following car can be disregarded.
g v0 r α
P
a) The velocity components are determined by using the center of instantaneous rotation Π. We obtain with ω = v0 /r, a = 2r sin α2 , β = π/2 α/2 = 75 and the conversion 1 m/s = 3.6 km/h Solution
v = ωa = 2v0 sin
ω
◦
−
α 2
= 15.53 m/s ,
v
y
vx = v cos β = 4.02 m/s ,
P
vy = v sin β = v 0 sin α = 15 m/s
v0 α/2
β a
r β
Π
x
b) The flight height h, flight distance d and flight time td follow from the equations for projectile motion (see page 32) h = v 2 sin2 β/(2g) = v 02 sin 2 α/(2g) = 11.47 m , α sin α/g = 12.29 m , 2 td = 2v sin β/g = 2v0 sin α/g = 3.06 s. d = v 2 sin2β/g = 4v02 sin 2
c) The minimum distance c follows from the distances covered during the flight time td . The following car and the stone arrive at the same time at the same position if: v0 tw = c + w ,
c = v 0 tw
− w = 79.45 m . v
v0
h c
w
and Acceleration Problem 4.16
In a gear, the shaft ① (radius 3a) and the ring ② (radius 6a) rotate with constant angular velocities ω1 and ω2 about the point 0. They drive at contact points B and C without slip a stepped shaft ③ (radii a and 2a).
97 P4.16
B a
ω2
2a P
A ③
6a
C ω1 0
a) Determine the magnitude of velocity v A and the acceleration ω 3 of the shaft ③. b) Determine the velocity v P (t) and the acceleration aP (t) of point P .
②
3a
①
a) The velocities of shaft ① and ring ② at the contact points follow from the angular velocities: Solution
vC = 3a ω1 ,
vB
vB = 6a ω2 .
The shaft ③ carries out a plane motion. Accordingly, its velocities at the contact points can be described by vC = v A
− aω
3
,
vA vC
−ω
1
2a a
A C
0
ω3 Π3
Equating the particular velocities leads to ω3 = 2ω2
③
3a
vB = v A + 2a ω3 .
vA = 2a(ω1 + ω2 ) ,
B
.
b) Point A undergoes a circular motion with the angular velocity
y P
ωA = v A /4a = (ω1 + ω2 )/2 .
ω3 t
A
If we choose a fixed coordinate system with the origin 0, the position vector of P is given by
ωA t
ey
0
x
ex
r P = [ 4a sin ωA t + 2a cos ω3 t]ex + [4a cos ωA t + 2a sin ω3 t]ey .
−
From that, the velocity and acceleration are determined by differentiation: vP = [ 4aωA cos ωA t
−
2 aP = [4aωA sin ωA
− 2aω sin ω t]e + [−4aω sin ω t + 2aω cos ω t]e , t − 2aω cos ω t]e + [ −4aω cos ω t − 2aω sin ω t]e . 3
2 3
3
3
x
x
A
2 A
A
A
3
2 3
3
3
y
y
Chapter 5
Kinetics of a Rigid Body
5
100
Principles of Linear and Angular Momentum
Spatial Motion: The motion of a rigid body is described by the Principle of Linear Momentum and the Principle of Angular Momentum . Principle of Linear Momentum: d p = F dt where
mac = F
or
F = sum of external forces,
m = total mass of the body, ac = acceleration of the center of mass, p = m v c = linear momentum.
Principle of Angular Momentum: dL(0) = M (0) dt where
M (0) L(0)
(0)
M
0
P dm
v
r 0P
= sum of external moments with respect to 0, = r 0P v dm = moment of momentum with respect to 0.
×
If the reference point 0 is space fixed or the center of mass, the angular momentum can be expressed by (0)
L
= Θ (0) ω
·
where
ω
= angular velocity,
Θ0 =
Θx Θxy Θxz Θyx Θy Θyz Θzx Θzy Θz
= inertia tensor
Euler’s Equations (Principle of angular momentum with respect to a body-fixed principal-axes system): Θ1 ω˙ 1 Θ2 ω˙ 2 Θ3 ω˙ 3
− (Θ − Θ ) ω − (Θ − Θ ) ω − (Θ − Θ ) ω
where
2
3
3
1
1
2
ω3 = M 1 , 3 ω1 = M 2 , 1 ω2 = M 3 2
1, 2, 3 = principal axes, Θi = principal moments of inertia, ωi = components of angular velocity.
101
Moments of Inertia
Moments of Inertia (see also volume 1, chapter 9) Axial Moments of Inertia:
Θx = (y 2 + z 2 )dm ,
c
Θy = (z 2 + x2 )dm , z
Θz = (x2 + y 2 )dm , Θa =
2
r dm = m
rg2
c
,
x
y
a
where rg = radius of gyration.
Θxy = Θyx = Θyz = Θzy = Θzx = Θxz =
Table of some moments of inertia:
slender rod
A
l
C
a
zxdm .
ml2 ΘA = 3
b b
thin disc
yz dm ,
mr2 m Θc = , Θb = (3r2 + l2 ) 2 12
l r
xydm ,
c
b
c
a
mr2 mr2 Θa = , Θb = 2 4
r C b C r
sphere
cuboid
c
A
− − −
ml2 Θc = , 12
C
cylinder
dm r
rs
Θa = Θc + mrc2 .
Parallel-Axis Theorem : Products of Inertia:
a
C
b C a
c
Θc =
2 mr 2 5
Θc =
m 2 (a + b2 ) 12
102
Equations of Motion, Conservation of Energy Law
Rotation about a Fixed Axis a
− a: La = Θa ω ,
Angular momentum
Θa ω˙ =
Principle of Angular Momentum Time integration yields
M a .
t
Θa ω
−Θ ω a
0
=
M a dτ = Ma .
t0
Plane Motion: Principles of linear and angular momentum (equations of motion) F 1
F 2 C
m¨ xc = m¨ yc =
ω
y
F x ,
F y ,
ΘA ˙ω =
F 3
x
M A
where A = fixed point or center of mass C . Time integration yields (principles of linear and angular impulse and momentum ) mx˙ c
− mx˙
c0
= F x ,
my˙ c
t
where
Fx =
− my˙
c0
= Fy ,
ΘA ω
t
F x dτ,
t0
Fy =
− T
0
A ω0
F y dτ,
t0
MA =
= MA ,
t
M A dτ.
t0
Work-Energy Theorem (see also chapter 2): T
−Θ
= U
Conservation of Energy Law (valid for conservative forces): T + V = T 0 + V 0 = const
Kinetic Energy (plane motion): T =
1 1 mvc2 + Θc ω 2 . 2 2
Special case Rotation about a fixed axis a
1 mvc2 , 2 1 T = Θa ω 2 . 2 T =
Special case pure Translation:
− a:
Translation Problem 5.1 A
block (mass m) slides downwards an inclined rough plane.
a
Determine the acceleration. Under what circumstances tilt over is excluded? Solution
As long as the block does not tilt, its motion is pure translation. With ω˙ = 0, y¨c = 0, the equations of motion are given by
: :
m¨ xc = mg sin α
C
C :
− µN ,
x C
y
α
mg
R = µN N d
Thus, it follows x ¨c = g (sin α
a µ
α
− mg cos α , 0 = d N − aµN .
0 = N
P5.1
b
b
g
103
− µ cos α) ,
d = a µ .
Tilt over is excluded for d
≤b
µ
≤ b/a .
Problem 5.2 A
homogeneous bar is hinged supported at point A of a vehicle and loosely rests at B.
P5.2
g B
2l m
Determine the reaction forces at A and B when the vehicle moves with an acceleration a.
45
A
c
◦
a
Solution
For pure translation with x ¨c = a, y¨c = 0 and ω = ˙ 0, the equations of motion read
√ → : ma = A − 2 B , √ ↑ : 0 = A + 2 B − mg , √ √ √ : 0= 2 lA − 2 lA − ( 2 c − l)B.
x
1 2
y
1 2
1 2
C
y
Remarks:
• •
x
l
C
B
mg Ax
B = mg 1
a + g
2 1
a l , g 4c
a l , g 2c
Ay = mg 1
2 1
For a = g follows B = 0, Ax = A y = mg. For a > g the bar lifts off at B.
y x
Ay
− √ − − √ −
This leads to Ax = mg
1 2
√ 2c
a l . g 4c
104 P5.3
Translation
Problem 5.3
A homogeneous disk (weight W = mg) of quadratic parabola shape is fixed in the position α by three pin-supported bars.
1
2
3b/8 α
Determine the acceleration of the disk and the forces in the bars immediately after the mounting by bar 3 is released.
α
3
W
b
C a
2a/5
Solution
When the mounting by bar 3 is released, the disk starts to move purely translationally. Therefore, the acceleration is the same for all points of the disk and it can be described by the acceleration of point B which rotates about the support A: 2
a = a B = v˙ et +
v en . ρA
α
v=0
2a/5 S 2 B
S 1
Immediately after the constraint is released, the velocity is still zero, i.e.,
3b/8
A
en
et
C
a = v˙ et ,
mg
and the equations of motion read
: :
C :
mv˙ = mg sin α , 0 = S 1 + S 2 0=
− mg cos α ,
3 3 aS 1 cos α + bS 1 sin α 5 8
− 52 aS cos α + 83 bS sin α . 2
2
From these three equations, we obtain v˙ = g sin α , S 1 =
mg (8cos α 16
− 5 ab sin α) ,
S 2 =
mg b (8 cos α + 5 sin α). 16 a
Remark: For tan α = 8a/5b the force in bar 1 is S 1 = 0. In this case, the action line of S 2 passes through C .
of a Rigid Body Problem 5.4 A
car with mass m brakes on a rough road without sliding.
l
105 P5.4
l
Determine the possible braking decelerations ab when the brakes are effective only to the front wheels or only to the rear wheels. The mass of the wheels shall be disregarded.
C
h
µ0
Solution
x ¨c =
−
When the brakes affect only the front wheels, we obtain with ab , y¨c = 0, ω = ˙ 0 from
→ : −ma = −H , ↑: 0 = N + N − mg , 0 = lN − lN − hH C : 1
b
1
ab
y
C x
2
mg
1
2
h
1
l
N 2
H 1 N 1
l
the forces H 1 = ma b ,
N 1 =
1 h m (g + a b ) , 2 l
N 2 =
− hl a ) .
1 m (g 2
b
The conditions that the front wheels do not slide and the rear wheels do not lift off lead to l 1 H 1 µ0 N 1 ab g = κ1 , h [2l/(µ0 h) 1]
≤
N 2
≤
≥ 0
ab
≤ g hl = κ
−
2
.
For µ0 h/l 1, we have κ1 κ2 , i.e. it results ab µ0 h/l 1 follows ab κ2 . When the brake is effective to the rear wheels, the force H 1 in the equations of motion must be replaced by H 2 . The resulting forces N 1 and N 2 then remain unchanged and H 2 we obtain H 2 = mab . Thus, from H 2 µ0 N 2 and N 2 0 follows
≥
≤
≤
≤
≥ l 1 a ≤ g = κ h [2l/(µ h) + 1] b
0
≤
3
,
ab
≤ g hl = κ
Because of κ3 < κ2 in this case always ab
2
≤
κ1 , while for
C mg N 2
N 1
.
≤ κ holds. 3
Remarks: The maximum deceleration is for front brakes always greater (κ1 , κ2 > κ3 ). The case µ0 h/l 1 does not occur for an ordinary car under real conditions. Therefore, κ2 is non-relevant.
• •
≥
106 P5.5
Moments of Inertia
Problem 5.5
The mass density of a graded shaft is given by ρ = ρ 0 (1 + κr2 ).
z
y r
x
Determine the moments of inertia Θx and Θy .
y
R
l/2
l/2
Solution
We first consider a thin circular disk element of thickness dx. With the mass z dm = ρ 2πr dr dx = 2πρ 0 (r + κr 3 )dr dx of a chosen ring, we obtain for its mass and its moment of inertia
R
∗
dm =
0 R
∗
dΘx =
r
dr
y R
π dm = ρ0 R2 dx(2 + κR2 ) , 2
R
2
r dm = 2πρ 0 dx (r3 + κr 5 )dr
0
= 2πρ 0 dx
0 6
4
R R +κ 4 6
=
π ρ0 dx R4 (3 + 2 κR2 ) . 6
Because of the radial symmetry of the disk, its axial moments of inertia with respect to x, y and z (axes through the center of mass of the disk) are simply related by dΘx = dΘy + dΘz = 2dΘy , i.e., ∗
∗
∗
∗
π ρ0 dx R4 (3 + 2 κR2 ) . 12
∗
dΘy =
To calculate Θx and Θy , we now integrate over the length l, where we use the parallel-axis theorem to obtain Θy : l/2
π Θx = dΘx = ρ0 R4 (3 + 2 κR2 ) 6 l/2 ∗
−
l/2
Θy =
(dΘy + x2 dm ) ∗
l/2
dx =
l/2
−
π ρ0 lR4 (3 + 2 κR2 ) , 6
∗
l/2
−
l/2 π π 4 2 = ρ0 R (3 + 2 κR ) dx + ρ0 R2 (2 + κR2 ) 12 2 l/2 π = ρ0 lR 2 2R2 (3 + 2κR2 ) + l2 (2 + κR2 ) . 24
−
l/2
x2 dx
l/2
−
Remark: For κ = 0 follow with m = ρ 0 πR 2 l the results of the homogeneous shaft: Θx = mR 2 /2 , Θy = m(3R2 + l2 )/12.
Moments of Inertia
107
Problem 5.6 Determine
the moments of inertia Θa of rings (density ρ) with different circular and half-circular cross sections of radius c = R/2. a
R
a c
A
R
a c
B
a
R c
C a
a
The moment of inertia is defined as r2 dm. We first choose an appropriate ring-shaped mass element of radius r and thickness dr. From the sketch follows Solution
a
dm = ρ 2πr2c cos ϕ dr ,
dr c
R
r = R + c sin ϕ ,
ϕ
dr = c cos ϕ dϕ ,
dm = 4πρc 2 (R + c sin ϕ)cos2 ϕ dϕ .
= 4πρc 2
ϕ2 (R ϕ1
a
+ c sin ϕ)3 cos2 ϕdϕ
ϕ2 3 [R ϕ1
cos2 ϕ + 3R2 c sin ϕ cos2 ϕ
+3Rc2 sin2 ϕ cos2 ϕ + c3 sin3 ϕ cos2 ϕ]dϕ
= 4πρc 2 R3 14 (2ϕ + sin 2ϕ) +3Rc2 ( 18
= 4πρc
2
2
3
− R c cos ϕ ϕ − sin 4ϕ) + c (− 3
1 32
In case A, we obtain with ϕ1 = ΘA a
c sin ϕ
r
Depending on the cross section, we must integrate over different domains: Θa = 4πρc 2
2c cos ϕ
−π/2, ϕ
2
3 1 3 cos ϕ
+
5 1 5 cos ϕ)
ϕ2 ϕ1
.
= +π/2, mA = 2π 2 ρc2 R
π 3 3π 3 R + Rc2 = m A R2 + c2 = 1.19 mA R2 . 2 8 4
Case B with ϕ1 = 0, ϕ2 = +π/2, mB = π 2 ρc2 (R + 4c/(3π)) (note: mB = m A /2 !) leads to
ΘB a
= 4πρc
2
π 3 3π 2 3 R + R2 c + Rc2 + c = 1.52 mB R2 . 4 16 15
Finally, in case C follows for symmetry reasons (mC = m A /2)) ΘC a
A
= Θ /2 = m C
3 R + c2 = 1.19 mC R2 . 4 2
P5.6
108 P5.7
Moments of Inertia
Problem 5.7 Determine
the moments of inertia Θx and Θy for a homogeneous cone of mass m.
z y
R x H
Solution
It is practical to use as mass element the sketched circular disk. With its radius y
a = x
R H
y
R
a
and mass
dx
R2 2 2 dm = ρπa dx = ρπ 2 x dx H
x
x y
its moment of inertia dΘx is given by 1 π dΘx = a2 dm = ρ 2 2
4
R H
x4 dx .
Thus, for the cone follows by integration Θx =
dΘx = ρ
π 2
R H
H
4
x4 dx = ρ
π 4 R H. 10
0
The moment of inertia of the circular disk element with respect to the y -axis is given by
dΘy
1 π = a2 dm = ρ 4 4
4
R H
x4 dx .
Therefore, we obtain for the cone by using the parallel-axis theorem
R H
Θy = [dΘy + x2 dm] = ρπ
π 1 2 = ρ R2 H R + H 2 5 4
2 H 0
1 4
R H
2
x4 + x4 dx
.
Finally, introducing the mass m = ρπR 2 H/3, the moments of inertia can be written as Θx =
3 mR2 , 10
Θy =
3 m 5
1 2 R + H 2 4
.
Moments of Inertia Problem 5.8 Determine
the moments of inertia Θx and Θxy for a homogeneous rectangular thin plate (mass m, thickness t a, b).
109 P5.8
y
b C
x
a
Solution
For the thin plate exists a direct relation between the mass moments of inertia and the area moments of inertia (see vol. 1, chapter 9). With dm = ρtdA and z x, y, it is given by
−
Θx = (y 2 + z 2 )dm = ρt Θxy =
xydm =
−ρt
y 2 dA = ρtI x ,
xydA = ρtI xy .
Consequently, also the transformation relations (see vol. 1, chapter 9) can be applied. In this way, we find for the axes x and y which are inclined by ϕ to the pricipal axes 1 and 2:
2
y C
−
Θx = Θxy =
−ϕ
1
x
Θ1 + Θ2 Θ1 Θ 2 + cos2ϕ , 2 2
−
Θ1
− Θ 2
2
sin2ϕ .
With ab3 mb2 Θ1 = ρt = , 12 12 a cos ϕ = , a2 + b2
√
2
cos 2ϕ = 2 cos ϕ
−
ma2 Θ2 = , 12 b sin ϕ = , a2 + b 2
√
a2 b 2 2ab 1= 2 , sin2ϕ = 2 sin ϕ cos ϕ = , a + b2 a2 + b2
−
we finally obtain m(a2 + b2 ) m(b2 a2 ) (a2 b2 ) ma2 b2 Θx = + = , 24 24 (a2 + b2 ) 6(a2 + b2 )
−
Θxy
m(b2 a2 ) 2ab = = 24 a2 + b 2
−
−
−
m(a2 b2 )ab . 12(a2 + b2 )
−
Remark: For a = b follows Θx = ma 2 /12 and Θxy = 0.
110 P5.9
Rotation
Problem 5.9
An angled homogeneous bar of weight W = mg is pin-supported at A.
A a g
Formulate the equation of motion.
W 2a
Solution
The motion of the system is a pure rotation about the fixed point A due to the moment of the weight. The weight acts at the center of mass C . Its coordinates follow from (see volume 1, chapter 2) 2m a 2 3 = a, xc = m 3 m a 2m + a 5 3 2 3 yc = = a m 6
l =
x2s + ys2 =
√ 29 6
x l C
A :
y
A l C ϕ mg l sin ϕ
ΘA ϕ = M ¨ A ,
where
yc
a.
To describe the motion, it is advantageous to introduce the angle ϕ, which characterizes the displaced position relative to the equilibrium position (where C is located below A). The principle of angular momentum then reads
xc
2m/3
m/3
m a2 (2a)2 2 5 ΘA = + + (a2 + a2 ) m = ma2 3 3 12 3 3 29 M A = mgl sin ϕ = mga sin ϕ . 6
−
−
√
This leads to the equation of motion 5 ma2 ϕ = ¨ 3
−
√ 29 6
mga sin ϕ
ϕ ¨+
√ 29
g sin ϕ = 0 . 10 a
Remarks: From ϕ(ϕ) ¨ the angular velocity ϕ(ϕ) ˙ can be determined by integration (see page 4). For small displacements (ϕ 1, sin ϕ ϕ), the equation of motion describes a harmonic vibration (see chapter 7).
≈
about a Fixed Axis
111
Problem 5.10
A circular disk (weight W = mg), which rotates with the angular velocity ω0 about the vertical axis, is put on a rough horizontal plane (coefficient of kinetic friction µ).
P5.10
ω0 W
After what time T and after how many turns u the disk comes to rest? Assume that the contact pressure is constant. Solution
r0
µ
The rotation about the fixed axis is described by Θ ω˙ = M .
Here, the moment M is determined from the friction forces distributed over the circular area. From dA = rdα dr ,
m = ρπr02 h
dN = ρghdA ,
dR = µdN
rdα
dr ϕ, ω
r0
r α
dR
h
follows (the friction moment is directed opposite to the angular velocity) M =
−
r0 2π
r dR =
−µρgh
2
r dα dr =
0
0
−
2 µr0 mg . 3
dN
Thus, with Θ = mr 02 /2 and ω˙ = ϕ, ¨ the equation of motion leads to ϕ = ¨
− 43 µg . r 0
Twice integration and considering the initial conditions ϕ(0) = 0, ϕ(0) ˙ = ω 0 yields ϕ = ω ˙ 0
− 34 µg t, r
ϕ = ω 0 t
0
− 32 µg t r
2
.
0
From the condition ϕ(T ) ˙ = 0, we finally obtain T =
3 ω0 r0 , 4 µg
3 ω02 r0 ϕ(T ) = 8 µg
3 ω02 r0 ϕ(T ) u = = . 2π 16π µg
112 P5.11
Rotation
Problem 5.11
A flywheel which initially rotates with speed n (rpm) about a fixed axis is brought to standstill in time tb by a constant braking moment M b . Determine the mass moment of inertia Θ and the number of revolutions u of the flywheel during braking.
Solution
Θϕ = ¨
During braking, the motion is described by
−M
ϕ = ¨
b
− M Θ
b
.
(Note that the braking moment acts opposite to the positive direction of rotation.) Twice integration and considering the initial conditions ϕ(0) ˙ = ω 0 = 2πn ,
ϕ(0) = 0
leads to ϕ = ˙
− M Θ
t + 2πn ,
ϕ =
− M 2Θ
t2 + 2πnt .
B
B
From the condition ϕ(t ˙ b ) = 0 (standstill) we obtain 0=
− M Θ
b
tb + 2πn
ϕb = ϕ(tb ) =
−
M b t2b + 2πnt b = πntb 2Θ
Θ=
M b tb , 2πn
u =
ϕb 1 = ntb . 2π 2
The problem can be solved easier by using the principle of angular impulse and momentum and the work-energy theorem. With ω(tb ) = 0 and ω0 = 2πn, they lead to tb
M dτ = Θω(tb)
0
T
− T
0
= W :
− Θω
0
− 12 Θω
2 0
:
=
−M t
b b
−M ϕ b
b
=
−Θω
0
u =
Θ=
M b tb , 2πn
ϕb 1 = ntb . 2π 2
about a Fixed Axis
113
Problem 5.12
On a homogeneous cylindrical roll of mass m1 , an inextensible rope is wound up, which is connected with a body ② of weight W 2 = m 2 g.
P5.12
r A
Determine the acceleration of body ② and the force S in the rope when the roll can rotate about A without friction. The mass of the rope can be disregarded.
g
m1
②
W 2
Solution We
separate the roll and body ② by cutting the rope, choose positive directions of motion and formulate the equation of motion for the roll
A :
A
B
r
Θϕ = r ¨ S
and the body
↓:
ϕ˙
S S
②
m2 x ¨ = m 2 g
−S .
m2
The 2 equations of motion containing 3 unknowns (¨ x, ϕ¨ and S ) must be suplemented by a 3rd equation describing the kinematic constraint. The velocity r ˙ ϕ of point B of the roll and the velocity x of ˙ body ② and the rope, respectively, must be equal: ˙ x = r ϕ˙
x
②
m2 g
¨ = r ϕ ¨. x
This leads with Θ = m 1 r2 /2 to x ¨ = g
2m2 , m1 + 2m2
S = m 2 g
m1 . m1 + 2m2
Remark: In the static case (roll is blocked) the force in the rope is S St = m 2 g, while in dynamics S < S St holds.
114 P5.13
Rotation
Problem 5.13 Two
homogeneous drums (weights m1 g, m2 g), initially rotating with angular velocity ω0 in the same direction, are placed on top of each other, such that they slide against each other (coefficient of dynamic friction µ).
②
r2
m2
ω0
µ m1
r1
After which time tR , the drums roll from one another and what are then their angular velocities?
①
ω0 Solution
During sliding the equations of motion
read ① ②
B :
A :
↑
:
Θ1 ϕ ¨1 = 2
r2
1
2
②
A
−r R , Θ ϕ ¨ = −r R , 0 = N − m g .
R
2
N R
2
With the friction law R = µN and Θ1 = m 1 r12 /2, ϕ˙1 Θ2 = m 2 r22 /2 and considering the initial conditions, it follows
− 2µg r 2µg =− r
ϕ ¨1 =
1
ϕ ¨2
m2g
ϕ˙2
m2 m1
m − 2µg t+ω r m 2µg =− t+ω . r 2
ϕ˙ 1 =
1
2
ϕ˙2
0
①
r1
B
,
1
2
0
The drums stop sliding and start rolling from each other when their velocities at the contact point are the same at time t = t R : r1 ϕ˙ 1 =
−r ϕ˙ 2
2
.
Note that positive angular velocities of the drums ① and ② lead to opposite directed velocities at the contact point! Thus, we obtain tR =
ω0 (r1 + r2 ) m2 2µg 1 + 1
and for the angular velocities m1 r2 ϕ˙ 1 (tR ) = ω 0 m2 mr11 , 1+ m2
−
ϕ˙ 2 (tR ) =
−
r 1 m1 1 r m ω0 2 2m1 . 1+ m2
−
Remark: For m1 /m2 = r2 /r1 both drums come to standstill.
about a Fixed Axis
115
Problem 5.14
At an elevator, the cable drum ② is driven by the wheel ① without slip.
P5.14
Θ2
r2
Determine the acceleration a of the elevator ③ (weight W = mg) and the cable force S , when the wheel is driven by a constant moment M 0 . The weight of the cable can be neglected.
②
S Θ1
r1
①
W ③
M 0
Solution
We separate the system and obtain for the different parts ①
②
③
A:
B :
↑
:
Θ1 ˙ω1 = M 0
②
ω2
−r H ,
Θ2 ˙ω2 = r 2 H
C
B
1
−r S , 2
ma = m v˙ = S
N
− mg .
Since the velocity of point C of the drum ② and the velocity of the elevator (inextensible cable) as well as the velocities of D and E (no slip) must be equal, the kinematic relations are given by r2 ω2 = v
r2 ˙ω2 = v˙ = a ,
r2 ω2 = r 1 ω1
r2 ω˙2 = r1 ω˙1 .
S
D
H
H ω1
E
①
A
M 0 S v
③
mg
Thus, we have 5 equations for the 5 unknowns (ω˙ 1 , ω˙ 2 , a, H , S ). Solving for a and S leads to M 0 mg r1 a= , Θ1 Θ2 m+ 2 + 2 r1 r2
−
S = m
M 0 + r1
Θ1 Θ2 + r12 r22 Θ1 Θ2 m+ 2 + 2 r1 r2
g .
Remark: For M 0 = r 1 mg we obtain a = 0 (statics) and S = H = mg.
116 P5.15
Rolling
Problem 5.15
A homogeneous drum (weight W = mg) rolls downwards a rough inclined plane (coefficient of static friction µ0 ).
W µ0
Determine its acceleration a. Under which circumstances pure rolling is possible?
α
Solution
With y¨c = 0 and x ¨ c = a, we obtain from the principle of linear momentum
: :
ma = mg sin α
y
−H ,
C :
C
x H
− mg cos α
0 = N
r
α mg N
and from the principle of angular momentum (with respect to the center of mass C )
ω
Θc ω˙ = rH .
After introducing Θc = mr2 /2 and the kinematic constraint (rolling without slip) rω = x˙ c
r ω˙ = x ¨c = a
ω˙ =
a , r
solving for the acceleration yields a =
2 g sin α . 3
The condition for non-slipping at the contact point of drum and plane (pure rolling) is given by H
≤ H
0
= µ 0 N .
With H = mg sin α
− ma = 31 mg sin α ,
it leads to
≥ 31
µ0
tan α .
N = mg cos α ,
and Sliding
117
Problem 5.16
A homogeneous bowling ball (r = 0.11 m) is placed with an initial velocity v0 = 7 m/s and an angular velocity ω0 = 10 s 1 in a rough bowling alley (kinetic friction coefficient µ = 0.15).
P5.16
ω0
−
r
v0
Determine the covered distance xr until the ball rolls without slipping and its final speed vr .
µ
Solution
We draw the free-body diagram and introduce positive directions for distance and angular velocity. After contact, the ball initially is slipping. Thus, the principles of linear and angular momentum yield ω
→: ↑:
C :
C
r mg
x
m¨ x =
−R , 0 = N − mg
N = mg ,
R
N
Θc ω˙ = rR .
We now introduce Θc = 2mr2 /5 and the friction law R = µN = µmg and subsequently integrate the 1st and the 2nd equation. Taking into account the initial conditions v(t = 0) = v0 , x(t = 0) = 0, ω(t = 0) = ω0 leads to
−
v = x = v ˙ 0
− µgt ,
x = v 0 t
− 21 µg t
2
,
ω=
5µg t 2r
−ω
0
.
When the ball is rolling, v and ω are related by v = rω , from which the time tr follows for onset of rolling: v0
− µgt
r
=
5 µ g tr 2
−ω
0
tr =
Finally, this leads to xr = x(tr ) = v 0 tr vr = v(tr ) = v 0
− 21 µg t
− µgt
r
2 r
= 9.17 m ,
= 4.69 m/s .
2(v0 + ω0 r) = 1.57 s . 7µg
118 P5.17
Principles of Linear
Problem 5.17
A cable drum on a rough surface is set into rolling motion (no slipping) by pulling with a force F .
m, ΘC
Determine the pulling angles α, for which the acceleration ac takes extreme values. What are the associated maximum forces, such that no slipping occurs? Given: W = mg, Θc = 3r12 m, r1 = 2r2 , µ0 .
r2 C
F
r1
α µ0
Solution
The principles of linear and angular momentum yield
→: ↑:
C :
−F cos α + H , 0 = N − mg + F sin α , Θc ω˙ = −r H + r F .
ω
mac =
1
mg r1 C
F
2
α
r2
vc
In connection with the kinematic condition for rolling, vc = r 1 ω
H N
v˙ c = a c = r 1 ˙ω ,
we obtain the acceleration and forces ac =
− mF cos α −Θrs/r 2
1
=
1+
F − 8m (2 cos α − 1) ,
r12 m Θc cos α r2 + r12 m r1 F H = F = (6cos α + 1), Θc 8 1+ 2 r1 m
N = mg
− F sin α .
The extreme values of ac follow from dac /dα = 0, i.e. sin α = 0, as α1 = 0
ac1 =
−F/8m , ,
α2 = π
ac2 = 3F/8m .
The associated maximum forces F i are calculated via the maximum force of static friction:
|H |(α ) = H i
0
= µ 0 N (αi )
F 1 =
8 µ0 mg , 7
F2 =
8 µ0 mg . 5
Remark: For α1 = 0, the drum moves to the left and for α2 = π to the right.
and Angular Momentum Problem 5.18
A homogeneous plate of weight W = mg is symmetrically supported. a) Determine the support reactions at A at motion initiation when the support B suddenly is removed. b) What distance d must be chosen, such that the vertical force atA does not change in comparison to the static case?
119 P5.18
2b
W
2h
B
A 2d
a) Immediately after removing support B the principles of linear and angular momentum (with respect to the fixed point A) yield Solution
→: ↑:
A :
rC
m¨ xc = A H , m¨ yc = A V
mg
y
− mg ,
C
ω
aC
h
α x
ΘA ˙ω = d mg ,
AH AV
d
where ΘA = Θc + mrc2 =
m m 2 (4b2 + 4h2 ) + m(d2 + h2 ) = (b + 4h2 + 3d2 ) . 12 3
Since the center of mass C is purely rotating about A, its acceleration components at initiation of motion (vc = 0) are given by 2 vC an = = 0 , rC
ac = a t = r c ˙ω ,
and we obtain with sin α = h/r c , cos α = d/r c x ¨c = a c sin α = h ω˙ ,
y¨c =
−ac cos α = −d ω˙ .
Solving for the support reaction leads to 3dh AH = W 2 , b + 4h2 + 3d2
b2 + 4h2 = W 2 . b + 4h2 + 3d2
AV
b) Before removing support B (statics), the vertical force at A is A V = W/2. Thus, from the condition that there is not change, it follows W b2 + 4h2 = W 2 2 b + 4h2 + 3d2 Remark: Because of d b 2 h.
√ ≥
√
1 d = 3
≤ b, the force A
V remains
b2 + 4h2 . unchanged only for
120 P5.19
Principles of Linear
Problem 5.19
An initially upright bar (weight W = mg) starts to move, where it slides without friction along the wall and the base.
ϕ W
Determine the angular velocity ϕ(ϕ). ¨ What forces are acting on the wall and the base? y
Solution As
long as the bar slides along the wall and the base, the principles of linear and angular momentum (with respect to C ) yield m¨ xc = B ,
l Θc ϕ = A ¨ sin ϕ 2
C :
l 2
B
C
yc
→: ↑:
mg
m¨ yc = A
− mg ,
,
−
l
ϕ x
xc
l B cos ϕ 2
l 2
A
where Θc = ml 2 /12. The kinematic quantities xc , yc and ϕ are related by l xc = sin ϕ , 2
yc =
l cos ϕ . 2
After differentiation one obtains l ϕ˙ cos ϕ , 2 l x ¨c = (ϕ ¨ cos ϕ 2 x˙ c =
2
− ϕ˙
sin ϕ) ,
y˙ c =
− 2l ϕ˙ sin ϕ ,
y¨c =
− 2l (ϕ¨ sin ϕ + ϕ˙
2
cos ϕ) .
Thus, we have 5 equations for the 5 unknowns (¨ xc , y¨c , ϕ, A, B). ¨ Solving for ϕ leads ¨ to ϕ = ¨
3g sin ϕ . 2l
The forces are determined by ¨c = m B = m x
l (ϕ ¨ cos ϕ 2
A = mg + m¨ yc = mg
2
− ϕ˙
sin ϕ) ,
− m 2l (ϕ¨ sin ϕ + ϕ˙
2
cos ϕ) .
and Angular Momentum
121
Here ϕ is ˙ required. We can calculate it from ϕ ¨=
dϕ˙ dϕ dϕ˙ = ϕ˙ dϕ dt dϕ
ϕ˙ dϕ = ˙ ϕ ¨ dϕ
by integration: 1 2 ϕ˙ = 2
3g ϕ ¨ dϕ + C = 2l
sin ϕ dϕ + C =
− 3g cos ϕ + C . 2l
The integration constant C follows from the initial condition ϕ(ϕ ˙ = 0) = 0 as C = 3g/2l, which leads to ϕ˙ 2 =
3g (1 l
− cos ϕ) .
Thus, the froces are given by B=
3 W (3cos ϕ 4
A =
1 W (9cos2 ϕ 4
− 2) sin ϕ , − 6 cos ϕ + 1) .
A B , W W 1
A/W B/W 30
60
◦
◦
90
◦
ϕ
ϕ1
−1
Remarks: ◦
• The force B is zero for ϕ = 0 and 3cos ϕ − 2 = 0 ϕ = 48.2 . • For ϕ > ϕ , the bar would loose contact with the wall at B. There0
1
1
1
•
fore, the results remain valid only if the support in B is such that it can transmit a tension force. 1 The force A is zero for cos ϕ = ϕ 70.5 . 3
≈
◦
122 P5.20
Principles of Linear
Problem 5.20
A homogeneous bar (weight W = mg), initially at rest in the sketched position, is released and slides down along a frictionless semi-circular path.
R
R
a) What are the velocity and acceleration of the center of mass in dependence on the position of the bar? b) Determine the forces at the contact points.
a) The center of mass C moves along a circular path with
Solution
radius l=
0
− R2
2
B
√ 3
R = R 4 2
C R/2
about 0. Thus, its velocity and acceleration are uniquely described by ϕ˙ and ϕ. ¨
ϕ l en
30
◦
R
mg A
Note that ϕ˙ and ϕ¨ are identical to the angular velocity and angular acceleration of the bar! The principle of angular momentum (pure rotation about 0)
0 :
Θ0 ¨ ϕ = mgl cos ϕ
yields with mR2 5 Θ0 = + ml2 = mR2 12 6 the angular acceleration
√
3 3 g ϕ = ¨ cos ϕ . 5 R From this result, using ϕd ˙ ϕ˙ = ϕdϕ ¨ and the initial condition ϕ(ϕ ˙ = 30 ) = 0, the angular velocity is determined by integration: ◦
√
ϕ
1 2 ϕ˙ = 2
30o
3 3g ϕ( ¨ ϕ)d ¯ ϕ = ¯ 5 R
ϕ = ˙
√
3 3g (2sin ϕ 5 R
sin ϕ
− 1) .
− 1 2
and Angular Momentum
123
b) To determine the forces, we apply the principle of linear momentum in direction of en and the principle of angular momentum with respect to C : : man = A cos 30 + B cos 30 mg sin ϕ , ◦
◦
− R2 B cos 30 . With a = l ˙ ϕ = g(2sin ϕ − 1) , Θc = mR /12 and the already known ϕ we ¨ obtain √ √ mg 3 mg 3 A = (28 sin ϕ − 9) + cos ϕ , B = (28 sin ϕ − 9) − cos ϕ . 10 3 10 3
Θc ϕ = ¨
C :
R A cos 30 2
−
◦
2
n
◦
2
9 10
A B , W W 1
A/W B/W
0.5
30
45
◦
◦
60
◦
75
90
◦
115
◦
◦
ϕ
The quantities ϕ ˙ and ϕ can ¨ alternatively be determined from the energy conservation law T + V = T 0 + V 0 . Choosing the zero level of potential energy at the lowermost point of the circular path and using vc = l ϕ = ˙ 3R ˙ ϕ/2, the different energy terms are given by
√
T 0 = 0 , V 0 = mg(R T =
o
− l sin 30 ) = mgR
1
1 1 5 mvs2 + Θc ϕ˙ 2 = mR2 ϕ˙ 2 , 2 2 12
V = mg(R
− l sin ϕ) = mgR
and we obtain ϕ˙ 2 =
− √
√ 3 3 5
g (2sin ϕ R
3 4
,
− √ 1
3 sin ϕ , 2
− 1) .
Differentiation with respect to time finally leads to 2ϕ˙ ϕ = ¨
√ 3 3g
5 R
2ϕ˙ cos ϕ
ϕ = ¨
√ 3 3g
5 R
cos ϕ .
124 P5.21
Equations of Motion
Problem 5.21
The displayed system consists of two homogeneous wheels of mass m that are connected by an inextensible cable with a body of weight W 1 = 5mg.
r
m
m
r
Determine the acceleration of the body and the cable forces, when the system freely moves and sliding occurs nowhere. The mass of the cable can be disregarded.
W 1
30
◦
Solution
We separate the system and write down the equations of motion for the different parts: ①
② ③
↓
:
A:
:
B :
5mv˙ 1 = 5mg
− S
1
Θ2 ˙ω2 = rS 1 mv˙ 3 =
ω2
v3 ω 3 ③
,
B
− rS , S − H − mg sin 30 2
◦
2
H
,
S 2
A
②
S 2
mg N
S 1 S 1 ①
v1
5mg
Θ3 ˙ω3 = rH ,
where mr2 Θ2 = Θ3 = . 2 With the kinematic relations (constraint by the cable) v1 = rω2 = v 3 = rω3
ω˙ 2 =
v˙ 1 , r
v˙ 3 = v˙ 1 ,
ω˙ 3 =
v˙ 1 r
we have 7 equations for the 7 unknowns (˙v1 , v˙ 3 , ω˙ 2 , ω˙ 3 , S 1 , S 2 , H ). Solving yields for the acceleration v˙ 1 =
9 g 14
and for the forces in the cable S 1 =
50 mg , 28
S2 =
41 mg . 28
Work-Energy Theorem
125
Problem 5.22
Two bodies (weights m1 g > m2 g) are connected by a cable which is passed over a wheel (moment of inertia Θ0 ). The coeffiΘ0 cient of kinetic friction between the r body and the inclined plane is µ. m2
Determine the velocity of the bodies in dependence of their position. Assume that they are initially at rest and that there is no slip between cable and wheel.
µ α
g
m1
Solution
The velocity can be determined easiest by using the workenergy theorem T
− T
0
= U .
x2
ϕ
With the kinematic relations (constraint by the cable, no slipping) x1 = x 2 = rϕ = x ,
R N m2 g α
,
x1 m1 g
x˙ 1 = x˙ 2 = r ϕ = ˙ x = v ˙
the kinetic energies follow as E k0 = 0
(initial position = rest) ,
1 1 1 1 m2 Θ0 E k = m1 ˙x21 + m2 ˙x22 + Θ0 ϕ˙ 2 = m1 v 2 1 + + 2 2 2 2 m1 m1 r 2
.
Using the friction law R = µN = µm2 g cos α, the work of the external forces is given by U = m 1 gx 1
− Rx − 2
m2 (m2 g sin α)x2 = m 1 gx 1 µ cos α m1
Thus, the work-energy theorem leads to
1 m2 Θ0 m1 v 2 1 + + 2 m1 m1 r 2
v(x) =
− 1
−
2gx
−
= m 1 gx 1
µ
m2 cos α m1
m2 m2 cos α sin α m1 m1 . m2 Θ0 1+ + m1 m1 r 2
µ
−
−
m2 sin α . m1
m −m
2 1
sin α
P5.22
126 P5.23
Work-Energy Theorem
Problem 5.23
The cabin of an elevator (weight W = mg) hangs at a cable drum, which is braked by a band brake (kinetic friction coefficient µ).
l µ
a) Determine the braking force F 0 , such that the cabin moves with constant velocity v0 . b) After which distance d, the cabin stops for a braking force F > F 0 ?
F
r1
0
r2
Θ0
v0
W
a) The braking moment M B , acting on the drum, is determined from the equilibrium condition at the lever Solution
A :
0=
−2r S +lF 2 1
µπ
and the belt friction law S 1 = S 2 e M B = r 2 S 2
− r S
2 1
= (1
−e
ω
Fl S 1 = 2r2 −
µπ
−
0 :
r1 S 3 = M B ,
S 1
F
S 1
:
S 2
S 2
S 3 v mg
F 0 =
r2
A
Fl ) . 2
S 3 = mg
0
S 3
The velocity of the cabin is constant, if the accelerations of the drum and the cabin are zero, i.e., if
r1
2r1 mg . (1 e µπ )l
− b) Now, we apply the work-energy theorem T − T
−
= U between initial state (0) and end state (stop) (1). With the kinematic relations v0 = r 1 ω0 and d = r 1 ϕd follows 1 1 v02 Θ0 2 2 E k1 = 0 , E k0 = mv0 + Θ0 ω0 = m+ 2 , 2 2 2 r1 ϕd F ld W = mgh M B dϕ = mgd (1 e µπ ) . 2r 1 0 1
−
Thus, from v02 Θ0 m+ 2 2 r1
−
− −
= mgd
we obtain by solving for d v02 (m + Θ 0 /r12 ) d = . (1 e µπ )F l/2r1 mg
−
−
−
µπ
−
− (1 − e
)
−
F ld , 2r1
0
Energy Conservation Law
127
Problem 5.24
A bowling ball (radius r) has to overcome a hight difference on the return path which consists of two one-eighth circle arcs (radius R = 5r).
P5.24
45
◦
R
r
A v0
45
R
◦
What speed v0 may the ball have, so that lift-off from the path in the upper sector A is prevented?
Solution
We first determine the velocity in sector A from the the energy conservation law v
T + V = T 0 + V 0 .
45 r
2
With Θc = 2mr /5, v = rω and h = 2R(1 cos 45 ) = 5r(2 2) follow (zero level of potential at depth of D)
√ −
T 0 =
v0
◦
−
T =
− h + r) = (5√ 2 − 4)mgr ,
h
R R(1
1 1 7 mv02 + Θc ω02 = mv02 , 2 2 10
V 0 = mg(R
C
◦
− cos45 ) ◦
ϕ
D
7 mv 2 , 10
V = mg 6r sin ϕ .
Solving for v leads to v 2 = v 02
−
√ 60 sin ϕ + 40 − 50 2 7
gr .
In sector A the centripetal acceleration of the center of mass C is an = v 2 /6r. Thus, from the equation of motion v
:
man =
−N + mg sin ϕ ,
mg
we obtain N (ϕ) =
−
mv02 mg + (102 sin ϕ 6r 42
√
− 50
2 + 40) .
∗
N an ϕ
◦
The minimum force N in sector A appears at ϕ = 45 . Therefore, so that lift-off at this point is prevented, the following condition must be fulfilled: ∗
◦
N = N (45 )
≥0
v0
≤
√ gr
√
2 + 40 = 2.43 gr . 7
128 P5.25
Conservation of
m
Problem 5.25
A homogeneous bar (weight W = mg) is hinge-connected with a body of mass m that can frictionless slide horizontally.
l
ϕ
a) Determine the angular velocity ϕ(ϕ) ˙ of the bar, if it is released from rest from a horizontal position? b) Determine the equation of motion.
a) The position of the center of mass C of the total system is calculated as 1 lm + lm 3 b = 2 = l. 2m 4
W
Solution ∗
b C C
∗
x1 xc
Since the system is not subjected to external forces in horizontal direction, there is no horizontal displacement of C . Thus, we have x1 =
−
1 4 l sin ϕ
x˙ 1 =
−
1 4
xc =
1 4
l sin ϕ ,
x˙ c =
1 4
lϕ˙ cos ϕ ,
yc =
1 2
l cos ϕ ,
y˙ c =
−
1 2
,
yc
C ϕ
∗
∗
l ˙ ϕ cos ϕ ,
x C
y mg
l ˙ ϕ sin ϕ .
We now determine the angular acceleration ϕ by ˙ using the energy con2 servation law T + V = T 0 + V 0 . With Θc = ml /12 follow V 0 = 0 , T 0 = 0 ,
V =
−mg
T =
1 2
1 2
l cos ϕ ,
mx˙ 21 +
ml2 ϕ˙ 2
=
1 8
=
2 2 1 ml ϕ˙ 48
1 2
2
2
m(x˙ s + y˙ s ) +
2 1 4 cos
(8
1 4
ϕ +
− 3cos
2
1 2
− 3cos
2
ϕ)
− g cos ϕ = 0
Θs ϕ˙
cos2 ϕ + sin2 ϕ +
ϕ)
and we obtain after substitution 1 l ˙ ϕ2 (8 24
2
ϕ = ˙
1 3
24g cos ϕ . l(8 3cos2 ϕ)
−
b) Differentiating the energy conservation law (line above) with respect to time leads to the equation of motion: l l ϕ˙ ϕ(8 ¨ 3cos2 ϕ) + ϕ˙ 3 cos ϕ sin ϕ + g ϕ˙ sin ϕ = 0 12 4 g ϕ(8 ¨ 3cos2 ϕ) + 3ϕ˙ 2 cos ϕ sin ϕ + 12 sin ϕ = 0 . l
−
−
Energy Law
129
Problem 5.26
A symmetric disk with a halfcircular boundary rolls without slipping on the flat ground.
P5.26
m, ΘC R
g
Determine the angular velocity in dependence of ϕ, if the body is released from rest at ϕ = 0. Calculate its maximum. Given: R, c = κR, m, ΘC = α mR2
c
Solution
To solve the problem, it is advantageous to apply the energy conservation law. To formulate the energies, we introduce a coordinate system and find
c cos ϕ Rϕ c
xc = Rϕ + c cos ϕ = R(ϕ + κ cos ϕ) , yc = c sin ϕ = κR sin ϕ , x˙ c = R ϕ˙ c ˙ ϕ sin ϕ = R ϕ(1 ˙ κ sin ϕ) , y˙ c = c ˙ ϕ cos ϕ = κR ˙ ϕ cos ϕ .
−
C
ϕ
y
ϕ C
x yc
−
Thus, the energies T 0 , V 0 in the initial position and in an arbitrary displaced position are given by T 0 = 0 , V 0 = 0 , 1 1 1 T = m(x˙ 2c + y˙ c2 ) + Θc ϕ˙ 2 = mR2 ϕ˙ 2 (1 2κ sin ϕ + κ2 sin2 ϕ 2 2 2 1 1 +κ2 cos2 ϕ) + α mR2 ϕ˙ 2 = mR2 ϕ˙ 2 (1 + κ2 2κ sin ϕ + α) , 2 2 V = mgy c = mgRκ sin ϕ ,
−
−
−
−
and the energy conservation law T + V = T 0 + V 0 leads to 1 mR2 ϕ˙ 2 (1 + κ2 2
ϕ(ϕ) ˙ =
− 2κ sin ϕ + α) − mgRκ sin ϕ = 0
2gκ sin ϕ . R(1 + κ2 2κ sin ϕ + α)
−
◦
The angular velocity takes its maximum at ϕ = 90 (lowest level of C ): ϕ˙ max =
2gκ . R[(1 κ)2 + α]
−
130 P5.27
Conservation of
Problem 5.27
A homogeneous circular disk (mass m, radius R) contains an excentric circular hole (radius r = R/3). The disk is released from rest from the displayed position.
g R r B
A m
Determine the angular velocity for the instant, when B reaches its highest position a) if the base is ideally frictionless, b) if the disk rolls on the rough base. Solution
The solution is found by using the energy conservation law. For that purpose, we need the position of the center of mass and and the mass moment of inertia. It is useful to determine them by calculating the difference of two circular disks: R2 π 9 m1 = 2 m = m , R π r2 π 8 r2 π 1 m2 = 2 m = m , R π r2 π 8
2
①
:
A1 = πR ,
②
:
A2 = πr2 ,
− −
1 m1 R 2 , 2 1 Θ 2 = m2 r 2 . 2 Θ1 =
This leads to 0 πR 2 xc = πR 2
2
− r · πr = − R , z c = 0 , − πr 24 Θc = [Θ + m xc ] − [Θ + m (xc + r) ] 9 9 1 = mR + mR − mR 16 8(24) 16 · 9 R R − m8 24 + 3 ·
1
2
1
2
2
2
2
2
z
2
① 2
2
=
2
A
C xc m1
m2 ②
B
x
311 mR2 . 576
a) In the initial state, the disk is at rest and, if the zero level of potential energy is chosen at level of A, we have V 0 = 0 ,
T 0 = 0 .
z B C
x
|xc|
Because the base is frictionless, there act no forces in horizontal direction. Therefore, the center of mass C experiences no velocity change in this direction, i.e. x˙ c = 0. When B reaches its highest position, C just reaches its lowest position. In vertical direction, the velocities of these points change at that instant their sign, and consequently,
Energy Law
131
z ˙c = 0 holds. Thus, the energies at that instant are V 1 =
1 T 1 = Θ c ω 2 2
−mg|xc| ,
and from the energy conservation law V 1 + T 1 = V 0 + T 0 , it follows 1 Θc ω 2 = mg xS 2
| |
ω=
±
2mg xc = Θc
| |
±
48 g . 311 R
b) At the initial state, we have as before V 0 = 0 ,
T 0 = 0 .
When the disk is rolling, the instantaneous center of rotation Π is located at the contact point between disk and base. When the center of mass reaches its lowest position (and B its highest position), its velocity has the horizontal direction and the magnitude x˙ S = ω(R xS ). Therewith, the energies at this moment are given by
−| |
V 1 =
1 1 T 1 = m x˙ 2c + Θc ω 2 , 2 2
−mg|xc| ,
and the energy conservation law leads to 1 2 ω [m (R 2
− |xc |)
2
B
+ Θ c ] = mg xc
| |
|xc|
and finally to ω=
±
2mg xc = m (R xc )2 + Θc
| | −| |
±
2g . 35 R
C
x˙ c ω
z x
Π
Remark: The angular velocity in case a) is higher than in case b). The reason for that lies in the fact that the kinetic energy in case b) is split into translational and the rotational parts, while in case a), there exists only a rotational part.
132 P5.28
Angular Momentum
Problem 5.28
At a pivoted massless arm, a motor is attached (point mass m1 ). Its motor shaft is rigidly connected with the center of mass C of a disk (mass m2 , moment of inertia Θ2 ).
C m2 , Θ2
a) What are the angular velocities of the arm and the disk, if the motor delivers during the time interval ∆t a constant torque M 0 and the system initially was at rest? b) Determine the work done by the motor.
m1 l
a) We separate the system to make the internal moment M 0 visible. The principles of linear and angular impulse and momentum then lead to Solution
① ②
ΘA ω 1 =
A :
↑
M0
C
lF t ,
F t ω1
Θ2 ω2 = M0 ,
C :
where
F n
①
M 0
m1
A
∆t
F n
F t
m2 vc = Ft ,
:
∆t
M 0 dt¯ = M 0 ∆t ,
M0 =
M 0
m2 , Θ2
− −
ω2
②
F t (t¯)dt¯ ,
Ft =
0
ΘA = m 1 l2 .
0
With the kinematic relation vc = lω1 follows by eliminating Ft ω1 =
∆t − (m M + m )l 0
1
2
2
,
ω2 = +
M 0 ∆t . Θ2
b) The work U done by the motor is determined from the kinetic energy of the system after time ∆t:
1 1 1 1 1 U = [(m1 + m2 )l2 ]ω12 + Θ2 ω22 = (M 0 ∆t)2 + . 2 2 2 (m1 + m2 )l2 Θ2 Remark: Since no external moment is present, the angular momentum of the total system remains zero: ΘA ω1 + m2 l2 ω1 + Θ2 ω2 = 0 .
Angular Momentum
133
Problem 5.29
On a disk ① (moment of inertia Θ1 ), which rotates with an angular velocity ω1 , a second disk ② (mass m2 , moment of inertia Θ2 ) rotates with ω2 about the shaft B (= center of mass of ②). When ② slides down along B and touches ①, both disks rub on each other such that ② comes to rest relatively to disk ① after time t0 .
P5.29
a
ω1 A
Θ1
ω2 m2 , Θ2 B ②
Determine the common angular velocity ω0 after time t0 . ①
Solution
We separate the systen and formulate the principles of angular impulse and momentum for ① (with respect to fixed axis A) and ② (with respect to its center of mass B):
A :
(Θ1 + m2 a2 )[ω0
B :
Θ2 [ω0
− ω ] = −M , 1
①
M (t) a m2
A Θ1
− (−ω )] = +M , 2
where
②
M (t)
t0
Θ2
M (t¯)dt¯ .
M =
B
0
Notice that the mass m2 of disk inertia of ①. Eliminating M leads to
② is
considered in the moment of
(Θ1 + m2 a2 )ω1 Θ2 ω2 ω0 = . Θ1 + Θ2 + m2 a2
−
The problem can be solved easier by applying the principle of angular momentum to the total system. Because there acts no external moment , conservation of angular momentum about the fixed axis A yields
A :
(Θ1 + m2 a2 )ω1
−Θ ω 2
2
= [Θ1 + (Θ2 + m2 a2 )]ω0 ,
i.e. we find the same result as above. Remark: For Θ2 ω2 = (Θ1 + m2 a2 )ω1 , the system comes to rest.
134 P5.30
Principles of Linear Momentum
Problem 5.30
A homogeneous circular disk (weight W 1 = mg) rotates with angular velocity ω0 about pin B of an initially resting bar (weight W 2 = 2mg, l = 2r), which is pin-supportet at A.
A M
ϕ
l
m
a) Determine the amplitude ϕ2 of the bar, if the disk is suddenly blocked at the bar. b) Determine the energy loss due to the blockade.
r
B
ω0
Solution
The angular velocity ω1 of the system immediately after blocking is determined from the conservation of angular momentum. With the moments of inertia of the disk and the blocked system ΘB = ΘA =
2 1 mr , 2 ( 12 mr 2 +
A ω1 disk blocked
ml2 ) + M l2 /3 =
2 43 mr 6
, B
it follows ΘB ω0 = ΘA ω1
ω1 =
ΘB 3 ω0 = ω0 . ΘA 43
The amplidute ϕ2 of the blocked system is calculated from the energy conservation after blocking: T 2 +V 2 = T 1 +V 1 . Choosing zero potential at the level of A, we obtain T 2 = 0 , V 2 =
l sin ϕ2 /2
−mgl sin ϕ − M g(l/2) sin ϕ , 2
2
2
ω02
1 3 mr T 1 = ΘA ω12 = , 2 4 43 V 1 = lmg lMg/2 = 4rmg ,
−
−
W 2 l sin ϕ2
−
W 1 position 2
−
what leads to sin ϕ2 = 1
ϕ2
position 1
3 rω 02 . 688 g
b) The energy loss is determined from the difference ∆T of kinetic energies immediately before and after blocking: ΘB ω02 ∆T = 2
−
ΘA ω12 mr 2 ω02 = 2 4
−
mr 2 ω02 3 10 = mr 2 ω02 . 4 43 43
and Angular Momentum Problem 5.31
A yo-yo (weight mg, moment of inertia ΘA ) moves at time t = 0 with speed v0 and angular velocity ω0 . Determine the velocities v1 and ω1 at time t1 when pulling the string with a force S (t) = S 0 t/t 1 . Solution
A :
P5.31
S (t)
ω0
A r v0
m, ΘA
The equations of motion read mv1
↓:
135
ΘA ω 1
− mv −Θ ω A
0
= Fx ,
0
= MA ,
where
t1
Fx =
0
A v
mg 1
0
MA =
ω0
− S (t¯)]dt¯ = mgt − 21 S t
[mg
t1
S (t)
0 1
,
1 rS (t¯)dt¯ = rS 0 t1 . 2
Solving for v1 and ω1 yields
− S 2mt
0 1
v1 = v 0 + gt 1
,
ω1 = ω 0 +
rS 0 t1 . 2ΘA
Problem 5.32
A homogeneous bar of mass m moves on a horizontal plane purely translationally with the velocity v0 .
A
Determine its angular velocity ω1 , when its end suddenly latches into the fixed bearing A.
v0
Solution
P5.32
m
l/2
C l/2
Initially, the angular momentum of the bar with respect to A is L0 = 2l (mv0 ). After latching, the angular momentum is L1 = ΘA ω1 = 31 (ml2 ω1 ). Since there acts no external moment with respect to A, the angular momentum is conserved: 1 2 1 ml ω1 = lmv0 3 2
ω1 =
3v0 . 2l
Remark: Latching leads to an energy loss ∆T : ∆T = T 0
− T
1
=
1 mv02 2
− 21 Θ
2 A ω1
=
1 1 mv02 = T 0 . 8 4
136 P5.33
Rotation of a
M 0
Problem 5.33
Two homogeneous bars of masses m and 2m are attached at a massless shaft which is driven by a torque M 0 . Determine the equation of motion and the support reactions at the bearings.
2a
m
2a
a
2m
2a
2a
Solution
We first determine the axial moment of inertia and products of inertia for the body-fixed system ξ , η, ζ
M 0
2m m 2 (2a)2 + a = 3 ma2 , 3 3 a = m 2a = ma 2 , 2
Θζ = Θξζ
ω, ω˙
C 2
η
ξ 2
−
−
Aη
Aξ
ζ
Θηζ = 0 .
C 1 ξ 1
Bξ
ξ Bη
With the external moments M ξ = 2aBη
− 4aA
η
,
M η = 4aAξ
− 2aB
ξ
,
M ζ = M 0 ,
it follows from the priciple of angular momentum with respect to the body-fixed system
ξ :
η :
− 2A ) = −ω˙ ma 2a(2A − B ) = ω ma
2a(Bη
η
ξ
ξ
2
2
2
M 0 = 3 ma2 ω˙
ζ :
Bη
− 2A 2A − B
η
ξ
=
− ma2 ω˙ ,
maω 2 , ξ = 2 M 0 ω˙ = . 3 ma2
The last equation is the equation of motion of the shaft. To determine the support reactions, the principle of linear momentum must be applied. Since the motion of the mass centers C 1 and C 2 is circular, the acceleration components of e.g. C 1 are given by ξ1 ˙ω in η-direction and ξ1 ω 2 in ξ-direction. Thus, we obtain in ξ- and in η-direction
−
−2m a ω
2
+ m(a/2)ω 2 = A ξ + Bξ ,
2m a ω˙
− m(a/2)ω˙ = A
η + Bη
and finally Aξ =
− 13 maω , 2
Aη =
2 M 0 , 9 a
Bξ =
− 56 maω , 2
Bη =
5 M 0 . 18 a
Rigid Body
137
Problem 5.34
A homogeneous plate (mass m) rotates with the constant angular velocity ω about a fixed axis.
P5.34
l
l a
a) Calculate the support reactions at A and B. b) Determine the additional masses m1 that must be attached at D and E , such that the support reactions are zero.
b
E C A
ω
B
D
a) Because the center of mass is located on the axis of rotation (acceleration of C is zero), ξ the equation of motion reduces to Solution
Aξ + Bξ = 0 .
α
With ω = ˙ 0 and
Aξ 2
Θξζ =
η Bξ
ω
ζ
2
m (a − b )ab − 12 a +b 2
2
(see Problem 5.8), it follows from the principle of moment of momentum M η = ωΘ ˙ ηζ + ω 2 Θξζ (notice the positive sense of rotation about the η-axis!)
η :
−lA
ξ
+ lBξ = ω 2 Θξζ .
Therewith we obtain Aξ =
−B
ξ
=
−
ω 2Θξζ mω 2 (a2 b2 )ab = . 2l 24l a2 + b 2
−
∗
b) Such that Aξ and Bξ are zero, the total product of inertia Θξζ (including m1 ) must vanish. Thus, with ξ sin α =
√ a b+ b 2
2
,
cos α =
√ a a+ b 2
2
the following condition must be fulfilled: ∗
Θξζ
−
−
α
m1 m
m1
b ζ cos α 2
m (a2 b2 )ab b = + 2m sin α 1 12 a2 + b2 2 m a2 m1 = 1 . 6 b2
−
b sin α 2
b cos α = 0 2
Remark: The masses m1 must be attached at both sides, such that the center of mass remains still located on the axis of rotation.
138 P5.35
Rotation
Problem 5.35
A homogeneous circular disk (radius r, mass m) is mounted obliquely (angle α) and with an eccentricity e to a rigid thin shaft. The system rotates with constant angular velocity ω0 .
a α
a m
e
r
ω0
Determine the forces in the bearings. Solution
To fully describe the motion of the rigid body, the principles of linear and angular momentum must be applied. The latter is advantageous2 y ly formulated with the aid of Euler’s α equations: x ω0 Θ1 ˙ω1 (Θ2 Θ3 )ω2 ω3 = M 1 , e 3 C 1 B A Θ2 ˙ω2 (Θ3 Θ1 )ω3 ω1 = M 2 , Θ3 ˙ω3 (Θ1 Θ2 )ω1 ω2 = M 3 . a a With
− − −
− − −
ω1 = ω 0 cos α,
ω˙ 1 = 0,
ω2 =
−ω
0
sin α,
ω˙ 2 = 0,
ω3 = ω˙ 3 = 0
and mr2 Θ1 = , 2
mr2 Θ2 = Θ 3 = , 4
it follows M 1 = 0 ,
M 2 = 0 ,
mr2 2 mr 2 2 M 3 = ω0 sin α cos α = ω0 sin 2α . 4 8
Here M 3 is the moment of the external forces about the principal axis 3. Its relation with the forces in the bearings is given by
C :
M 3 = aB
− aA
aB
−
mr 2 2 aA = ω0 sin 2α . 8
Now, we use the principle of linear momentum. Since C moves circularly with constant angular velocity, its acceleration a = eω02 is directed to the ratation axis. Thus,
| |
↑:
m eω 2 = A + B .
Solving these two equations leads to
mω02 A = 8e 16
−
r2 sin2α , a
mω02 r2 B= 8e + sin2α . 16 a
Stress Resultants
139
Problem 5.36
A homogeneous bar (mass m) on a frictionless horizontal base is accelerated by a force F acting at point A in length direction of the bar.
P5.36
F A c
l
Determine the distribution of the normal force N along the bar.
Solution
The acceleration of the bar (in length direction) is a = F / m. To calculate the normal force, we first introduce the body-fixed coordinate x and then cut the bar at an arbitrary location x above and/or below the point A. For the lower cut (x < c), the principle of linear momentum yields with m = m ¯ x/l
N (x)
m ¯ x
m ¯
m ¯ x ¯ a = F = F . N (x) = m m l
N (x)
F c
x
For the upper cut (x > c), we obtain N (x) + F = m ¯ a =
N (x) =
m ¯ F m
−F (1 − xl ) .
The normal force is linearly distributed and has a jump of magnitude F at x = c. This leads to the below displayed graph of N (x).
c F l
c
l
F
140 P5.37
Stress Resultants
Problem 5.37
A pin-supported homogeneous beam (mass m, length l) is initially at rest. Starting at time t = 0 it is subjected to a constant force F , acting perpendicularly to its longitudinal axis.
l F
m
a x
Determine the stress resultants (M , V , N ) in the beam domain a x l at time t > t0 . Neglect the weight of the beam.
≤ ≤
ϕ A
Solution
Since the motion of the beam is a pure rotation, its acceleration and velocity are uniquely described by ϕ¨ and ϕ. ˙ The principle of angular momentum with respect to A yields
A:
ΘA ϕ = a ¨ F
ϕ = ¨
a F . ΘA
Integration in conjunction with the initial condition ϕ(0) ˙ = 0 leads to the angular velocity ϕ(t) ˙ =
a F t. ΘA
Now, we introduce body-fixed coordinates, cut the system at a position x > a and introduce the stress resultants. Here, we first restrict ourselves to the shear force V and the bending moment M . For the free part of the beam the principle of angular momentum with respect to the ¯ center of mass C and the principle of linear momentum in z -direction yields
¯ : C
:
¯ c¯ ϕ = Θ ¨ m ¯ z¨ c¯ =
−M (x) − V (x)
l
2
−r
¨ c ¯ ϕ =
− l +2 x ϕ¨ .
Therewith and with m ¯ = (1 shear force:
V M
−V (x) .
From the kinematics (circular motion) follows the acceleration z¨ C ¯: z¨ c¯ =
m, ¯ ¯ Θc¯
−x,
x
¯ C
1 (l 2
− x)
rc¯ = 21 (l + x)
z
−
x m l2 ) m and ΘA = , we obtain the l 3
l + x l + x a F 3 a F V (x) = m ¯ ϕ = ¨ m ¯ = 2 2 ΘA 2 l
− 1
x l
2
.
in moving Bars, Beam and Arcs
141
¯ c¯ = Introducing this result into the 1st equation of motion leads with Θ 1 ¯ (l x)2 to the bending moment m 12
−
−Q l −2 x − Θ¯ ϕ¨ 3 x = − a F 1 − 4 l x = −a F 1 − l
M (x) =
c ¯
− − 2
2
x 1+ l 1 x 1+ 2 l
m l2 12
1
x l
3
a F ΘA
.
The normal force N is obtained by using the principle of linear momentum in x-direction: ¯ m
:
m ¯x ¨c¯ =
−N (x) .
¯ C
2
Here, x ¨c¯ = rc¯ ϕ˙ is the centripetal acceleration. After introducing the already known quantities, this leads to
−
N x
N (x) = m ¯ rC ˙2 ¯ ϕ = m (1
9 M 02 t2 = 2 m l3
M 0 t ΘA
x l
1
2
− x)
rc¯ = 21 (l + x)
z
− − x x + l ) l 2
1 (l 2
2
.
Contrary to the bending moment and the shear force, the normal force increases with the square of time t. The graphs of normal force, shear force and bending moment are sketched below for the special case a = l/3.
M (x)
V (x)
− 4 9
14 81
N (x)
Fl
+
F
a x
−
1 2
F
−
5 9
4F 2 t2 9ml
a
F
x
a F 2t2 + 2ml
x
142 P5.38
Stress Resultants
Problem 5.38
A homogeneous beam (mass m, length l) is lifted from a horizontal base by a force F = 4 mg.
g
F
l/12
Determine the stress resultants immediately after lift-off. l/2
Solution
Immediately after lift-off, the beam still has a horizontal position. The principles of linear and angular momentum then yield with Θc = ml2 /12
↑:
C :
m ¨ yc = F
− mg
Θc ϕ = F ¨
l 12
l/12 y
ϕ = ¨
F
ϕ C x
y¨c = 3g ,
l/2
mg
Fl g =4 . 12 Θc l
We now cut the beam at position x, introduce the stress resultants and 7 consider first a cut left to the point where F applies (x < 12 l). In this case the principles of linear and angular momentum yield for part ①
↑:
C 1 :
m1 y¨C = 1
−m g − V 1
ΘC ¨ ϕ = M 1 1
−
1
m1 , ΘC 1
,
C
①
x V 1 . 2
m1 g x (l
With x m1 (x) = m , l
ΘC
1
M 1 (x)
C 1
V 1 (x)
− x)/2
1 1 x3 2 = m1 x = m 12 12 l
and the kinematic relation y¨C = y¨C 1
−
1 (l 2
− x) ϕ = ¨ 3g −
1 (l 2
−
g x x) 4 = g 1 + 2 l l
we obtain the resultants V 1 (x) =
− −
−m (g + y¨ 1
C 1 )
=
x x 2 mg 1+ , l l
x x M 1 (x) = V 1 + ΘC ¨ ϕ = mgl 2 l mgl x 2 x = 3+2 . 3 l l 1
−
,
2
x 1 x3 g 1+ + m4 l 12 l l
in moving Bars, Beam and Arcs
For a cut right to the point where F applies (x > the same way now for part ②
↑:
C 2 :
M 2 C
m2 y¨S =
−m
ΘC ¨ ϕ =
−M − V l −2 x .
2
2
2
g + V 2 ,
2
7 l), 12
2
we obtain in
V 2
C 2 m2 , ΘC 2 ②
m2 g
x
l x/2
With l
− xm ,
1 m2 = ΘC = m2 (l l 12 x x y¨C = y¨C + ϕ = g ¨ 3+2 , 2 l 2
2
it follows
2
2
1 (l x)3 x) = m, 12 l
−
2
x l
2
− x V M (x) = − 2
−x
− − − −
V 2 (x) = m 2 (g + y¨C ) = 2mg 1 l
−
ΘC ¨ ϕ = 2
143
2+
mgl 1 3
x , l
x l
2
7+ 2
x l
.
The graphs of V (x) and M (x) are plotted below: 155 mg 72
V
M x
x
7 l 12
mg − 133 72
−0, 474 mgl Remarks:
• •
•
At the point of application of F , the shear for has a jump: ∆V = F = 4 mg. d M Also in the dynamic case, the relation V = dx holds. This follows from the principle of angular momentum for a beam element of length d x by considering that the moment of inertia is small of higher order. The acceleration y¨ is positive for all points: y(x) ¨ = y¨C + (x l/2)ϕ ¨ = g(1 + 4 x/l) > 0 , i.e., the beam actually completely lifts off from the base.
−
144 P5.39
Stress Resultants
Problem 5.39
A chimney stack of mass m and length l is blown up at the base and falls over.
m
Assuming a hinged support at the base and a constant mass distribution, determine the stress resultants during the motion. At which point xB and at which angle ϕB the maximum transmissible moment M B = mgh/100 is exceeded? Calculate the normal force at this point.
h x
Solution
We first determine the angular acceleration ϕ and ¨ angular velocity ϕ˙ of the chimney. With ΘA = mh2 /3, the principle of angular momentum leads to
¨ A : ΘA ϕ = mg
h sin ϕ 2
ϕ = ¨
ϕ h (1 2 C mg
h 2
3g sin ϕ . 2h
˙2 = 0 Aϕ
− cos ϕ)
A
To calculate ϕ it ˙ is practical to use the energy conservation law:
−mg h2 (1 − cos ϕ) + 21 Θ
ϕ
h sin ϕ 2
ϕ˙ 2 = 3
g (1 h
− cos ϕ) .
Now we cut the system, introduce the stress resultants and formulate the equations of motion for the cut part:
: :
¯ : C
− mg ¯ sin ϕ , m¨ ¯ x = −N − mg ¯ cos ϕ , 1 Θ ϕ = M ¨ + (h − x)V . 2 ¯ yC m¨ ¯ = V
¯ C
¯ C
¯ C
The unknown accelerations x¨C yC ¯ , ¨ ¯ therein can be calculated from the known quantities ϕ ¨ and ϕ˙ 2 :
− =−
2
− −
−
−
Thus, with m = ¯
y
¯ mg ϕ N x + h x 2
−
h+x g 3 x xC ˙ = 3 (1 cos ϕ) = g 1+ (1 ¯ ϕ 2 h 2 h h+x 3 g 3 x xC ¨= sin ϕ = g 1+ sin ϕ . ¯ ϕ 2 2 h 4 h
x ¨C ¯ = y¨C ¯
m ¯
V M
h
− x x = 1 − x h h
m,
− cos ϕ) ,
−
m(h ¯ x)2 mh2 ΘC = 1 ¯ = 12 12
−
x h
3
,
in moving Bars, Beam and Arcs
145
we obtain the stress resultants
− − − − x h
V (x) = m ¯ (¨ yC ¯ + g sin ϕ) = mg 1 = N (x) = =
mg sin ϕ 1 4
−m¯ (¨x
¯ + C
−mg
1
M (x) = ΘC ¨ ¯ ϕ
x h
1
3
3 x 1+ sin ϕ + sin ϕ 4 h
x , h
g cos ϕ)
− − − − − − − − − − x h
3 x 1+ (1 2 h
h 1 2
mh2 = 1 12
x h
cos ϕ) + cos ϕ ,
x [ m¨ ¯ yC ¯ sin ϕ ] ¯ + mg h
3
mgh = sin ϕ 1 4
3 g sin ϕ 2 h x h
2
h mg 1 2
x h
2
3 x 1+ + 1 sin ϕ 4 h
x . h
The bending moment has its maximum, where the shear force is zero. From V = 0 follows
− − 1
x h
1
3
x h
=0
xB =
h 3
(The 2nd solution x = h is not of interest because the bending moment is zero at this point). This leads to the maximum bending moment M max = M (xB ) =
1 mgh sin ϕ . 27
From the condition M max = M B the angle ϕB follows: mgh mgh sin ϕB = 27 100
sin ϕB =
27 100
◦
ϕB = 15.66 .
Finally, the normal force at xB and at the angle ϕB is obtained as
h N , ϕB 3
2mg 3
−
3 4 (1 2 3
− cos ϕ
=
−
B)
=
− 23 mg (−2 + 3 cos ϕ
B)
=
+ cos ϕB
−0.59 mg .
146 P5.40
Stress Resultants
Problem 5.40
A homogeneous arc of mass m rotates with the angular velocity ω and angular acceleration ω˙ about the vertical axis.
r
2r
Calculate the bending moments and shear forces at location A.
ω, ω˙ A
Solution We
introduce body fixed coordinates ξ, η, ζ and cut the system at A. Since the center of mass C rotates along a circular path, its acceleration components are aξ =
−ηs ˙ω ,
aη =
−ηsω
2
r cos ϕ
r
dm
ζ
dϕ
.
C
r sin ϕ
ϕ
ηC
Using the density ρ and the cross section as auxiliary quantities its distance ηc from the ζ -axis follows as
V η M ξ
r(πrρA) π ηc = = r. 2rρA + πrρA 2+π
2r
ω V ξ
η
M η
ξ
Therewith, the principle of linear momentum yields the shear forces V ξ = ma ξ =
−mrω˙ 2 +π π ,
V η = ma η =
−mrω
2
π . 2+π
The bending moments are given by the principle of angular momentum: M ξ = ωΘ ˙ ξζ
2
− ω Θ
ηζ
M η = ωΘ ˙ ηζ + ω 2 Θξζ .
,
With dm = ρArdϕ = mdϕ/(2 + π), we obtain Θξζ = 0 and Θηζ = =
−
ηζ dm =
−
− 2(12 ++ππ) mr
2
mr 2 2+π
π
(1 + cos ϕ)(2 + sin ϕ)dϕ
0
.
Thus, it follows M ξ =
2(1 + π) mr 2 ω 2 , 2+π
M η =
− 2(12 ++ππ) mr ω˙ . 2
Chapter 6
Impact
6
148 Impact
Impact: A sudden collision of two bodies leading to a sudden change of their motion is called Impact . Notation:
v1 , v2
contact tangent plane y
= velocities at contact points P before impact, ①
v1x , v2x = velocity components at points P in direction of the contact normal before impact,
②
C 1
v1x P v2x
t
F x (t¯) dt¯ = linear impulse.
C 1
F x
x
F x
0
Assumptions:
contact normal F y C 2
∗
Fx =
x
v2
v1
v 1x , v2x = velocity components at points P in direction of the contact normal after impact,
C 2
F y
∗
• The impact duration t is neglegible small. • Position changes of the bodies during impact are neglegible small.
Other forces (e.g. weight) are small compared with the impulsive forces at the contact point and can be neglected during impact.
• Impact Hypothesis: e=
− vv −− vv 1x
2x
1x
2x
=
separation velocity − relative relative approach velocity
where e = coefficient of restitution: 0 e 1 e = 1 : elastic impact (no energy loss), e = 0 : plastic impact (v 2x = v 1x , no separation of bodies).
≤ ≤
Solution of plane impact problems: Application of the principles of linear and angular impulse and momentum to each body (see p.102): m(v cx ΘA
− vc ) = (ω − ω) = x
Fx ,
MA
m(v cy
− vc ) = y
Fy ,
Impact
149
A = fixed point or center of mass C . The equations are complemented by the impact hypothesis and the kinematic relations between the velocities at the contact points and the velocities of the centers of mass and the angular velocities.
Remarks:
If the bodies are smooth , the direction of linear impulse (due to contact) coincides with that of the contact normal.
•
• If the bodies are sufficiently rough (no slip during
contact), the velocity components in tangential direction at the contact points after impact are equal.
Impact
Remarks y
v 1 , v 2 in direction of contact normal,
v2
direct
v1
x
impulse always in direction of contact normal (F y = 0).
y
oblique
v 1 , v 2 not in direction of v2
v1
x
contact normal.
x
centers of mass located on contact normal.
y
central
C 1
C 2
y
eccentric C 1
x
C 2
centers of mass not located on contact normal.
Direct central impact:
y
m1 v1 + m2 v2 e m2 (v1 v2 ) , v1 m1 + m2 S 2 v2 m1 v1 + m2 v2 + e m1 (v1 v2 ) S 1 v2 = , m1 + m2 1 e 2 m1 m2 ∆T = (v1 v2 )2 = energy loss. 2 m1 + m2
−
v1 =
− −
−
−
x
150 P6.1
Impact
Problem 6.1
A point mass m1 , suspended from a wire, is released from rest at height h1 and collides at A with a mass point m2 = 2m1 which initially is at rest. After impact mass m1 swings back to a height of h1 /2.
m1
Solution
m2
The velocities of m1 and m2 immediately before impact are
v1 =
A
h1
Determine the coefficient of restitution e and the velocity of m2 immediately after impact.
2gh 1 ,
v2 = 0 .
Thus, denoting the velocities after impact by v1 , v2 , the principle of linear impulse yield ① ②
m1 (v 1
→: →:
−v )= 1
−
F ,
②
v1 , v 1
v2
m1
m2 v 2 = +F
①
F
F
m2
Eliminating F and using the impact hypothesis e =
− v v− v 1
2
1
leads to v 1 = v 1
m1 e m 2 , m1 + m2
−
v2 = v 1 (1 + e)
m1 . m1 + m2
The velocity v1 can be determined from the energy conservation law for m1 after impact: 1 h1 m1 v 21 = m 1 g 2 2
v1 =
gh 1 =
v √ . 2 1
Thus, from the two equations above we obtain e =
m1 m2
m − m√ + 2m 1
2
,
2
− √
m1 v 2 = v 1 1 m2
1 . 2
h1 /2
v1
m1
m2 v2
of Point Masses
151
Problem 6.2 A
ball is dropped from height h0 onto a flat surface and bounces back five times (coefficient of restitution e = 0.85).
P6.2
h0
a) Determine the height h5 the ball reaches after the last impact. b) Determine the energy loss for each impact.
h5
a) If the positive direction of velocity is chosen upwards, the ball has immediately before the 1st hit the velocity Solution
v1 =
−
2gh 0 .
h0
h1
Denoting the velocity immediately after impact by v 1 , the impact hypothesis leads to e=
−
v v
v1 =
−e v
1
= e
h2 h5
2gh 0 .
From the energy conservation law follows the new height, reached after the first hit: 1 mv 21 = mgh 1 2
v 21 h1 = = e 2 h0 . 2g
In the same way we find for all further hits hi = e 2 hi
1
−
,
and therefore h5 = e 2 h4 = e 4 h3 = . . . = e 10 h0 = 0.197 h0 .
b) The energy loss during an impact can be determined from the difference in potential energy calculated from the heights before and after impact. Choosing zero potential at the surface we obtain with V i 1 = mgh i 1 and V i = mgh i (the kinetic energy is zero when reaching the heights) −
−
∆V i = V i
1
−
2
− V = mg(h − h ) = (1 − e ) mgh i
i−1
i
i−1
.
Remark: The energy loss can alternatively be determined from the difference in kinetic energies. Note that the energy loss decreases with the number of hits.
152 P6.3
Impact
Problem 6.3
At a quality test, balls ① fall from height H on a rigid smooth plate ② which is inclined by α = 20 against the horizontal.
①
◦
H
Which distance d must have a collecting container ③, so that only the balls with a coefficient of restitution e 0.8 reach the container?
③
≥
Solution
②
α
d
That balls with e 0.8 reach the container, the distance d must exactly be equal to the flight distance of balls with the coefficient of restitution e = 0.8. The balls ① hit plate ② with the velocity v1 =
≥
2gH .
Using the displayed coordinate system (x = contact normal), its components are
v1x =
1
v1y
1
x
v1y α
v1x v1
−v cos α , = −v sin α .
v 1y
v1 β
γ v 1x
y
Since the impulse in y-direction is zero (smooth plate), v 1y = v 1y holds. The impact hypothesis leads then with v2x = v 2x = 0 (rigid plate) to e =
− vv
1x
v 1x =
1x
−e v
1x
= e v1 cos α .
Thus, for the limit case e = 0, 8 follows for the angle β tan β =
|v | = |v | = 1 tan α |v | e v e 1y
1x
1y
◦
β = 24.46
1x
and for the launching angle γ to the horizontal ◦
γ = 90
◦
− α − β = 45.54
.
With the equations for the projectile motion (see page 32) we obtain the flight distance d of balls with e = 0.8: d =
1 2 1 v 1 sin 2γ = (v 21x + v 21y )sin2γ g g
= 2H (e2 cos2 α + sin2 α)sin2γ = 1.364 H .
of Point Masses
153
Problem 6.4
Two old cars (no ABS) move along the highway with same speed v0 in a short distance b when a traffic jam gets in sight in distance c. Both cars make a full brake and get sliding, car ① a shock moment ∆t later than car ②.
P6.4
c
a) When and where both cars clash? Determine their velocities after the clash (coefficient of restitution e). b) Does car ② come to stand ahead of the jam? Given: v0 = 30 m/s (=108 km/h), m1 = 1.5 m, m2 = m, ∆t = 1 s , µ1 = µ2 = 1/2, e = 0, 8, b = 25m, c = 150 m. Neglect the lengths of the cars.
v0 m2
②
b
v0 m1
①
a) We introduce the coordinate s and start counting time when car ② slames on brakes. During sliding the acceleration of both cars is the same: a 1 = a 2 = µg. Considering the initial conditions and the shock delay, the velocities and covered distances are given by Solution
−
v2 = v 0 v1 = v 0
− µgt , − µg(t − ∆t) ,
s2 = v 0 t s1
− = v t − 0
2 1 µgt 2 1 2 µg(t
, 2
− ∆t) − b .
∗
The collision time t and the associated position and velocities are found from the condition s1 (t ) = s 2 (t ): ∗
s
∗
b + µg(∆t)2 /2 t = = 5.6 s , µg∆t 2 1 s = s 2 (t ) = v 0 t µgt = 91.17 m , 2 ∗
∗
∗
∗
∗
∗
∗
∗
v2 = v 2 (t ) = v0 v1 = v 1 (t ) = v0
v2
c
b
v1
∗
− − µgt = 2.56 m/s , − µg(t − ∆t) = 7.46 m/s . ∗
∗
The velocities after collision are calculated by using the formulas on page 149 for central impact: ∗
∗
∗
∗
m1 v1 + m2 v2 + e m1 (v1 v2 = m1 + m2 m1 v1 + m2 v2 e m2 (v1 v1 = m1 + m2 ∗
∗
−
− v ) = 25.4 m/s , − v ) = 14.1 m/s . 2
∗
∗
2
∗
b) With v2 , v 1 and s the positions for standstill would be ∗∗
s2
v 22 = s + = 65.8 m , 2µg ∗
∗∗
Because s2 > c, car
② would
∗∗
s1
v 21 = s + = 20.3 m . 2µg ∗
crash into the jam.
154 P6.5
Impact
Problem 6.5
A bullet (mass m) with the velocity v hits centrally an initially resting frictionless supported board of mass M . It M penetrates the board and has thereafter the velocity v¯. C m v v¯ a) Determine the velocity w¯ of the board after the penetrating shot. b) How much energy is needed for producing the penetration hole? Given: M = 10 m, v¯ = v/2.
a) The velocity w ¯ of the board is determined by using the conservation of linear momentum law with the known bullet speeds v and v¯: Solution
mv = m¯ v + M w ¯.
→:
Solving for w leads ¯ to w ¯=
m (v M
− v¯) = 101 v2 = 20v .
b) The energy needed for producing the penetration hole is calculated from the energy loss of the system. Since the potential energy does not change, only the kinetic energy before and after penetration must be considered: T =
1 mv 2 , 2
T =
1 1 m¯ v 2 + M w ¯2 . 2 2
The difference yields the energy loss
− T = 21 m(v − v¯ ) − 21 M w¯ 2
∆T = T
1 = m (v 2 2
2
− v¯ ) −
m (v M
− v¯)
2
2
− − −
1 = mv 2 1 2 =
2
29 1 mv 2 . 40 2
v¯ v
2
m 1 M
v¯ v
2
− −
1 = mv 2 1 2
1 4
1 1 10 4
of Point Masses
155
Problem 6.6
Two cars (point masses m1 , m2 ) collide at an intersection. From a radar measurement the m1 velocity v1 before collision and from the m2 sliding track the velocity v 1 and angle γ v 1 γ after collision are known. v2 v1 m1
Determine the velocities v2 , v2 and the angle β of the 2nd car by assuming that no energy is absorbed during impact. Given: m1 = m2 = m, v1 , v1 = v1 /3, α = 45 , γ = 60 . ◦
β
α v2 m2
◦
Solution
Before and after collision the cars have the velocities v1 , v2 , v 1 and v 2 . They are related by the conservation of linear momentum law
→: ↑:
m1 v1 + m2 v2 cos α = m 1 v 1 cos γ + m2 v 2 cos β , m2 v2 sin α = m 1 v 1 sin γ + m2 v 2 sin β ,
which, after inserting the given quantities, leads to
√ v + v 2/2 = v /2 + v cos β , √ √ v 2/2 = v 3/2 + v sin β . 1
2
1
2
1
2
2
Conservation of energy during impact requires 1 1 1 1 m1 v12 + m2 v22 = m1 v 21 + m2 v 22 2 2 2 2
v12 + v22 = v 21 + v 22 .
Herewith we have three equations for the three unknowns v2 , v 2 and β . To solve them, we first eliminate β : v 22 = v 22 (cos2 β + sin2 β )
√ − v v − v v √ 2/2) √ + (v /2 + v 3/4 − v v 6/2) √ √ √ = v + v + v + v v 2 − v v − v v (1 + 3) 2/2 = (v12 + v22 /2 + v 21 /4 + v1 v2 2 2 2
2 1
2 1 2 1
2 2
1 1
2 1
2 1
1 2
1 1
2 1
Introduction into energy conservation then leads to v (v − 2v ) √ √ √ = 0.144 v . v 2 − v (1 + 3) 2/2 = v + v − v v = 0.953 v 1
v2 = v 22
1 2 1
1
1
1
1
2 2
2 1
2
1
.
and subsequently to
√
v 2 sin β = v 2 2/2
−v
1
√
3/2
sin β =
−0.196
β =
◦
−11.3 .
P6.6
156 P6.7
Impact
Problem 6.7
A point mass (mass m1 ) is shot upwards a rough inclined plane (coefficient of kinetic friction µ) with an initial velocity v0 . In point B it hits a resting point mass m2 , which after this 1st impact bounces against a f C spring (stiffness k) at the end g of the smooth path BC . k B a) Determine the velocity v1 of the point mass m1 immediately before the impact. b) Calculate the coefficient of restitution e if the maximum compression of the spring is f .
m2
a A
smooth
v0
m1
α
µ
rough
a
a
α
◦
Given: m1 = m2 = m = 0.1kg, g = 9.81 m/s, α = 30 , v0 = 6 m/s, µ = 0.5, k = 400 N/m, a = 1 m.
a) Application of the work-energy theorem V A + T A = V B + T B + U AB to the mass m1 between the points A and B yields with R = µ N = µ m1 g cos α and Solution
T A = 21 m 1 v02 ,
V A = 0,
V B = m 1 ga sin α,
U AB =
T B = 21 m 1 v12
−R a = −µ m ga cos α 1
m1 g
the velocity v1 =
− v02
v
m1
2ga(sin α + µ cos α) = 4.21 m/s .
N
R α
b) The velocity of m2 immediately after impact follows with v2 = 0 and m1 = m 2 as v2 =
m1 v1 + m2 v2 + em1 (v1 m1 + m2
−v ) = 1+e v 2
2
1
.
(a)
Now we apply the energy conservation law V B + T B = V C + T C to m2 between the points B and C . With V B = 0, T B = 21 m2 v 22 , V C = m 2 ga(1 follows v2 =
c 2 f + 2ga(1 m2
− cos α) = 3.55 m/s .
Introduction into (a) finally leads to e = 2
v 2 v1
− cos α) +
3.55 − 1 = 2 4.21 − 1 = 0.69 .
2 1 cf , 2
T C = 0
Central
157
Problem 6.8
A bowling ball ① hits with speed v1 the pin ②. Assume that all surfaces are smooth and that the impact is partially elastic (coefficient of restitution e).
P6.8 ②
m2 Θc
Determine the velocities of pin and ball immediately after impact. Given: v1 = 7 m/s, m1 = 4.5kg, m2 = 1.5 kg, Θc = 0.013 m2 m2 , α = 10 , c = 0.02 m, e = 0.9.
C
①
c α
m1
v1
◦
Solution
The impact is eccentric. Since the surfaces are smooth, the linear impulse F acts in the line of impact and thus, the impulse laws lead to
① ②
→: :
m1 (v 1
1
m2 v 2 = F ,
Note that the weights can be neglected during impact and that the ball cannot move vertically. Using the impact hypothesis e =
−
②
y x
Θc ω 2 = c F .
C :
vP 1x P v1x
− v ) = −F cos α ,
− −
m2 Θc
ω2
v2
①
v1 , v 1
P F
m1
α
P
C
F
N
v P 2x P v2x
and P v1x = v1 cos α ,
P v2x = 0 ,
vP 1x = v 1 cos α ,
v P 2x = v 2 + c ω 2
we obtain by solving for F
F = (1 + e)
m2 v1 cos α 2 = 14.41 kgm/s . 1 + m2 /m1 + c2 m2 /Θc
Thus, the velocities follow as v 1 = v 1
−
F cos α = 3.8 m/s , m1
v2 =
F = 9.6 m/s . m2
Note that the pin after impact has an angular velocity: ω 2 = c F /Θc = 13.2 1/s .
c α
158 P6.9
Eccentric Impact
Problem 6.9
A point mass m1 impinges with the velocity v1 eccentrically a homogeneous bar (mass m2 ), which initially rests on a smooth plane.
m2
Determine the velocities of the point mass and the bar after an ideal elastic impact. Given.: m2 = 2m1 , b = l/4.
l
C b
m1 v1
Solution
The principle of impulse an momentum applied to the point mass ① and the bar ② yield (with v2c = 0, ω2 = 0) ① ②
→: →:
m1 (v 1
②
− v ) = −F , 1
ω2
m2 v 2c = F ,
Θc ω 2 = b F .
C :
C
v1 , v 1 ①
F
F
v 2S B
These equations are complemented by the impact hypothesis e =
− v −v v 1
2B
=1
1
and the kinematic relation v 2B = v 2c + b ω 2 . Thus, we have five equations for the five unknowns (v 1 , v 2c , v 2B , ω 2 , F ). With Θc = m 2 l2 /12 we obtain
m1 12b2 1+ 2 m2 l v 1 = v 1 m1 12b2 1+ 2 m2 l v 2c =
m1 (v1 m2
−
1 =
+1
− v ) = 158 v 1
1
,
v − 15 , 1
ω2 =
b 8 v1 m2 v 2c = . Θc 5 l
Remark: If the impact is purely elastic the impact hypothesis can be replaced by the energy conservation law. In our case it reads 1 1 1 1 m1 v12 = m1 v 21 + m2 v 22c + Θc ω 22 . 2 2 2 2
Plastic Impact
159
Problem 6.10
A homogeneous angleshaped body (mass m, dimensions a, t a) slides along a smooth plane and impinges with its edge plastically against an obstacle H .
P6.10
m
t
v
a
Calculate the minimum impact speed v, so that the body tilts over.
a
H
Solution
Since the impact at H is plastic , the edge of the angle does not separate from H . Therefore, tilting can be regarded as a pure rotation about the fixed point H . With the distance a/4 of the center of mass, the moment of angular momentum LH before impact (= moment of linear momentum with respect to H ) and LH after impact are given by LH =
a (mv) , 4
C a/4
H
LH = ΘH ω
where ΘH = 2
m a2 2 3
ma2 = . 3
Since there acts no angular impulse with respect to H (weight can be disregarded during impact), the moment of momentum is conserved: LH = L H
a ma2 mv = ω 4 3
ω=
3 v . 4 a
The body can only tilt over, if the center of mass reaches the highest possible position. The minimum velocity required for that follows from energy conservation, applied to the C motion after impact: 1 ΘH ω2min = mg 2
√ − a 2 4
a 4
C
a/4
This finally leads to vmin =
√ − 8 ( 2 3
1)ga = 1.05
√ ga .
√ a 2
4
160 P6.11
Impact
Problem 6.11
At a hammer mill, the thumb C of the rotating flywheel (moment of inertia ΘA , angular velocity ω1 ) hits the resting hammer (moment of inertia ΘB ) which is pivoted at B.
ΘA C
ΘB
ω1 A
B r
a
a) Determine the angular velocities ω1 of the flywheel and ω2 of the hammer shortly after impact (coefficient of restitution e) b) Calculate the energy loss for the case ΘB /ΘA = a 2 /r2 .
a) The principles of angular impulse and momentum for ① and ② yield Solution
A:
ΘA (ω 1
− ω ) = −r F , 1
ΘB ω2 = a F .
B :
ω1 , ω 1 ①
C F
A r
F
With the impact hypothesis e =
− vv −− vv 1C
2C
1C
2C
ω2 ②
C
B a
and the kinematic relations v1C = r ω1 ,
v1C = rω1 ,
v2C = 0 ,
v2C = a ω 2
we obtain r 2 ΘB 1 e 2 a ΘA , ω 1 = ω 1 r 2 ΘB 1+ 2 a ΘA
r a . ω2 = ω 1 2 r ΘB 1+ 2 a ΘA
−
(1 + e)
b) The energy loss is calculated from the difference of kinetic energies before and after impact. With r2 ΘB = a 2 ΘA follows
−
1 ∆T = ΘA ω12 2 1 = ΘA ω12 [4 8
1 1 ΘA ω 21 + ΘB ω 22 2 2
− (1 − e) − (1 + e) ] = 41 (1 − e )Θ 2
2
2
2 A ω1
.
Remark: In case of an ideal elastic impact (e = 1) we would obtain for the given datai ω1 = 0 and ω2 = ω 1 r/a.
of Rotating Bodies
161
Problem 6.12
A springy pivoted hammer (torsional spring constant cT ) can rotate in a horizontal plane. It consists of a homogeneous bar of mass m2 and the hammer head (point mass m3 ). The hammer is hit by a ball of mass m1 .
P6.12
a
a
2a A
ϕ cT
m2
m1
v m3
Determine the required speed v of the ball so that the hammer for a given coefficient of restitution e just reaches a maximum angle ϕ0 . Given: m1 = m 2 = m 3 = m, a, cT , e = 21 .
Solution
Since the hammer initially is at rest, the angular momentum of the system with respect to the fixed point A before impact is given by the moment of linear momentum LA = a m1 v. Immediately after impact the hammer (moment of inertia ΘA ) has the angular velocity ω and the ball (mass m1 ) the velocity v. Thus, conservation of angular momentum leads to
a m1 v = ΘA ω + a m1 v .
A:
With the impact hypothesis e=
ωa
−v
v
follows m1 (1 + e)a ω= v where ΘA + m1 a2
m2 (4a)2 16 ΘA = + (2a)2 m3 = ma2 . 12 3
Immediately after impact (no spring deformation) the energy of the hammer-spring system is solely given by its kinetic energy 21 ΘA ω 2 . At maximum angle ϕ0 the potential energy is just 21 cT ϕ20 while the kinetic energy is zero. Thus, from the energy conservation law after introducing ω and the given data we obtain the required velocity: 1 1 ΘA ω 2 = cT ϕ20 2 2
v=
ω=
2
cT ϕ0 ΘA
cT ΘA + m1 a 19 ϕ0 = ΘA m1 (1 + e)a 18
3cT ϕ0 . m
162 P6.13
Impact
Problem 6.13
A soccer ball (mass m, moment of inertia Θc ) hits with speed v0 horizontally the rough post of the goal. The impact (coefficient of restitution e) is central under the angle α to the goal line.
post goal line α ω0
Determine the required spin (angular velocity ω0 ) such that the ball crosses after impact the goal line if during impact there is no slip (static friction!).
v0
r
C m, ΘC
Solution The
principles of impulse and momentum, applied to the ball, yield m(vx
: :
−
− v ) = −F , 0
mvy = H ,
C :
Θc (ω
−ω )= 0
−
H
F
ω0 , ω
rH .
With the impact hypothesis (rigid post) e =
α F
r A C
H
y x
vx v0
and the condition for static friction (velocity of contact point A in ydirection is zero after impact) v Ay = v y
−rω = 0
we obtain for v x and vy vx =
−e v
0
,
vy =
rω 0 . r2 m 1+ Θc
The ball only crosses the goal line if v ξ = v x sin α + v y cos α > 0 . This leads for the require spin to ω0 >
ev0 r
2
1+
r m Θc
tan α .
ξ
y vy η
α
x α
vx
of Rotating Bodies Problem 6.14
The motion of a homogeneous door (mass m, width b, thickness t b) is limited by a stopper B.
163 P6.14
A
In which distance c from the door hinge A the stopper must be fixed, that the impulse A at the hinge is zero during impact?
c
b m B
t
Solution
We assume that before impact the door has the angular velocity ω. Then the principles of impulse and momentum read m(vcx
− vc
↑:
m(vcy
− vc ) = A
A :
ΘA (ω
− ω) =
where vcx =
= Ax
y
b ω, 2
Ax
y
ω b 2
− −
→:
x)
Ay
B, C
,
B
cB ,
v cx =
b ω, 2
vcy = v cy = 0 .
Using the impact hypothesis e=
− vv
Bx
Bx
=
ω − cω =− cω ω
and solving the equations for the hinge impulses leads to
Ay = 0 ,
ΘA Ax = (1 + e)ω c
−
mb 2
.
From the condition Ax = 0 with ΘA = mb 2 /3 finally follows c =
2ΘA 2 = b. mb 3
c
164 P6.15
Impact
Problem 6.15
The double pendulum, consisting of two homogeneous bars, is struck horizontally in point D by a linear impulse F .
a) Determine the impulsive forces in A and B, and the state of motion of the lower bar immediately after impact. b) Under which circumstances the magnitudes of angular velocities after impact are related as ω2 = 2 ω1 ?
| |
A m1
c
D g
a
F
B m2
b
| |
a) We separate the two bars, draw the free-body diagram and formulate the priciples of linear and angular momentum (note that the weights can be neglected during impact): Solution
①
m1 v 1 = F
A:
ΘA ω1 = c F
←:
m2 v 2 = B ,
②
− − −
←:
C 2 :
A
A a/2 v1
A
①
F
B
B,
B
aB ,
b/2
C 2
v2
where ΘA = 31 m1 a2 ,
ΘC = 2
b
ω2
ΘC ω 2 = (b/2)B 2
ω1 c a C 1
②
2 1 m b 2 12
.
Since the velocities of both bars at point B must be equal after impact, and v1 and ω 1 are simply related, the kinematic relations read a ω 1 = v 2 + (b/2) ω 2 ,
v 1 = (a/2) ω 1 .
Solving these six equations yields
− A = 1 v1 = 4
c 1 F, a 1 + m1 /m2
B , m2
ω1 = 4
B , am2
B=
c F , a 1 + m1 /m2
v2 =
B , m2
ω2 =
b/2 B. ΘC
2
b) Inserting the quantities into the condition ω 2 = 2 ω 1 leads to
| |
b/2 B B=8 ΘC am2
2
bam2 = 16 ΘC
2
| |
a =
4 b. 3
Remark: Note that A = 0 if c is chosen such that the bracket is zero.
Eccentric Problem 6.16
A bar (mass m1 ) hits under the angle α = 45 with speed v a resting homogeneous quadratic plate of mass m2 . v
m1
we denote the velocities of bar ① with v and of the plate ② with w, the principles of impulse and momentum lead to
y
m1 (v cy
↑: ②
x)
=
m2 w cx = F ,
→:
2
e=
− v v− w Ax
4
√ 2
C 1 :
A
l
C 2 :
Θ2 ω 2 = 0 ,
√ = v/ 2. Hence, using the impact
Ax
Ax
where vAx
v = , 2
vAx = v Cx
√
−
√ 2 4
lω 1 ,
wAx = w Cx
we obtain v v cx = 2
√
w cx =
√ v2
m1 2e m2 , m1 5+ 2 m2
v cy =
m1 m2 m1 , 5+2 m2
w cy = 0 ,
3+ 2
−
x
Θ1 ω 1 = 4 l F ,
m2 wcy = 0 ,
where Θ1 = m1 l /12 and vcx = vcy hypothesis
√ 2
C 2
ω1
−F , ↑:
②
F
C 1
− vc ) = 0 , y
A F
l/2
①
− vc
a
l
Solution If
m1 (v cx
C 2
α
Determine the velocities of the bar and the plate after impact (coefficient of restitution e). Assume that all surfaces are smooth .
→:
P6.16
m2
◦
①
165
√ v2 ,
ω1 =
v 6(1 + e) , l 5 + 2 m1 m2
2(1 + e)
ω2 = 0 .
166 P6.17
Impact
Problem 6.17
A homogeneous circular disk slides frictionless along a smooth plane and hits under angle α = 45 with speed v and angular velocity ω a rough rigid boundary (coefficient of restitution e = 1/2).
m
◦
2nd impact
ω
B
r v α A 1st impact
Determine the magnitude and direction of the velocity after the second impact at the opposite boundary. Assume that the disc does not slide during impact at the contact point. Solution
The first impact is described by the principles of impulse and momentum m(v cx
C :
x
the impact hypothesis e =
ω, ω
− vc ) = −H , Θc (ω − ω) = −rH ,
→:
− vv
Ay
Ay
=
− vvcc
1
C
1
r A
y
F 1
x
y
H 1
y
and the condition of no sliding at point A v Ax = v cx + r ω = 0 . From the first two equations follows ω = ω
− mr (vc − v c Θc x
x ) .
Hence, with Θc = mr 2 /2 and vcx = v cos α =
√ v2 ,
−v sin α = − √ v2
vcy =
the velocity components are obtained as v cx =
√
rω 2 1 + vcx = ( 2v r ω) , 3 3 3
−r ω = − − Remarks: • For ω > √ 2v/r follows vc
v cy = x
−evc
y
=
√ 2 4
< 0, i.e. after impact
the disc moves to the left.
• The impulse F can be determined from the prin1
ciple of impulse and momentum in y-direction.
v.
Eccentric
We denote the velocities after the second impact with two dashes. Then the principles of impulse and momentum read
→:
m(vcx
C :
− vc
Θc (ω
x)
=
−H
− ω) = r H
2
F 2
,
2
167
H 2
.
B C r
y
ω, ω
x
With the impact hypothesis e=
− vv
By
=
By
− vvcc
y y
and the condition of no sliding at B (consider signs!) v Bx = v cx
−rω = 0
follow
√
rω 2 1 v cx = r ω = + v cx = ( 2 v 3 3 9 v sy =
−evs
y
=
−
√ 2 8
− rω) ,
v.
Hence, mangnitude and direction of the velocity are given by v=
=
2
B
2
v cx + v cy C
2 2 + 81 64
v2
−
√ 2 2 9
2
vrω +
2
r ω , 81
β
v Cy y
√ | vc | 9 2 tan β = = √ . rω vc 8( 2 − v )
v vCx
x
y
x
Remarks:
•
For ω = 0 we obtain v tan β = 9/8 β = 48.4 .
√ • For ω = 2 v/r
=
0.24 v
and
◦
the disk rebounds orthogonal at A and therefore also at B.
Chapter 7
Vibrations
7
170 Free
The following formulas and problems are restricted to vibrations of linear systems with one degree of freedom.
1. Free undamped vibration The equation of motion (vibration equation ) x ¨ + ω 2 x = 0 has the general solution x(t) = A cos ωt + B sin ωt = C cos(ωt
− α)
where x
ω
x(t)
= circular frequency,
ω 2π 1 2π T = = f ω f =
√ C = A
2
+ B2
B α = arctan A
C
= frequency,
π
= period of vibration,
2π
ωt
α
= amplitude, ωT = 2π
= phase angle.
Remarks:
•
The constants A, B, C and α follow from the initial conditions x(0) = x 0 , x(0) ˙ = v 0 as A = x 0 ,
v0 B= , ω
C =
x20 +
v0 ω
2
,
α = arctan
v0 . x0 ω
•
A system which is described by the differentiial equation above is also called a harmonic oscillator.
•
If the position coordinate is not counted from the equilibrium position, the equation of motion takes the form x ¨ + ω 2 x = ω 2 xst and its solution reads x(t) = D cos(ωt
− ψ) + x
st
.
Vibrations
171
Examples of single mass vibration systems All coordinates are counted from the respective static equilibrium position.
spring-masssystem
eigenfrequency ω
diff. equ.
system
k m
m¨ x + kx = 0
k m
x
simple pendulum (small displacements)
g l
lϕ ¨ + gϕ = 0
l ϕ
m x
massless bar with end mass
EA
massless beam with end mass
EI
m
l
l Θ
k m
m¨ x + kx = 0
with k = EA/l k m
m x
l
massless shaft with end disk
GI T ψ
m¨ x + kx = 0
with k = 3EI/l3
¨ kT ψ = 0 Θψ +
kT Θ
with kT = GI T /l
Spring constants Springs in parallel:
k k1
k2
Springs in series:
k2
ki
1 = k
1 ki
k3 m
m k1
k =
∗
∗
k3
m
k
∗
m
∗
Spring Compliance: The inverse spring stiffness is called spring compliance (flexibility): c = 1/k. The compliance of an elastic system can be determined by loading it by a virtual force 1“ at the position of the ” vibrating mass and calculating the dicplacement δ in direction of the force. Then c = δ and k = 1/δ .
172
Damped Vibrations
2. Free damped Vibration a) Dry (Coulomb) Friction: The solution of the equation of motion x ¨ + ω 2 x = is given by
2
∓ω r
for
x˙ ≷ 0
> 0 . x˙ < Amplitude decrease: x(ωt) x(ωt + 2π) = 4r.
x(t) = C cos(ωt
− ϕ) ± r
f u ¨r
−
b) Viscous Damping: The solution of the equation of motion x ¨ + 2ξ x˙ + ω 2 x = 0 reads for ζ = ξ/ω < 1 (underdamped system):
ξt
−
x(t) = C e
cos(ωd t
− α)
ξ
=
damping coefficient,
ζ = ξ/ω
=
damping ratio,
ζ 2
=
circular frequency of damped vibration,
T d = 1/f d = 2π/ω d
=
period.
ωd = ω
− 1
x
C e
ξt
−
C
x(t) C cos α ωd t
x(t + T d )
ωd T d =2π
The amplitude decay is characterized by the logarithmic decrement δ = ln
x(t) 2πξ = = x(t + T d ) ωd
For weak damping ζ
2πζ . 2 1 ζ
−
1 follows δ ≈ 2πζ .
ζ = ξ/ω = 1 (critical damping): ζ = ξ/ω > 1 (overdamped system):
ξt
−
x(t) = (A1 + A2 t) e ξt
−
x(t) = e
where
.
(A1 eµt + A2 e
µt
−
µ = ω
− ζ 2
) 1.
Forced Vibrations
173
3. Forced vibrations The equation of motion for harmonically excited vibrations can be written as 1 2ζ x ¨ + x˙ + x = E x0 cos Ωt ω2 ω where
E =
Ω
=
circular frequency of excitation,
η= Ω ω δ ζ = ω
=
frequency ratio,
=
damping ratio,
Ex0
=
excitation amplitude,
1
excitation through a force or via a spring,
2Dη
excitation via a damper,
η2
excitation through unbalanced rotation.
It has the steady state solution (particular solution) x = x 0 V cos(Ωt
− ϕ) .
Here are V =
− (1
tan ϕ = Remarks:
•
E η 2 )2 + 4ζ 2 η 2
2ζη 1 η2
−
= magnification, frequency response ,
= phase angle .
The general solution of the equation of motion is composed of the solution of the homogeneous differential equation (decaying motion, see page 172) and the particular solution.
• For undamped vibrations (ζ = 0) the amplitude tends for η → 1 (Ω → ω) to infinity (resonance ). • For weakly damped systems (ζ 1) at resonance (η ≈ 1) the maximum magnification is: V ≈ E/2ζ . • An excitation with η < 1 is called subcritical and with η > 1 supermax
critical.
•
The phase angle ϕ represents the delay of the response x relative to the excitation.
174 P7.1
Free undamped
Problem 7.1 Two
drums support a homogeneous beam of weight W = mg. They rotate with different coefficients of kinetic friction µA , µB in opposite directions. W Show that the horizontal motion of A µ B the beam is a harmonic vibration and A µB Ω Ω determine the natural frequency ω. l Solution
We first determine the equilibrium distance a of point C of the beam and the support forces. From the equilibrium conditions and the friction law
→:
Aeq H
eq = B H ,
Aeq V
↑:
eq A : BV l = mga , Aeq H = µ A AV ,
+
eq BV
C AH AV mg a l
= mg,
l
− a mg , l
eq BV =
C
a mg . l
AH AV
Now we consider an arbitrary displacement x from the equilibrium position. The support reactions then are l
a l
mg
BH BV
x
− (a + x) mg , B = a + x mg = B + x mg , l l l l − (a + x) x = µ = A − µ mg , B = B + µ
AV = AH
BV
eq BH = µ B BV
follow a/l = 1/(1 + µA /µB ) and Aeq V =
BH
eq V
V
A
eq H
l
A
l
H
eq H
B
x mg . l
eq Thus, with Aeq H = B H the equation of motion is given by
→:
mx ¨ = A H
− B
H =
A + µB )
−(µ
x mg , l
which leads to the differential equation for harmonic vibrations x ¨ + (µA + µB )
g x = 0 . l
The natural frequency is given by ω=
(µA + µB )
g . l
Remark: In case of µA = µ B = µ, the result simplyfies to ω =
2µg/l.
Vibrations
175
Problem 7.2 A
small homogeneous disk (mass m, radius r) is attached to a big disk (mass M , radius R). In the positions a) and b), both systems are in the equilibrium position. g R R M A Determine the natukT r M kT ral frequencies ω for m A a both systems. Assur a me small rotational amplitudes. a) b)
Solution
To formulate the equation of motion it is advantageous in both cases to apply the principle of angular momentum. case a) We obtain
A :
ΘA ¨ ϕ =
kT ϕ A
kT (ϕ − ϕeq ) a A
a ϕ
a)
ϕ mg
b)
mg
−k ϕ − mga sin ϕ . T
For the assumed small amplitudes (ϕ 1) this equation reduces with sin ϕ ϕ to the differential equation for harmonic vibrations, which provides the natural frequency:
≈
ϕ + ¨ ω2ϕ = 0
with
kT + mga ωa2 = ΘA
ωa =
kT + mga . ΘA
case b) Since the torsion spring is in the equilibrium position already stretched by ϕeq = mga/kT , its moment due to a displacement ϕ is given by kT (ϕ ϕeq ). Thus, we obtain in this case
−
ΘA ¨ ϕ =
ϕ + ¨ ω2ϕ = 0
−k
eq
− ϕ ) − mga cos ϕ = −k ϕ + mga(1 − cos ϕ) , which for small amplitudes, i.e. cos ϕ ≈ 1, reduces to A :
T (ϕ
with
ωb2 =
T
kT ΘA
ωb =
kT . ΘA
Introducing the moment of inertia
M R2 mr2 ΘA = + + ma2 2 2
the natural frequencies can be written as ωa =
2(kT + mga) , M R2 + m(r2 + 2a2 )
ωb =
2kT . M R2 + m(r2 + 2a2 )
P7.2
176 P7.3
Free undamped
Problem 7.3
The system shown consists of a homogeneous disk (mass m, radius r), a block (mass M ) and a spring (stiffness k). The mass of the string and of the pulley B can be neglected.
r
Determine the equation of motion and the period of the vibration. Specify the solution for the case that point A is horizontally displaced by a from equilibrium and then released with zero initial velocity.
g
m k
A
B M
Solution We
separate the system and introduce coordinates with their origin at the equilibrium position of the system. With this choice, the weights of the disk and the block must not be considered in the free-body diagram. Thus, the equations of motion are given by ① ②
↓: →:
A :
M ¨ x =
−S
1
②
kxA
x , 2
2rϕ = x
S 1
S 1
− kx
,
A
①
Π
− rS .
x
2
If we use the kinematic relations (Π= instantaneous center of rotation) xA =
ϕ
xA
,
m¨ xA = S 1 + S 2 ΘA ¨ ϕ = rS 1
S 2
x ¨A =
¨ x , 2
ϕ = ¨
x˙ A
x ¨ 2r
x˙
and introduce the moment of inertia ΘA = mr 2 /2 we obtain, by solving for x, the equation of motion x ¨+
k 4M +
3 2
m
x=0
ω=
where
k 4M +
3 2
m
is the natural frequency. It is related with the period of the vibration by T =
2π ω
T = 2π
k 4M +
3 2
m
.
From the general solution x = A cos ωt + B sin ωt with xA = x/2 and the initial conditions xA = a, x˙ A = 0 follows x(t) = 2a cos ωt .
Vibrations
177
Problem 7.4 The
system shown consists of two clamped beams (negligible masses, bending stiffnesses EI i ), a spring (spring constant k3 ) and a block of mass m.
P7.4
k3
l
Determine the eigenfrequency ω.
EI 1
l EI 2
m
1. approach: We separate the system and formulate the equation of motion for the block: Solution
m¨ x =
↓:
−F − F − F 1
2
3
F 3 F 1
.
F 2 F F = F 1 +F 2 +F 3
The forces F i and the displacements wi of the beams and the spring, respectively, are related by (see volume 2, table of elastic lines)
x
F 1 l3 F 2 l3 F 3 w1 = , w2 = , w3 = . 3EI 1 3EI 2 k3 Therefore, from the condition w1 = w 2 = w 3 = x follows 3EI 1 3EI 2 m¨ x+ + + k3 x = 0 . l3 l3 Hence, we obtain for the eigenfrequency
1 3EI 1 3EI 2 + + k3 . ω = m l3 l3 2
2. approach: Since all spring ends undergo the same displacement x, the springs are in parallel: k3
k1
k1
k2
k3
m
With the spring constants F 1 3EI 1 = 3 , x1 l
k2 =
F 2 3EI 2 = 3 x2 l
we obtain according to page 171 3EI 1 3EI 2 k = k 1 + k2 + k3 = 3 + 3 + k3 l l and therefore 1 3EI 1 c 3EI 2 ω2 = = + + k3 . m m l3 l3 ∗
∗
k
∗
m
m
k1 =
k2
178 P7.5
Free Undamped
Problem 7.5
The system consists of a hinge supported beam, reinforced by three bars, and of a block of mass m. Determine the eigenfrequency for vertical vibrations. Assume that the mass of the beam and the bars is negligible.
a
a m
EI
EA
EA
EA
a/2
Solution
The elastic system of beam and bars can regarded as a system consisting of two springs in parallel, one representing the beam, one representing the bars. To determine the spring constant kB of the beam we subject it by the unit force F B = 1 at the location of the block. It produces the F B = 1 deflection (see volume 2, chapter 4) 1 (2a)3 wB = . wB 48EI Hence, the spring constant is given by F S = 1 1 48 EI 6 EI f S kB = = = . 1 wB (2a)3 a3 2 3 To find the spring constant kS of the bar system, we apply a force F S =1 at the bar 1. It causes the displacement (see volume 2, chapter 6) ¯i2 li S ¯i = forces in bars, li = lengths of bars) . (S f S = hS = EA With 5 a ¯1 = 1 , ¯2 = S ¯3 = 5 , S S l1 = , l2 = l 3 = a 2 2 2 we obtain 1 5 2 5 5 5 a 2 a f S = ( 1) +2 a = 1+ EA 2 2 2 2 2EA 4EA kS = . (2 + 5 5)a
·
√
−
−
· √
√
· √ √
√
From the spring constants of the two springs in parallel follows the constant k of the equivalent single spring: ∗
6 EI EA + . l3 (1 + 2)l Thus, the eigenfrequency is given by ∗
√
k = k B + kS =
k 1 = m l ∗
ω=
√
1 EAl2 6 EI + ml 1+ 2
.
Vibrations
179
Problem 7.6 For
the two systems ① and ② the equivalent spring constants for vertical vibrations of the body of mass m shall be determined. The mass of the beams can be neglected.
P7.6
EI
k
m 2a EI
2EI
①
3a 2EI
②
k m Solution
In system ① all three springs are directly connected with the vibrating body of mass m and experience the same displacements. Thus, the springs are in parallel: ∗
k =
ki .
The spring constants kL and kR of the left and the right beam follow from the end displacement of a cantilever beam (stiffness EI , length l) subjected to a unit force 1“ ” 1 l3 1 1 w= = k and c = l , EI 3EI k as w 1 3EI 1 3(2EI ) kL = = , k = = . R cL (2a)3 cR (3a)3
·
Hence, the stiffness of the equivalent spring is 43 EI 43EI + 72ka3 k = k L + kR + k = +k = . 72 a3 72a3 ∗
In system ② both beams are in parallel. Their equivalent spring with ¯ in series with the spring with constant k. Thus, the total constant k is equivalent constant k is calculated as follows: ∗
¯ = k L + kR = 43 EI , k 72 a3 1 1 1 72a3 1 = ¯ + = + k k 43EI k k
kL
∗
43EI
m
EI 72a + 43 k 3
k
∗
k
k
∗
43EIk k = = 43EI + 72ka 3
k¯
kR
m
m
.
Remark: The second system has a smaller stiffness and therefore vibrates with a lower frequency.
180 P7.7
Free Undamped
Problem 7.7
A wheel (mass m, moment of inertia ΘC , radius r) rolls without slipping on a flat surface. On its top, the wheel carries a horizontally guided beam (mass M ) and moves it without slipping.
k2
M k1
B
C r
m, ΘC
Determine the eigenfrequency of the system. Neglect the mass of the guiding roll B.
Solution
We separate the system and introduce the coordinates x1 , x2 and ϕ, measured from the equilibrium position. Then the equations of motion are ①
mx ¨1 =
→:
1 1
ΘC ϕ ¨=
C :
②
−k x − H + H ,
M x ¨2
→:
A
A
x2
H A
D
−r H − rH , = −k x + H .
k1 x1
H A
A x1 mg
D
2 2
Mg
②
H D
A
k2 x2 B
ϕ C r
①
D
With the kinematic relations x1 = rϕ ,
x2 = 2x1
x ¨1 = r ϕ ¨,
x ¨2 = 2¨ x1
we now have five equations for the five unknowns x1 , x2 , ϕ, H A and H D . Solving for x1 (t) yields the equation of motion x ¨1
ΘC m + 4M + 2 +(k1 + 4k2 ) x1 = 0 r
or in standard form x¨1 + ω 2 x1 = 0
with the eigenfrequency ω=
k1 + 4k2 . ΘC m + 4M + 2 r
Vibrations Problem 7.8
For a connecting rod of weight mg, the periods T a and T b are measured for the hangings a) and b).
a)
P7.8
b)
a
Determine the moment of inertia Θc and the distance a of the center of mass C .
181
C l
C b
Solution
The equation of motion for small amplidudes for the hanging a) is given by
A
ΘA ϕ ¨ + mgaϕ = 0 ϕ
and the respective period follows as
mg
(2π)2 4π 2 ΘA = = . ω2 mga Analogous for hanging b) results T a2
C B
4π 2 ΘB = . mgb With the parallel axis theorem T b2
ΘA = Θc + ma2 ,
ΘB = Θc + mb2
and with a + b = l we obtain T a2 mga Θc = 4π 2
− ma
2
and
T b2 g 4π 2 l a= l. (T a2 + T b2 )g 8π 2 l
−
−
Remarks: A swinging rigid body (here a connecting rod) which cannot be regarded as point mass at a massless cord, is called a compound pendulum or physical pendulum .
• •
Analogeous to a simple pendulum, the equation of motion of a compound pendulum is often written as ϕ ¨+
g lred
ϕ=0
with
lred =
ΘA , ma
where a = distance between pivot A and center C of mass.
182 P7.9
Free Undamped Vibration
Problem 7.9
A body (mass m) is pulled by a spring (stiffness k) to a smooth horizontal path. In vertical position m a the spring is stretched by the distance a and its length in the unstretched state is l.
a) Which condition must be fulfilled by the horizontal displacement, that the vibration about the equilibrium position is harmonic? b) Determine the period for this case.
l
a) The equation of motion is given by
Solution
→:
k
mx ¨ =
with
mg
−F sin ϕ
F k = k ∆k ,
x
k
sin ϕ =
x (l + a)2 + x2
F k ϕ
and the spring elongation ∆k = a +
(l + a)2 + x2
− (l + a)
.
N
x
l+a The vibration is harmonic if the equati(l+a)2 +x2 2 on of motion is of the type x¨ = ω x. ϕ Consequently, F k sin ϕ must be linear in x! This is only possible if the term x2 in the square roots can be neglected in comparison with (l + a)2 , i.e. if the condition
−
x2
2
(l + a)
|x| l + a
is fulfilled. Therefore, the displacement x must always be sufficiently small. In this case, in a first approximation, ∆k = a = const and sin ϕ = x/(l + a) hold and the equation of motion is given by mx ¨+
ka x = 0 . l + a
b) We write the equation of motion in its standard form x ¨ + ω2x = 0
with
ω2 =
k a m l + a
and obtain with ω = 2π/T for the period T = 2π
m(l + a) . ka
Free Damped Vibration
183
Problem 7.10
A simple pendulum is connected with a spring (stiffness k) and a dashpot (damping coefficient d).
P7.10
g a
a) Determine the damping coefficient d such that vibrations are possible. Assume small amplitudes. b) What damping ratio ζ must be chosen such that the amplitude is reduced to 1/100 of its initial value after 20 full cycles. Calculate the corresponding period T d .
k a d m
a) Assuming small amplitudes, i.e. cos ϕ 1, sin ϕ ϕ, the equation of motion follows from the principle of angular momentum: Solution
≈
≈
A ϕ
ΘA ¨ ϕ =
A :
−F a − F 2a − mg2aϕ . k
d
F k
Introducing ΘA = m (2a)2 ,
F k = k a ϕ ,
Fd = d 2a ϕ˙
we obtain d k g ϕ + ¨ ϕ˙ + + ϕ = 0 m 4m 2a
where ξ=
d , 2m
ω2 =
F d
mg
ϕ¨ + 2ξ ϕ + ˙ ω 2 ϕ = 0 ,
k g + . 4m 2a
To ensure vibrations, the system must be underdamped, i.e. ξ < ω: d < 2m
k g + 4m 2a
d<
gm 2 km + 2 . a
b) The necessary damping ζ ratio follows with xn+20 = xn /100 from the logarithmic decrement (see page 172): 20
2πζ xn = ln = ln 100 2 x n+20 1 ζ
−
This leads to the period T d =
ω
2π 1 ζ 2
−
≈ 2πω = 2π
ζ =
4am . ak + 2gm
1
40π ln100
2
= 0.037 . +1
184 P7.11
Free Damped Vibration
Problem 7.11
Determine the eigenfrequency for the displayed system with viscous damping. The mass of the cantilever beam and the bar can be neglected.
l2 , EA
l1 , EI
k m d
Solution We
replace the stiffnesses of the beam, the bar and the spring by an equivalent stiffness k . When the body (mass m) is displaced by x from its equilibrium position, it is F k loaded by a spring force F k = k x and a damping force F d = dx. ˙ Thus, the equation of motion is given by m ∗
∗
x
↓:
m¨ x =
−F − F k
.
d
F d
The stiffness of the equivalent spring follows from the spring stiffnesses in parallel of the beam and the bar k12 = k 1 + k2
k1
k12
k2
where
k k 3EI EA k1 = 3 , k2 = m m l1 l2 and the spring stiffnesses in series of the spring k12 and the spring k as
1 1 1 = + k k12 k 3EI EA k + l13 l2 k = . 3EI EA k + 3 + l1 l2
∗
∗
k12 k
k m
Hence, we obtain x ¨ + 2ξ x˙ + ω 2 x = 0
∗
m¨ x + d x˙ + k x = 0
where d ξ= , 2m
∗
k ω = . m 2
Therefore, the eigenfrequency is given by ωd =
− ω2
− k m
∗
δ 2 =
d 2m
2
.
∗
m
Free Damped Vibration
185
Problem 7.12
A bar (weight mg, length l) which is connected with a spring (spring constant k) vibrates in a viscous fluid about point A. The viscous drag force F d is proportional to the local velocity (proportionality factor α).
P7.12
A g l
a) Derive the equation of motion under the assumption of small amplitudes. b) Calculate the critical value α , separating vibrations from a creep motion.
k
∗
a) An element of length dx of the bar is subjected to the drag force Solution
A
x
dF d = α v(x) dx = αx ˙ ϕ dx . l cos ϕ ≈ l ϕ
Thus, considering small amplitudes ϕ 1 (sin ϕ ϕ, cos ϕ 1), the principle of angular momentum yields
≈
A :
≈
ΘA ¨ ϕ =
−
−mg l2 ϕ−kl ϕ 2
l
dx dF d
mg
≈
klϕ
l lϕ sin ϕ ≈ 2 2
αx2 ϕ˙ dx .
0
l sin ϕ ≈ lϕ
Evaluating the integral and introducing ΘA = ml 2 /3 leads to the equation of motion
αl 3k mg ϕ + ¨ ϕ˙ + 1+ ϕ=0 m m 2kl where ξ=
αl , 2m
ω2 =
3k mg 1+ m 2kl
ϕ ¨ + 2ξ ϕ˙ + ω 2 ϕ = 0
.
b) Damped vibrations are separated from an aperiodic motion by the critical damping ξ = ω
or
ζ = 1 .
From this condition follows ∗
α l = 2m
3k mg 1+ m 2kl
∗
α =
12km l2
1+
mg . 2kl
186 P7.13
Free Damped Vibration
Problem 7.13
Determine for the displayed underdamped system
R2 r2
a) the equation of motion for vibrations around the equilibrium position, b) the circular frequency for the case r2 = R 2 /4, k2 = 2k1 , m2 = 4m1 and c) the solution x1 (t) for the initial conditions x1 (0) = 0, x˙ 1 (0) = v 0 . The rope and rolls can be considered as massless.
g
x1
R1 m2 k2
a) We introduce all coordinates with their origin at the equilibrium position. Then the weights of the bodies must not be considered. From the kinematic relations at the rolls ① and ②
m1 k1
d
Solution
x3 = 2x1 ,
x2
x3
m2
x ¨2 = 2
① ②
Π 1:
A:
m1 x ¨1 = m2 x ¨2
r2 x ¨1 . R2
x2
−S − F − F , = S − F , 4
1
k1
d
k2
R1 S 4 = 2R1 S 2 , r1 S 1 = R 2 S 2 .
In conjunction with the spring and damper laws F k1 = k 1 x1 ,
S 2
S 1
Now we separate the system and formulate the equations of motion for the bodies (mass m 1 , m2 ) and the rolls. Since the rolls are massless, the latter are reduced to the equilibrium conditions:
↓: ↑:
②
x3 R2 = x2 r2
it first follows r2 x2 = 2 x1 , R2
m1
A
F d = d x˙ 1 ,
F k2 = k 2 x2
m2
①
x3
S 3 Π1 x1
S 4
F k2
x1 m1 F k1
F d
Free Damped Vibration
187
we now have 9 equations for the 9 unknowns x1 , x2 , S 1 , S 2 , S 3 , S 4 , F k1 , F d and F k2 . Solving for x1 leads to the differential equation
m1 + m2
r 2 2 R2
2
x ¨1 + d x˙ 1 + c1 + c2
r 2 2 R2
2
x1 = 0
or in standard form x ¨1 + 2ξ x˙ 1 + ω 2 x1 = 0 where d
2ξ =
m1 + m2
r 2 2 R2
2
r 2 k1 + k2 2 R2 ω2 = r 2 m1 + m2 2 R2
,
2
2
.
b) The circular frequency of the underdamped vibration is calculated from ωd = ω 2 ξ 2 . For the parameters r2 = R2 /4, k2 = 2k1 , m2 = 4m1 we obtain
−
d 2ξ = 2 m1
d2 ξ = , 16 m21 2
ω2 =
3 k1 , 4 m1
and it follows ωd =
3 k1 4 m1
−
d2 . 16 m21
c) The general solution for the vibration of an underdamped system reads ξt
−
x1 (t) = e
(A cos ωd t + B sin ωd t) ,
from which it follows by differentiation ξt
−
x˙ 1 (t) = e
[( A ξ + B ωd )cos ωd t
−
− (A ω + B ξ)sin ω t] . d
d
The initial conditions lead to x1 (0) = 0
x˙ 1 (0) = v 0
A = 0 ,
−A δ + B ω
d
and we finally obtain x1 (t) =
v0 e ωd
ξt
−
sin ωd t .
= v 0
B=
v0 , ωd
188 P7.14
Forced
v0
Problem 7.14
A car (mass m), simplified modelled by a springmass-system, drives with constant horizontal velocity v0 through sineshaped periodic bumps (amplitude U 0 , wave length L).
m
x
U 0
k
L
a) Derive the equation of motion of the car in vertical direction and determine the exciting frequency Ω. b) Determine the vertical amplitude x0 of the car in dependence of the velocity v0 . c) Calculate the critical velocity vc (resonance!).
a) We denote the vertical displacement of the car by x and describe the shape of the bumps by u. Then from Newton’s law follows Solution
↑:
m¨ x =
−k (x − u) .
With the position of the car s = v 0 t in horizontal direction, the function u is represented as u = U 0 cos
m x
u
k u
L
U 0 s
2πs 2πv0 t = U 0 cos = U 0 cos Ωt , L L
and we obtain the equation of motion and the exciting frequency mx ¨ + k x = k U 0 cos Ωt
where
Ω=
2πv 0 . L
b) The steady state solution of the equation of motion is of the type of its right-hand side. Thus, the ansatz x = x 0 cos Ωt leads with ω 2 = k/m to the amplitude of the steady state vibrations x0 =
U 0 2 = 1 Ω2 1 ω
−
−
U 0 . 2 2 4π v0 m L2 k
c) The amplitude x 0 tends to infinty when Ω approaches ω (resonance): Ω2 = ω 2
4π 2 vk2 L2
m =1 k
vk =
L 2π
k . m
Vibrations
189
Problem 7.15
A pressure gauge consists of a piston ① (mass m1 , cross section A), a bar ② (mass m2 ), a thin needle ③ (mass m3 ) and a spring (stiffness k).
P7.15
m1
①
p
m2
②
a m3
l
k
③
a) Determine the eigenfrequency of the system. b) Calculate the amplitude Q0 (small displacements!) of the needle tip in the steady state case if the pressure is given by p = p 0 cos Ωt.
a) We separate the system and make all acting forces visible. Then, we have for the parts ① + ② and ③ Solution
→:
(m1 + m2 )¨ x =
x
①
−F − kx + p(t)A ,
F ϕ
ΘB ¨ ϕ = aF .
B :
②
F ③
Hence, with ΘB = m 3 l2 /3 and the kinematics x = aϕ
kx
p
B
x ¨ = a ϕ ¨
follows the equation of motion
m3 l 2 m1 + m2 + 3a2
x ¨ + kx = p 0 A cosΩt .
For the eigenfrequency, we directly obtain from this equation ω=
k 2
m1 + m2 +
m3 l 3a2
.
b) The steady state solution is described by an ansatz (of the right hand side type) x = x 0 cos Ωt . Substituting into the differential equation yields p0 A x0 = . Ω2 k 1 ω2 With the leverage we obtain for the amplitude of the needle tip
−
Q0 = x 0
l = a
1 1
−
p0 A l . 2 k a Ω ω2
190 P7.16
Forced
Problem 7.16
A homogeneous wheel of mass m is attached to a spring (spring constant k). The wheel rolls without slipping on a rough surface which moves horizontally according to u = u 0 cos Ωt.
①
→:
C :
x = u + rϕ
x ϕ
r
②
k(x − u) C
H
N
¨ = u ¨ + rϕ = ¨ x
u
mg C
With the kinematic relation
m
C r
①
kx
−k x + H , ϕ = ¨ −r H .
m
k
mg
mx ¨ = ΘC
②
x
a) The equations of motion read in case ①
C r
u
a) Derive the differential equations for the vibrations in case ① and ②. b) Determine the amplitudes of the steady state vibrations.
Solution
k
N
−u Ω 0
2
ϕ
r
H
cos Ωt + rϕ ¨
and ΘS = 21 mr2 we obtain the differential equation for forced vibrations x ¨+ In case
→:
2 k x = 3 m ② the
− 13 u Ω 0
2
cosΩt .
equations of motion are given by
mx ¨ =
−k(x − u)x + H ,
ΘC ϕ = ¨
C :
−r H
and the kinematic relation again reads x = u + rϕ
x ¨ = u ¨ + rϕ = ¨
−u Ω 0
2
cos Ωt + rϕ ¨
which leads to the differential equation
2 k 1 k x ¨+ x = 2 3 m 3 m
2
− Ω
u0 cos Ωt .
b) The steady state solution is of the type of the right-hand side. Thus, the ansatz x = x 0 cos Ωt leads to the amplitudes case ①
Ω2 , 2k/m 3Ω2
|x | = u | 0
0
−
|
2
case ②
|x | = u 0
0
|2k/m − Ω | . |2k/m − 3Ω | 2
Note that resonance in both cases occurs for Ω2 = 2k/3m, but that the amplitudes for Ω2 2k/3m and for Ω2 2k/3m are very different.
Vibrations
191
Problem 7.17
In a soil compactor (housing weight mg), the drive (mass M ) is resiliently mounted. In the drive two unbalances (each mass m2 /2) m counter-rotate with constant number n of revolutions per minute. n m2 /2 e M
How must the spring and damper be designed so that the device runs in resonance and the base plate does not lift-off from the ground (small damping ratio!)?
d
k/2
k/2
Solution
We replace the drive by the displayed model. With the displacement x of the drive from the equilibrium position we obtain for the unbalanced mass x2 = x + e cosΩt
x ¨2 = x ¨
2
− eΩ
e
cosΩt .
−
−dx˙ −kx+S cosΩt ,
Ωt S
x2
M − m2 d
The equations of motion in vertical direction for both masses are given by: (M m2 )¨ x =
m2
m2 x ¨2 =
x k
−S cos Ωt .
Introduction of x¨2 and elimination of S yields x ¨ 2ξ x˙ + + x = x 0 η 2 cosΩt , 2 2 ω ω
where ω 2 = k/M , ξ = d/2M , x0 = em 2 /M , η = Ω/ω. The particular solution (steady state) reads x = x 0 V cos(Ωt
− ϕ) ,
V =
− (1
η2 , 2 2 2 2 η ) + 4ζ η
ζ =
ξ . ω
Resonance occurs for small damping ratio ζ at η Ω = ω
πn = 30
k M
k=
πn 30
≈ 1:
2
M.
Furthermore, in resonance the magnification is V 1/2ζ . No lift-off occurs, if the following condition at maximum spring displacement xmax (then the damping force is zero!) is fulfilled:
≈
N = (M + m)g
− kx
Hence, it follows kx 0 (M + m)g 2ζ
≥
max
= (M + m)g
− kx2ζ ≥ 0 . 0
kx 2
≥ 2(M kx+ m)g .
ζ
0
(M +m)g kx 2 N
P7.17
192 P7.18
Forced
Problem 7.18
A single story frame is modelled by a rigid beam of mass m which is supportet by clamped massless beams (stiffness EI ) m and damped by a dashpot. Due to an earthquake, the ground moves with a horizontal acceleration u ¨E = b 0 cos Ωt EI l which is known from measurements. d EI
Determine the maximum amplitude of the steady state vibrations by assuming small amplitudes and a weakly damped system.
45
◦
u¨E x − uE
Solution
We separate the system and replace the vertical beams by equivalent springs with the spring constant k = 3EI/l3 . When the beams are deflected by x uE , the elongation of the diagonal, assuming small amplitudes, is
m
F k x
F d F k
−
∆ = (x
−u
F k
F d ∆
uE
√
E )/ 2 .
45
Hence, the spring forces and the damping force can be written as F k = k(x
F k
˙ = d (x˙ F d = d ∆
−u
E) ,
− u˙
◦
√
E )/
2 .
Then, the equation of motion of the horizontal beam is given by
→:
m¨ x =
−
√ 2 2
F d
− 2F
k
d m¨ x + (x˙ 2
Thus, the relative displacement y = x m¨ y +
d y + ˙ 2ky = m b0 cos Ωt 2
where ω2 =
2k , m
ζ =
d 2
1 , 8k m
−u
E
− u˙
E ) + 2k(x
−u
E)
is described by
1 2ζ y ¨ + y + ˙ y = y 0 cos Ωt , ω2 ω
y0 =
m b0 . 2k
The maximum amplitude A occurs for resonance, i.e. for η = ω/Ω In case of weak damping (ζ 1) it is given by A = y 0 V max
≈
√ b
y0 0 =2 2 2ζ d
m3 . 4k
= 0.
≈ 1.
Vibrations
193
Problem 7.19
A wheel is connected with a rigid frame via a spring and a damper. When the frame experiences a displacement u = v0 t, the initially (t = 0) resting wheel starts rolling without slip. a) How must the damper be designed so that for free vibrations critical damping occurs? Which form has the equation of motion in this case? b) Determine the solution for the given initial conditions.
g
u = v 0 t m Θ d
k
r
a) We count x and ϕ from the equilibrium position of the wheel at t = 0. Then the equations of motion Solution
→:
m¨ x =
−d(x˙ − u)˙ − k(x − u) − H ,
C :
lead with u = v ˙ 0 and x = r ϕ by eliminating H and ϕ¨ to the differential equation Θ m+ 2 x ¨ + d x˙ + k x = d u˙ + k u . r
u
ϕ
x
It can be written in the standard form 2
Θ ϕ = r ¨ H
mg
2
m, Θ
x ¨ + 2ξ x˙ + ω x = 2 ξ v0 + ω v0 t , ˙ d(x˙ − u)
where d 2ξ = , m + Θ/r2
k(x − u)
C
H
k ω = . m + Θ/r2 2
N
The condition for critical damping of free vibrations is given by ξ/ω = 1, or ξ 2 = ω 2 . This leads for the damping coefficient to d2 k = 4 (m + Θ/r2 )2 m + Θ/r2
d = 2
k (m + Θ/r2 ) .
In this case, the equation of motion takes the following form: x ¨ + 2ω x˙ + ω 2 x = 2 ω v0 + ω 2 v0 t .
b) The solution of the differential equation is composed of the solution of the homogeneous differential equation (= free vibration at critical
P7.19
194
Forced
damping) ωt
−
xh = (A1 + A2 t) e
and the particular solution xp . To find the latter one, we choose an ansatz of the type of the right hand side: xp = a + b t . Introducing it into the differential equation and comparing the coefficients yields a = 0 and b = v0 . Thus, the general solution is given by ωt
−
x(t) = x h + xp = (A1 + A2 t) e
+ v0 t .
The constants A1 and A2 are determined from the initial conditions: x(0) = 0
A1 = 0 ,
x(0) ˙ =0
A2 =
−v
0
.
Hence, we obtain the specific solution x(t) =
ωt
−
−v t e 0
+ v0 t = v 0 t (1
ωt
−
−e
).
It can be seen that the motion of the wheel exponentially approaches the motion of the frame. For ωt 1 the motion of the frame and the wheel are the same.
Chapter 8
Non-Inertial Reference Frames
8
196
Non-Inertial Reference Frames
Fixed and moving Reference Frame It is often advantageous to describe the motion of a point P not in reference to a fixed coordinate system (x,y,z ) but in reference to a moving system (ξ , η , ζ ) . Kinematics of a point for a translating and rotating reference system v
= v f + v r ,
a = af
P
z
ω
ζ
+ ac + ar ,
r 0P
r0
where absolute velocity
v
,
absolute acceleration
a
,
fictitious velocity
v f
relative velocity
vr
fictitious acceleration
af
relative acceleration
ar
ξ
x
η
0
y
= r˙ 0 + ω = r∗ ,
×r
0P
,
¨ 0 + ω ˙ =r = r ∗∗ ,
×r
0P
+ω
0P
× (ω × r
0P )
,
0P
Coriolis acceleration ac = 2ω vr . and () = time derivative with respect to the fixed system (x,y,z ), ()∗ = time derivative with respect to the moving system (ξ , η , ζ ) .
×
·
Remarks:
•
The equations simplify for a pure translation of the reference system (ω = 0).
•
The Coriolis acceleration
ac is
orthogonal to
ω
and
vr .
Equation of motion in a moving reference system In addition to the real forces F acting on the point mass, the fictitious force F f and the Coriolis force F c appear in the equation of motion: mar = F + F f + F c where F f F c
−ma = −m[¨r + ω˙ × r = −ma = −2mω × v . =
f
c
0
r
0P
+ω
× (ω × r
0P )] ,
Translation of the Reference Frame Problem 8.1
Point A of a simple pendulum (mass m, length l) moves with a constant acceleration a0 obliquely upwards.
P8.1
a0
A
197
α g
Derive the equation of motion and determine the force in the wire.
l m
Solution We
introduce the ξ, η coordinate system that moves translatoric with point A. Then the equations of motion in the moving system read mξ ∗∗ = F ξ + F f ξ ,
mη ∗∗ = F η + F f η
where, with af ξ = a0 , and af η = 0, the (real) forces and fictitious forces are given by F ξ = S sin ϕ mg sin α , F f ξ = mafξ = ma0 ,
− −
−
−
ξ
A x ϕ α y
S
η ma0
mg
F η = S cos ϕ + mg cos α , F f η = mafη = 0 .
− −
The relative accelerations arξ , arη follow from the coordinates of the point mass in the moving system through differentiation (note that the time derivatives of ϕ in the moving and the fixed system are the same: ϕ∗ = ϕ) ˙ ξ = l sin ϕ , η = l cos ϕ , v = ξ ∗ = l ˙ ϕ cos ϕ , v = η ∗ = l ˙ ϕ sin ϕ ,
− = η ∗∗ = −lϕ ¨ sin ϕ − l ˙ ϕ
rη
rξ
arξ = ξ ∗∗ = l ϕ¨ cos ϕ
2
− l ˙ϕ
sin ϕ ,
arη
2
cos ϕ .
Introducing them into the equations of motion yields lϕ¨ cos ϕ lϕ¨ sin ϕ
−
2
− lϕ˙ − l ˙ϕ
sin ϕ = 2 cos ϕ =
−S sin ϕ − mg sin α − ma −S cos ϕ + mg cos α .
0
,
These are two equations for the unknowns ϕ and S . Solving for ϕ and subsequently for S leads to the equation of motion and the force in the wire: lϕ + ¨ g sin(α + ϕ) + a0 cos ϕ = 0 , S = m[l ˙ ϕ2 + g cos(α + ϕ)
− a sin ϕ] . 0
Remark: For ϕ = ¨ 0 the equation of motion reduces to tanϕ0 = a 0 /g cos α
−tanα characterizing the equilibrium position of the pendulum.
198 P8.2
Translation of the Reference Frame
Problem 8.2
A point P moves on a plate with constant relative velocity vr and initial condition ϕ(0) = 0 along a circular path. The plate moves rectilinear with the velocity v = a 0 t.
vr P r
ϕ
v(t)
Determine the magnitude of absolute velocity and acceleration of P as functions of angle ϕ. Solution We
introduce the fixed x, ycoordinates and the translatoric moving reference frame ξ, η. In the moving system, introducing ϕ∗ = ϕ˙ = vr /r, the components of the relative velocity and acceleration are
η y r
fixed
x
0
ϕ
vr P ξ
vrξ = ξ ∗ = vr sin ϕ , v = η ∗ = v cos ϕ ,
−
rη
arξ arη
r
= ξ ∗∗ =
vr ϕ∗ cos ϕ =
− − = η ∗∗ = −v ϕ∗ sin ϕ = − r
vr2 cos ϕ , r vr2 sin ϕ . r
The reference frame undergoes a translation in x-direction with velocity v = a0 t and acceleration a, where time t on account of ϕ = ϕt ˙ = vr t/r can be replaced by ϕ. Accordingly, the absolute velocity and acceleration are given by vx = a 0 t + vrξ = a 0 t
− v sin ϕ = av r ϕ − v sin ϕ , 0
r
r
r
vy = v rη = v r cos ϕ , ax = a 0 + arξ = a 0 ay = a rη =
−
−
vr2 cos ϕ , r
vr2 sin ϕ . r
Thus, the magnitudes of velocity and acceleration follow as v(ϕ) =
a(ϕ) =
vx2 +
vy2
a2x +
a2y
=
=
a20 r 2 2 ϕ vr2
a20 +
vr4 r2
2 r
− 2a r ϕ + v ,
−
0
vr2 2a0 cos ϕ . r
Rotation of the Reference Frame Problem 8.3 A
point P moves on a disk with constant relative velocity vr along a circular path. The disk rotates with constant angular velocity ω about A.
199 P8.3
vr
P r
Determine the absolute velocity and the absolute acceleration of P .
ω
e
A
Solution
We introduce the moving coordinate system ξ, η, ζ with its origin in the center 0 of the disk. Relative to this system, point P undergoes a circular motion. With the relative velocity vr and the magnitude ar = vr2 /r of the relative acceleration and its direction (from P to 0) we can write vr ar
η
vr r 0P
0 y
e A
= v r ( eξ sin ϕ + eη cos ϕ) , vr2 = (eξ cos ϕ + eη sin ϕ) . r
−
P ϕ ω
x
−
With ˙ ω
= ω eζ , r 0 = e eη ,
= 0 , r 0P = eξ r cos ϕ + eη r sin ϕ , ˙ 0 = e ω eξ ¨ 0 = a0 = e ω 2 eη r r
ω
−
−
we obtain = r˙ 0 + ω r0P = (e + rω sin ϕ)eξ + rω cos ϕeη , (ω r 0P ) af = a0 + ω = e ω 2 eξ + rω 2 [eζ (eζ eξ cos ϕ) + eζ (eζ eη sin ϕ)] = (e + r cos ϕ)ω 2 eξ rω 2 sin ϕeη , ( eξ sin ϕ) + eζ eη cos ϕ] ac = 2ω v r = 2ωv r [eζ = 2ωv r (eξ cos ϕ + eη sin ϕ) . v f
× − × × − × × − − × ×− −
× ×
×
Thus, the absolute velocity and acceleration are found as v
= v f + v r =
a = af
=
− =−
−[e + (v
r
+ rω)sin ϕ]eξ + (vr + rω)cos ϕeη ,
+ ar + ac
vr2 vr2 2 [eω + (rω + + 2ωv r )cos ϕ]eξ [rω + + 2ωv r ]sin ϕ eη r r vr vr [eω 2 + r(ω + )2 cos ϕ]eξ r(ω + )2 sin ϕ eη . r r 2
2
−
−
ξ
200 P8.4
Rotating
Problem 8.4
On the rotating earth (radius R = 6370 km) a point P moves with speed vr = 150 km/h northwards.
N vr P
Determine the magnitudes and directions of the fictitious acceleration and Coriolis acceleration at latitude ϕ = 30◦ .
R
S
Solution
The earth-fixed system ξ , η , ζ rotates with the angular velocity ω = 2π/(24 3600) 73 10−6 s−1 around the ζ axis. If we neglect the motion of the earth around the sun (¨ r 0 = 0), we obtain with
·
ω r 0P
˙ ω vr
eζ
ω
vr P
ζ
,
ξ
0
= R cos ϕ eη + R sin ϕ eζ ,
ϕ η
= 0 , =
af
af
=ω
ac
≈ ·
= ω
for
=
N
−v sin ϕ e r
and
η
+ vr cos ϕ eζ
S
ac
× (ω × r 2
0P )
= ω 2 R[eζ
× (e × cos ϕ e ) + e × (e × sin ϕ e )] η
ζ
ζ
ζ
ζ
−ω R cos ϕ e , = 2ω × v = 2ωv [e × (− sin ϕ e ) + e × (cos ϕ e )] η
r
r
η
ζ
ζ
ζ
= 2ωv r sin ϕ eξ . Thus, the magnitudes of the accelerations for ϕ = 30◦ are af = ω 2 R cos ϕ = (73)2 10−12 6370 103 cos 30◦ = 0, 029 m/s2 ,
·
·
ac = 2ωv r sin ϕ = 2 73 10−6 150
· ·
·
·
· 3,16 · sin 30◦ = 0, 003 m/s
2
.
The fictitious acceleration is perpendicular to the axis of rotation of the earth and the Coriolis acceleration points tangential to the latitude to the west. Remarks:
•
With reference to the moving system, the motion of point P is a pure circular motion.
•
The Coriolis acceleration has its maximum at the north pole.
Reference System
201
Problem 8.5 Along
the upper part of an angled rod which rotates with angular velocity ω, a knuckle (weight W = mg) may fritionless slide.
P8.5
s1 B
s0 A
After which time t1 the knuckle touches the rod end B if it is released at A with zero relative velocity?
m
α ω
Solution
We introduce the rod-fixed rotating reference system ξ, η, ζ and the coordinate s. Then we obtain with s
= ω eζ ,
ω
ζ
= s cos α eξ + s sin α eζ ,
r 0P
= v r cos α eξ + vr sin α eζ
vr
vr
ω
0
r 0P
P
α
ξ
¨ 0 = 0 for the fictitious force and Coriolis force: and ω˙ = 0, r =
2
−mω × (ω × r ) = mω s cos α e , F = −2mω × v = −2mωv cos α e . Thus, the equation of motion in s−direction reads ζ : m¨s = −mg sin α + mω s cos α s¨ − κ s = −g sin α with κ = ω cos α. F f
0P
c
r
r
η
2
ξ
N
2
2
s
mω 2 s cos α ξ
α
From the general solution of this differential equation s(t) = A cosh κt + B sinh κt +
g sin α κ2
in conjunction with the initial conditions s(t = 0) = s0
A = s 0
B = 0 ,
vr (t = 0) = s(t = ˙ 0) = 0 s(t1 ) = s1 follows 1 t1 = arcosh κ
s1 s0
− −
g sin α κ2 g sin α κ2
α , − g sin κ 2
s1 = A cosh κt1 +
.
F c
g sin α κ2
mg
202
Translation and Rotation
ζ, ζ
∗
P8.6
η
Problem 8.6
The cantilever of a whirlgig moves about the xaxis with the time-dependent angle ϑ(t). At its end a circular disk is fixed which rotates with con- z stant angular velocity ω0 about an axis perpendicular to the cantilever.
0
ϑ
∗
ω0
r ϕ
y
η P ξ
∗
ξ l
Determine for point P of the disk the absolute velocity and absolute acceleration by using
x
a) the cantilever-fixed systems ξ, η, ζ and b) the disk-fixed systems ξ ∗ , η ∗ , ζ ∗ . Solution
The absolute velocity and acceleration are determined from the general relations = r˙ 0 + ω ¨ 0 + ω ˙ a = r v
×r ×r
+ vr , (ω 0P + ω
0P
× ×r
0P ) +
2ω
×v
+ ar .
r
a) For the coordinate system ξ, η, ζ with the unit vectors we first have r0
= l eη ,
ω
= ϑ˙ eξ ,
r 0P
= r cos ϕ eξ + r sin ϕ eη .
From these relations, by introducing (e˙ i = ω ˙ξ e
˙η e
= 0 ,
= ϑ˙ eζ ,
˙ζ e
=
eξ , eη , eζ
−ϑ˙ e
η
×e ) i
,
ϕ = ω ˙ 0,
we obtain ˙0 r
= l ϑ˙ eζ ,
¨0 r
= l ϑ¨ eζ
− l ϑ˙ e 2
η
˙ ω
,
= ϑ¨ eξ .
With reference to this system, the movement of P is a circular motion with constant angular velocity ω0 , i.e. we have vr
= r ω0 ( sin ϕ eξ +cos ϕ eη ) ,
ar
−
=
2 0
−r ω (cos ϕ e +sin ϕ e ) . ξ
Introducing these relations with
× r = r ϑ˙ sin ϕ e , ω × (ω × r ) = −r ϑ˙ sin ϕ e ω
0P
˙ ω
ζ
0P
2
η
,
×r ω×v
0P r
= r ϑ¨ sin ϕ eζ ,
= r ω0 ϑ˙ cos ϕ eζ
η
of the Reference Frame
203
finally leads to the results v
=
a =
−r ω sin ϕ e + r ω cos ϕ e + (l + r sin ϕ)ϑ ˙ e 0
−
0
ξ
r ω02 cos ϕ eξ
−
η
ζ
˙2
ω02 sin
(l + r sin ϕ) ϑ + r
+ (l + r sin ϕ) ϑ¨ + 2 r ω0 ϑ˙ cos ϕ
eζ
ϕ
.
,
eη
b) For the coordinate system ξ ∗ , η ∗ , ζ ∗ from the representations r0
= l (sin ϕ e∗ξ + cos ϕ e∗η ) , = ϑ˙ (cos ϕ e∗ξ
− sin ϕ e∗ ) + ω
ω
η
= r e∗ξ ,
r 0P
∗,
0 eζ
with
∗ = ω e∗ + ϑ˙ sin ϕ e∗ , 0 η ζ
∗=
˙ξ e
∗=
˙ζ e
˙η e
−ϑ˙ (sin ϕ e∗ + cos ϕ e∗ ) ,
0 eξ
ϕ = ω ˙ 0
η
ξ
∗ + ϑ˙ cos ϕ e∗ , ζ
−ω
follow the relations
−l ϑ˙ (sin ϕ e∗ + cos ϕ e∗ ) + l ϑ¨ e∗ , ˙ = (ϑ¨ cos ϕ − ϑ˙ ω sin ϕ)e∗ − (ϑ¨ sin ϕ + ϑ˙ ω cos ϕ)e∗ . ω ˙0 r
= l ϑ˙ e∗ζ ,
¨0 r
2
=
0
ξ
η
0
ξ
ζ
η
Since P with reference to the system ξ ∗ , η ∗ , ζ ∗ is at rest, we now have v r = ar = 0 . With
×r ˙ ×r ω ω × (ω × r ω
0P
= r ω0 e∗η + r ϑ˙ sin ϕ e∗ζ ,
= r(ϑ¨ sin ϕ + ϑ˙ ω0 cos ϕ)e∗ζ , r(ω˙ 02 + ϑ˙ 2 sin2 ϕ)e∗ξ r ϑ˙ 2 sin ϕ cos ϕ e∗η 0P ) = +r ω ϑ˙ cos ϕ e∗ 0P
−
−
0
ζ
we then obtain v
= r ω0 e∗η + ϑ˙ (l + r sin ϕ) e∗ζ ,
a =
r ω02
˙2
− ϑ ( l + r sin ϕ) sin ϕ
− ϑ˙ ( l + r sin ϕ) cos ϕ e∗ 2
∗
eξ
¨ l + r sin ϕ) + 2 r ω0 ϑ˙ cos ϕ + ϑ(
η
∗
eζ .
Remark: The representations in a) and b) can be transformed into each
other using
eξ
= e∗ξ cos ϕ
− e∗ sin ϕ , η
eη
= e∗ξ sin ϕ + e∗η cos ϕ .
204 P8.7
Translation and Rotation
Problem 8.7
At the lever ①, rotating with constant angular velocity ω about the fixed axis AA, a cylinder ② is mounted. The cylinder rotates about the axis BB parallel to AA with constant angular velocity ω ∗ with respect to the lever. Fixed at the cylinder is a tube where a point P moves with speed v(t) with respect to the cylinder.
B ζ
A z , ζ
∗
②
P η
η
a) the space fixed system x, y, z , b) the lever-fixed system ξ, η, ζ , c) the cylinder-fixed system ξ ∗ , η ∗ , ζ ∗ .
ξ
∗
∗
y ϕ
①
Determine the absolute velocity and absolute acceleration of P by using
s
A
r b
B a x
ϑ ξ
a) With the angular velocity ω2 = ω + ω ∗ of the cylinder with respect to the space-fixed system and with Solution
ϑ = ω ∗ t ,
ϕ = ωt ,
ϕ + ϑ = ω 2 t ,
the position vector of P in the space-fixed system x, y, z is given by = (a cos ϕ + r cos(ϕ + ϑ))ex + (a sin ϕ + r sin(ϕ + ϑ))ey + (b + s)ez
r P
= (a cos ωt + r cos ω2 t)ex + (a sin ωt + r sin ω2 t)ey + (b + s)ez . From its derivatives, by considering s = v, ˙ follow vP aP
=
−(aω sin ωt + rω sin ω t)e + (aω cos ωt + rω cos ω t)e + ve , = −(aω cos ωt + rω cos ω t)e − (aω sin ωt + rω sin ω t)e + v˙ e . 2
2
2
2 2
b) The general relations for = r˙ 0 + ω ¨ 0 + ω ˙ aP = r v P
×r ×r
x
2
2
x
v P
and
+ v r , (ω 0P + ω
2
2
2 2
aP are
given by
0P
× ×r
0P ) +
2ω
×v
r
+ ar .
For the ξ, η, ζ system we have ω
= 0 , r0 = r˙ 0 = r¨0 = 0 , ϑ = ω ∗ t = (a + r cos ω ∗ t)e + r sin ω ∗ t e + (b + s) e ,
= ω eζ ,
r 0P
˙ ω
ξ
η
ζ
y
2
z
y
z
of the Reference Frame
× r = ω(a + r cos ω∗t)e − rω sin ω∗ t e , ∗ ∗ ∗ ∗ v = −rω sin ω t e + rω cos ω t e + v e , ω × (ω × r ) = −ω (a + r cos ω ∗ t)e − ω r sin ω ∗ t e , 2ω × v = −2rωω ∗ sin ω ∗ t e − 2rωω ∗ cos ω ∗ t e , ∗ cos ω ∗ t e − rω ∗ sin ω ∗ t e + v˙ e , a = −rω ω
0P
η
r
ξ
η
ξ
2
0P
ζ
2
ξ
r
η
η
2
r
ξ
2
ξ
η
ζ
and by considering ω + ω ∗ = ω 2 we obtain v P aP
−rω sin ω∗ t e + (aω + rω cos ω∗t)e + ve , = −(aω + rω cos ω ∗ t)e − rω sin ω ∗ t e + v˙ e .
=
2
2
ξ
2
2 2
η
2 2
ξ
ζ
η
ζ
c) For the ξ ∗ , η ∗ , ζ ∗ system we have = ω 2 e∗ζ ,
= 0 , ∗ ∗ a sin ω ∗ t e∗ + b e∗ , r 0 = a cos ω t eξ η ζ aω ∗ sin ω ∗ t e∗ξ aω ∗ cos ω ∗ t e∗η , r˙ 0 = ¨ = aω ∗2 cos ω ∗ t e∗ + aω ∗2 sin ω ∗ t e∗ , r ω
˙ ω
−
− −
0
− ξ
η
= r e∗ξ + s e∗ζ , = v e∗ ,
r 0P vr
ζ
× r = rω e∗ ) = −rω ω × (ω × r 2ω × v = 0 . ω
0P
2 η
0P
2 ∗ 2 eξ ,
r
Thus, it follows v P aP
−aω∗ sin ω∗ t e∗ − (aω∗ cos ω∗t + rω )e∗ + ve∗ , = −(aω ∗ cos ω ∗ t + rω ) e∗ + aω ∗ sin ω ∗ t e∗ + v˙ e∗ .
=
2
ξ
2
2 2
ξ
2
η
ζ
η
ζ
205
206 P8.8
Translation and Rotation
Problem 8.8 In
the frictionless channel of a disk, rotating with the angular velocity ω, a slider of mass m is fixed at springs.
η c m
ξ
ω
Formulate the equation of motion with respect to the moving system ξ, η. Determine the force exerted from the channel on the slider.
A
h
c
Solution
The coordinate system carries out a circular motion about A. With (ζ to ξ, η)
η
⊥
ω
¨0 r
˙ ω
= ω eζ ,
= 0 ,
2
= a0 =
−hω e
ξ
,
r 0P vr
P
m
= η eη ,
= η eη
r 0P
A
h
ξ
0
the fictitious acceleration and Coriolis acceleration are given by af
= a0 + ω˙ 2
ac
ξ
r
0P
+ω
2
2
× (ω × r ) = −hω e + ηω [e × (e × e )] = −hω e + ηω [e × (−e )] = −hω e − ηω e , = 2ω × v = 2ωη (e × e ) = −2ωη e . ×r
2
0P
ζ
ξ
ζ
2
η
ξ
2
ξ
: :
mη =
−2cη + mηω
0 = N + mhω 2 + m2ωη
η
η 2cη
−
2
η
ξ
ξ
In the equations of motion must be considered in addition to the external forces (spring force 2cη, channel force N ), the fictitious force F f = maf and Coriolis force F c = mac . Thus, with ξ = 0 we obtain for the equation of motion and the channel force
−
ζ
m2ωη
mhω 2
N mηω 2
ξ
η +
−
N =
−mω(h ω + 2η ) .
2c m
ω 2 η = 0 ,
Remark: The equation of motion has the solution (see page 170) η(t) =
A cosΩt + B sinΩt with the angular frequency Ω = 2c/m ω 2 . For ω 2 = 2c/m the slider rotates with the disk without vibrating.
−
of the Reference Frame
207
Problem 8.9 A
little sphere (point mass m) oscillates frictionless in a circular channel of a horizontal disk which rotates about an axis through A with constant angular velocity Ω.
P8.9
Ω A e
m
a ϕ
Determine the circular frequency ω of the sphere if small amplitudes ϕ are assumed.
Solution
We introduce the rotating ξ, η, ζ - coordinate system with its origin 0 at the center of the disk. Then we can write vr
Ω = Ω eζ , ˙ = 0 , Ω
a0 r 0P
¨0 = =r
2
−eΩ e
ξ
η
,
m
r 0P
= a cos ϕ eξ + a sin ϕ eη .
Ω
ϕ
e
0 The relative velocity can be expresA sed by the relative angular velocity ϕ∗ where ( )∗ denotes the time derivative relative to the moving system): vr = aϕ ∗ vr = aϕ∗ sin ϕ eξ + aϕ∗ cos ϕ eη .
ξ
·
−
Thus, the fictitious forces
F f
and
−ma − mΩ × (Ω × r
F c
= m(eΩ2 + aΩ2 cos ϕ)eξ
F f
=
F c
+mΩ a sin ϕ eη , = 2mΩ vr = 2mΩaϕ∗ (eξ cos ϕ + eη sin ϕ) .
0 2
−
0P )
are
×
With the tangential relative acceleration art = aϕ∗∗ , the equation of motion in tangential direction is obtained as 2 mΩ a sin ϕ
:
maϕ∗∗ = m(aΩ2 + 2aϕ∗ Ω) sin ϕ cos ϕ a rt m[eΩ2 + (aΩ2 m ∗ η +2aϕ Ω) cos ϕ]sin ϕ
−
N
=
2
−meΩ
sin ϕ .
ϕ
2maϕ Ωsin ϕ m(eΩ2 + aΩ2 cos ϕ) ∗
2maϕ Ωcos ϕ ∗
ξ
Assuming small amplitudes (sin ϕ harmonic vibrations
≈ ϕ), this leads to the equation for
eΩ2 ∗∗ ϕ + ϕ = 0 . a Hence, the circular frequency of the oscillations is ω=
e/a Ω .
208 P8.10
Translation and Rotation
Problem 8.10
The displayed system rotates about the vertical axis with constant angular velocity ω. The two bodies are connected by an inextensible cable and they can frictionless slide along the parts of the elbow.
r m1 m2
Determine the reaction force acting on m1 and the path r(ϕ) of m1 , if at time t = 0 the initial conditions are given by r(0) = r0 and r∗ (0) = 0.
ω, ϕ
Solution
We introduce the rotating elbow-fixed coordinate system ξ , η , ζ and draw the free-body diagrams for W 1 = m1 g ζ F c both bodies with the external forces, reacF f η tion forces and fictitious force and Coriolis ξ S force. There is no relative motion of m1 in N 2 r ζ - and in ξ-direction. Thus, the reaction N 1 S forces follow directly from the respective ’equilibrium conditions’ as N 2 = F c = 2mωr ∗ .
N 1 = W 1 ,
W 2 = m2g
The equations of motion m2 ζ 2∗∗ = S
− W
2
m1 r ∗∗ = F f
,
2
− S = m rω − S 1
lead with ζ 2∗∗ = r ∗∗ (inextensible cable!) to (m1 + m2 )r ∗∗
2
− m ω r = −W 1
2
.
This inhomogeneous differential equation has the solution m2 g r(t) = r h + rp = Aeλt + Be−λt + where m1 ω 2
λ = ω
m1 . m1 + m2
The initial conditions r∗ (0) = 0
A
− B = 0 ,
r(0) = r 0
r0 = A + B +
m2 g m1 ω 2
lead to A = B = 21 [r0 m2 g/(m1 ω 2 )] and herewith, considering ωt = ϕ, we finally obtain the path equation
−
r(ϕ) = r0
−
m2 g cosh ϕ m1 ω 2
m1 m1 + m2
+
m2 g . m1 ω 2
Remark: For r0 m1 ω 2 = S = m2 g we have ’equilibrium’ (r ∗∗ = 0) in
this position, which, however, is unstable: for any small diplacement (disturbance) the system starts to move!
of the Reference Frame Problem 8.11
In the radial channel of a wobbling disk, which rotates with constant angular velocity ω, slides frictionless a knuckle of mass m.
ζ
P8.11
m
η
α
a) Determine the force K (r) that must exerted to the knuckle in channel direction, that it moves according to the law r(t) = r0 sin ωt (the weight shall be neglected). b) Determine the lateral contact force N η (r) between the knuckle and the channel.
r
209
ξ
η ω ξ
m
a) We use the rotating reference system ξ , η , ζ and draw the freebody diagram by considering the fictitious force and Coriolis force Solution
F f F c
=
2
ζ r 0P
−m ω × (ω × r ) = mrω cos α e = −2mω × v = −2m ω v cos α e , 0P
r
ξ
r
where vr = r ∗ = r 0 ω cos ωt. With ar = r0 ω 2 sin ωt = direction reads
−
N η K
F c η
,
ξ
F f N α
η
2
−ω r the equation of motion in channel
mar = K + mrω 2 cos2 α . It leads to the force K : K (r) = ma r
− mrω
2
cos2 α =
2
−mω r (1 + cos
2
α) .
b) The lateral contact force N η is calculated from the ’equilibrium condition’ (no relative acceleration in η-direction) N η = 2mωv r cos α = 2mω 2 r0 cos α cos ωt or with cos ωt =
−
2
1
2
sin ωt =
−
N η (r) = 2mω cos α
r02
−
r2 .
1
(r/r 0 )2 as
210 P8.12
Translation and Rotation
Problem 8.12
The arm of an automatic assembly machine starts moving from rest with constant acceleration a0 along a straight track. At the ar same time, the arm begins to rotate with constant angular velocity ω0 and the slider s P P begins to move towards point A with constant relative acceleration ar . The ϕ initial positions of the arm and the slider are given by ϕ0 = 0 and s0 . a0 A ω0 Determine the absolute velocity and acceleration of the slider P in dependence of time t.
Solution
We use the fixed coordinate system x,y,z , where the x-axis coincides with the track. In addition, we introduce the moving coordinaP ξ te system ξ, η, ζ , where the ξ-axis rotates y η r AP with the arm. Then, the general equations ω0 ϕ for the absolute velocity and acceleration are A ζ x z r AP + v r , v = v A + ω ˙ rAP + ω (ω rAP ) + 2 ω vr + ar . a = aA + ω
× ×
× ×
×
We start counting time t from the beginning of the motion. then, by using the given accelerations, angular velocity and initial conditions, we obtain = a 0 ex = ω 0 eζ ,
aA ω ar
=
v0
˙ ω
−a e
= 0, vr
r ξ
= a0 t ex , =
−a t e r
ξ
r AP
=(
−
2 1 2 ar t
+ s0 )eξ ,
and 2 1 2 r 2 0
× r = ω(− a t + s )e , 2 ω × v ) = −ω (− a t + s )e . ω × (ω × r ω
AP
AP
0
1 2
η
r
2
Taking into account the relations ω0 t, we finally obtain
=
0
ξ
=
eξ cos ϕ
−2 ω
0
ar t eη ,
− e sin ϕ, where ϕ = η
1 2
2
− a t]e + [−a t sin ω t + ω (− a t + s )]e , a = [a cos ω t − ω (− a t + s ) − a ]e − [a sin ω t + 2 ω a t]e . v
= [a0 t cos ω0 t
ex
r
0
0
r
2 0
ξ
2 1 r 2
0
0
0
r
0
ξ
0
r
0
0
η
0
r
η
Chapter 9
Principles of Mechanics
9
212
D‘Alembert’s Principle
It is often advantageous to determine the equations of motion not by using Newton’s axioms (principles of linear and angular momentum) but by using equivalent laws which are called Principles of Mechanics .
Formal Reduction of Kinetics to Statics Rewriting Newton’s law of the motion for a point mass (or the center of mass of a rigid body) in the form
ma =
F
− F
ma = 0
and introducing D‘Alembert’s inertial force (pseudo force, fictitious force ) F I
=
−ma
leads to the ‘dynamic equilibrium condition of forces’
F
+ F I = 0
Accordingly, a point mass or the center of mass of a rigid body moves such that the sum of external forces F and the inertial force F I is equal to zero. In case of a plane motion of a rigid body, the pseudo moment M IA = ΘA ˙ω must be taken into account in the ’dynamic equilibrium condition of the moments’. Instead of the equations of motion according to page 102 we then obtain the ’equilibrium conditions’
−
− F x
max = 0 ,
F y
− ma
y
= 0 ,
where A= fixed point or center of mass. Remarks:
M A
−Θ
ω A ˙
=0
•
The inertial force (pseudo force) and the pseudo moment are directed opposite to the positive acceleration and angular acceleration, respectively.
•
When solving problems, the pseudo forces and pseudo moments must be drawn into the free-body diagram with the respective sign.
d‘ALEMBERT’s Principle A system moves such that for a virtual displacement the sum of the works δU of the external forces (moments) and δU I of the pseudo forces (pseudo moments) vanishes at all times: δU + δU I = 0
Lagrange Equations
of the 2nd Kind
213
• Virtual
displacements are infinitesimally small, fictitious and kinematically admissible.
Remarks:
•
The work done by constraint forces is zero for rigid constraints.
•
D‘Alembert’s principle may preferably be applied for systems with several constraints, if the constraint forces shall not be determined.
•
In case of statics, the principle reduces to δU = 0 (see volume 1, chapter 7).
LAGRANGE Equations of the 2nd kind The motion of a system with n degrees of freedom is described by d dt where
− ∂T ∂ q ˙j
∂T = Q j ∂q j
( j = 1, . . . , n)
T
=
kinetic energy,
q j
=
generalized coordinates,
q ˙j
=
generalized velocities,
Qj
=
generalized forces.
For conservative forces (having a potential) the equations of motion simplify to d dt
−
∂L =0 ∂q j
where
L = T
=
Lagrangian,
V
=
potential energy.
Remarks:
∂L ∂ q ˙j
− V
( j = 1, . . . , n)
•
The numbers of generalized coordinates (i.e. equations) and of degrees of freedom are equal.
•
Generalized coordinates are linearly independent and may be lengths or angles.
•
Generalized forces act in the direction of generalized coordinates and may be e.g. forces or moments.
214 P9.1
Dynamic
Problem 9.1
Two blocks of weights W 1 = m1 g and W 2 = m2 g are suspended by a rope drum (moment of inertia ΘA ). The block of mass m1 slides frictionless on an inclined plane. Determine the angular acceleration of the drum and the force in rope 2 by using dynamic equilibrium conditions.
ΘA
r2
g
A
r1
1
m1
2
m2 α
Solution
We separate the system, introduce coordinates x 1 ,x2 , ϕ, describing the motion of the system, and draw the free-body diagram. Since we solve the problem by applying dynamic equilibrium conditions, the inertia forces mi x ¨i and the pseudo moment must be considered; they point in negative coordinate directions. Then, the equilibrium conditions yield
: ② ↓: ①
③
A:
S 1
− m x¨ − m g sin α = 0 , m g−m x ¨ − S = 0 , −r S + r S − Θ ¨ϕ = 0 . 1 1
2
2
x1
S 1
m1 ¨ x1
m1 g
③
S 2
①
N α
2
2
ΘA ¨ ϕ A
1
2 2
1 1
ϕ
②
x2 m2 g m2 ¨ x2
A
Using the kinematic relations x1 = r 1 ϕ
x ¨1 = r 1 ¨ ϕ,
x2 = r 2 ϕ
x ¨2 = r 2 ¨ ϕ,
we obtain the angular acceleration ϕ = ¨
r2 m2 r1 m1 sin α g r12 m1 + r22 m2 + ΘA
−
and the force in the rope S 2 = m 2 (g + r2 ¨ ϕ) = m 2 g
r1 (r1 + r2 sin α)m1 + ΘA . 2 2 r1 m1 + r2 m2 + ΘA
Remark: For r2 m2 > r1 m1 sin α the drum rotates clockwise, for
r2 m2 < r1 m1 sin α it rotates counterclockwise. In the special case r2 m2 = r 1 m1 sin α, the system is in static equilibrium: ϕ¨ = 0.
Equilibrium Problem 9.2 The
displayed pendulum consists of a homogeneous cylinder of mass m1 = 2m and a point mass of weight G = m2 g = mg which is rigidly fixed by a massless rod. The cylinder rolls without slip on the rough plane.
215 P9.2
m1 r l
Formulate the equation of motion.
m2
Solution
We choose the reference system x, y in the undisplaced position. The center of mass C lies in the distance
rϕ
x
lm1 2 e= = l m1 + m2 3
l
y
m1 g
C H
ϕ
e (m1 + m2 )x¨c
N
from the point mass m2 . In the displaced position we draw all external forces, inertial forces and the pseudo moment into the free-body diagram. The ’equilibrium conditions’ then read
3
sin ϕ
cos ϕ
ΘC ¨ ϕ
(m1 + m2)y¨c l
3
m2 g
2 l sin ϕ 3
→ : −H − (m + m )¨xc = 0 , ↓ : (m + m )g − N − (m + m )¨yc = 0 , l l 2l l ¨ 0 C : m g sin ϕ + H cos ϕ − r − N sin ϕ − m g sin ϕ − Θc ϕ = 3 3 3 3 1
1
2
1
2
1
2
2
where
m1 r 2 l + m1 Θc = 2 3
2
+ m2
2l 3
2
2 = m r + l 2 . 3 2
For an angle change ϕ, the center of the cylinder is displaced by rϕ to the left. Thus, we find for the center of mass xc = yc =
−rϕ + 3l sin ϕ l cos ϕ 3
−rϕ¨ + 3l ¨ϕ cos ϕ − 3l ϕ˙ sin ϕ , l l y¨c = − ¨ ϕ sin ϕ − ϕ˙ cos ϕ . 3 3 2
x ¨c =
2
Solving the equations yields ϕ(l ¨ 2 + 4r2
2
− 2lr cos ϕ) + lrϕ˙
sin ϕ + gl sin ϕ = 0 .
216 P9.3
d‘Alembert’s
Problem 9.3
A wheel (weight m1 g, moment of inertia ΘA ) and a block (weight m2 g), both on inclined planes, are connected by a rope. The wheel rolls without slip while the block slips frictionless.
g
m1 , ΘA r
m2
Determine the acceleration of the block applying d’Alembert’s principle. Neglect the masses of the rope and the pulley.
A
α α
Solution
Since the constraint forces (force in the rope, contact forces) need not to bee determined, it is advantageous to apply d’Alembert’s principle. To describe the motion we choose the coordinates xi , ϕ. In addition to the real forces, the inertial forces mi x ¨i and the x1 ϕ pseudo moment ΘA ϕ ¨ (acting x2 opposite to the chosen coordinaΘA ¨ ϕ tes) are drawn into the sketch of A x2 the system. Then, d’Alembert’s m2 ¨ m1 ¨ x1 α principle requires that the virm2 g α m1 g tual work of all forces vanishes: δU + δU I = 0 This leads to
−m x¨ δx − m g sin αδx − Θ 1 1
1
1
1
¨ A ϕδϕ
With the kinematic relations
+ m2 g sin αδx 2
− m x¨ δx 2 2
δx 1 = δx2 = rδϕ = δx
x1 = x 2 = rϕ = x
x ¨1 = x ¨2 = r ϕ = ¨ x ¨
we obtain
−
m1 x ¨
− m g sin α − Θr
A 2
1
x ¨ + m2 g sin α
− m x¨ 2
δx = 0 .
Since δx = 0, the expression in the bracket must vanish. Thus,
x ¨ = x ¨2 = g
(m2
− m )sin α . 1
m1 + m2 +
ΘA r2
2
= 0 .
Principle
217
Problem 9.4
Two drums are connected by a rope and carry blocks of weights m1 g and m2 g. Drum ② is driven by the moment M 0 .
P9.4
ΘA ΘB r1
Determine the acceleration of block ① using d’Alembert’s principle. Neglect the mass of the ropes.
①
ϕ1
We introduce the inertial forces mi x ¨i and the pseudo moments ΘA ϕ ¨1 , Θ B ¨ ϕ2 . They act in opposite directions to the chosen positive coordinate directions. Then, d’Alembert’s principle requires
ΘA ϕ ¨1
x1
−m (g + x¨ )δx + m (g − x¨ )δx +M δϕ − Θ ϕ ¨ δϕ − Θ ¨ ϕ δϕ 0
2
A
1
2
2
1
B
2
B
M 0
ϕ2
which leads to 1
ΘB ϕ¨2
A
δU + δU I = 0 ,
1
②
m2
m1
Solution
1
B
M 0
A r2
r1
m1 g
m2 g
m1 ¨ x1
m2 ¨ x2
x2
2
2
= 0.
With the kinematic relations x1 = r 1 ϕ1 x2 = r 1 ϕ2 r2 ϕ1 = r 1 ϕ2 we obtain
−
x ¨1 , r1 δx 1 δϕ 1 = , r1 ϕ ¨1 =
r2 x ¨1 , r12 r2 δϕ 2 = 2 δx 1 , r1 ϕ ¨2 =
−
r2 r2 r2 M 0 x ¨1 + r1 r1 r12
m1 (g + x ¨1 ) + m2 g
− Θr
A ¨1 2 x 1
−
r2 x ¨1 , r1 r2 δx 2 = δx 1 r1 x ¨2 =
r22 ΘB r14
x ¨1 δx 1 = 0.
Since δx 2 = 0, the expression in the curly bracket must vanish. Thus, the acceleration of block ① is
1 x ¨1 = g
m r r M −m − r r m g 2 2
2
0
1 1
12
1
m2 r2 1+ m1 r1
2
ΘA r22 ΘB + + m1 r12 m1 r14
.
218 P9.5
d‘Alembert’s Principle
Problem 9.5
The displayed system shows a lever (weight m1 g, moment of inertia ΘD ), pivoted at D. Attached at the lever ends are a turnable wheel ②, rolling along the rigid half-circular cylinder ①, and a counterweight W 3 = m 3 g. Determine the eigenfrequency small displacements ϕ.
②
g
r1 ϕ
m2 , Θ2 r2
①
D m1 , ΘD l3 l
m3
for
Θ2 ¨ ϕ2 We will derive the equation m2 (r1 +r2 )ϕ¨ of motion by applying two different approaches. First we use d‘Alembert’s ΘD ¨ ϕ ϕ2 principle. For this purpose, in additiϕ m2 g on to the external forces, all inertial D forces and pseudo moments are drawn m1 g into the free-body diagram in the dis- m l ¨ 3 3ϕ placed position. The virtual work of all forces and moments must vanish (nom3 g tice, sin ϕ ϕ): Solution
≈
−Θ
ϕδϕ D ¨
2
2
2
−m gl ϕδϕ + m g 3
2
2 3 3
¨ − m l ¨ ϕδϕ − Θ ¨ϕ δϕ − m (r + r ) ϕδϕ
3
1
2
1
2
− l 2
l3 ϕδϕ + m2 g(r1 + r2 )ϕδϕ = 0 .
Because the wheel rolls, the kinematic relation reads
ϕ ¨2 = (1 + r1 /r2 ) ϕ ¨,
r2 ϕ2 = (r1 + r2 ) ϕ
δϕ 2 = (1 + r1 /r2 ) δϕ .
Introducing these quantities and considering δϕ = 0, we obtain the equation of motion
− − − − − −
ΘD + Θ 2
r1 1+ r2
+ m3 l3
m1
2
+ m2 (r1 + r2 )2 + m3 l32 ϕ ¨
l 2
l3
m2 (r1 + r2 ) gϕ = 0 ,
from which the eigenfrequency is directly found as m3 l 3
ω=
ΘD + Θ 2
m1
l 2
r1 1+ r2
l3
2
m2 (r1 + r2 ) 2
+ m2 (r1 + r2 ) +
m3 l32
g.
Lagrange
Equations of the 2nd Kind
219
In the second approach we apply the Lagrange equations of the 2nd kind. The system is conservative and, by choosing the zero potential at the level of D, we obtain V = m1 g
− l 2
l3 cos ϕ + m2 g(r1 + r2 )cos ϕ
− m g l cos ϕ , 3
3
1 1 1 1 T = ΘD ϕ˙ 2 + m2 (r1 + r2 )2 ϕ˙ 2 + Θ2 ϕ˙ 22 + l32 m3 ϕ˙ 2 , 2 2 2 2 L = T
− V
.
With the kinematic relations r2 ϕ2 = (r1 + r2 )ϕ
r2 ϕ˙ 2 = (r1 + r2 )ϕ˙
follow
− −
∂L r1 = ΘD + m2 (r1 + r2 )2 + Θ 2 1 + ∂ ϕ˙ r2
d ∂L dt ∂ ϕ˙
2
= ΘD + m2 (r1 + r2 ) + Θ 2
∂L l = m1 ∂ϕ 2
r1 1+ r2
l3 + m2 (r1 + r2 )
−m l
+
2
3 3
Thus, from d ∂L dt ∂ ϕ˙
2
m3 l32
ϕ˙ ,
+ m3 l32 ϕ ¨,
g sin ϕ .
∂L =0 ∂ϕ
with sin ϕ ϕ (small displacements) we obtain the already known result for the equation of motion
≈
− − −
ΘD + Θ 2
r1 1+ r2
+ m3 l 3
m1
2
l 2
2
+ m2 (r1 + r2 ) + l3
m3 l32
ϕ ¨
m2 (r1 + r2 ) gϕ = 0 ,
and accordingly for the eigenfrequency. It can be seen that vibrations are only possible if the nominator in the square root is positive, i.e. if m3 l 3 > m 1
− l 2
l3 + m2 (r1 + r2 ) .
Remark: The system has only one degree of freedom; its position can
uniquely be described by the generalized coordinate ϕ.
220 P9.6
Equation of Motion
Problem 9.6
A pivoted homogeneous bar (mass m) is held by springs and carries an attached point mass m/4. Determine the equation of motion for small displacements from the equilibrium position by using: a) the energy conservation law, b) d’Alembert’s principle, c) the angular momentum theorem.
k
m/4
m
k
A 2l
l
Solution
The motion of the system is a rotation about point A, which appropriately is described by the rotation angle ϕ. Thereby, assuming small displacements, each spring experiences a length change 2lϕ. ϕ a) The total energy 1 1 E = V + T = 2 k(2lϕ)2 + ΘA ϕ˙ 2 2 2
2lϕ
A
ΘA
of the conservative system must be constant at all times. This leads to dE =0 dt
2k(2lϕ)2lϕ + ˙ ΘA ϕ˙ ϕ = ¨ 0
b) When applying d’Alembert’s principle δU + δU I = 0, the work of the pseudo moment ΘA ¨ ϕ (opposite to the positive direction of motion) must be taken into account:
−Θ
¨ A ϕδϕ
− 2l 2 k(2lϕ)δϕ = 0
ΘA ϕ ¨ + 8kl 2 ϕ = 0 .
2k(2lϕ)
ΘA ¨ ϕ
ϕ 2lϕ
A
ΘA ϕ ¨ + 8kl 2 ϕ = 0 .
c) Application of the angular momentum theorem with respect to the fixed point A directly yields 2k(2lϕ)
A :
ΘA ϕ = ¨
−2l 2 k(2lϕ)
ΘA ϕ ¨ + 8kl 2 ϕ = 0 .
ϕ 2lϕ
ΘA
The results are (as expected) in all cases the same. With
1 2 1 1 1 5 ΘA = m (2l)2 + m l2 + ml 2 = ml2 3 3 3 3 4 4 the equation of motion can also be written as ϕ ¨+
32 k ϕ = 0 . 5m
A
Lagrange
Equa Eq uati tio on of the 2nd Kind
Problem 9.7
A symmetri symmetricc disk with a hal halffcircular boundary rolls without slip on the flat surface.
221 P9.7
m, ΘC R
g
Deriv Der ivee th thee eq equa uati tion on of mo moti tio on us usin ing g th thee Lagrange formalism. Given: R, c = = κR κR,, m, ΘC = α mR2
c
C
Solution
The system is conservativ conservativee and has one degree of freedom. To describe the motion we choose as generalized coordinate the angle ϕ. Then, with the kinematic relations
Rϕ xc c ϕ
− − − −
−
yc
C
xc = Rϕ + c sin ϕ = R(ϕ κ sin ϕ) x˙ c = Rϕ ˙ + cϕ ˙ cos ϕ = Rϕ ˙ (1 κ cos ϕ) , yc = c cos ϕ = = κR κR cos ϕ , y˙ c = cϕ ˙ sin ϕ = κR κR ˙ ϕ ˙ sin ϕ .
− −
x
y
−
we can formulate the energies, the Lagrangean and the required derivatives: V =
−mgy
c
=
−mgκR cos ϕ ,
1 1 1 T = m(x˙ 2c + y˙ c2 ) + Θc ϕ˙ 2 = mR2 ϕ˙ 2 (1 κ cos ϕ)2 2 2 2 1 +(κ +( κ sin ϕ)2 + α = mR2 ϕ˙ 2 1 2κ cos ϕ + κ2 + α , 2 1 L = = T T V = mR Rϕ ˙ 2 1 2κ cos ϕ + κ2 + α + 2gκ 2gκ cos ϕ , 2 ∂L 1 = mR 2Rϕ ˙ 1 2κ cos ϕ + κ2 + α , ∂ ϕ˙ 2 d ∂L 1 = mR 2Rϕ ¨ 1 2κ cos ϕ + κ2 + α + 4κR 4κR ˙ ϕ ˙ 2 sin ϕ , dt ∂ ϕ˙ 2 ∂L = mR κR κR ˙ ϕ ˙ 2 sin ϕ κg sin ϕ . ∂ϕ
− −
− −
− −
−
Substituting these expressions into
−
d ∂L dt ∂ ϕ˙
−
∂L =0 ∂ϕ
yields the equation of motion ϕ ¨(1
2
− 2κ cos ϕ + κ
+ α) + κϕ ˙ 2 sin ϕ + κ
g sin ϕ = 0 . R
222 P9.8
Lagrange
Problem 9.8
A homogeneous bar (mass m, length l) is attached to a thread of length l/ l/2, 2, which is pinned at point A. The mass of the thread is negligible.
A
g
l/22 l/ l
Derive Deri ve the equ equati ations ons of mo moti tion on using the Lagrange formalism.
m
Solution
The system system is conserv conservati ative ve and has two degrees of freedom. As generalized coordinates we choose ϕ1 and ϕ2 and assume the zero level of the potential at the height of A. With
x
A ϕ1 ϕ2 C
xc = (l/ l/2)(sin 2)(sin ϕ1 + sin ϕ2 ) , yc = (l/ l/2)(cos 2)(cos ϕ1 + cos ϕ2 ) , x˙ c = (l/ l/2)( 2)(ϕ˙ 1 cos ϕ1 + ϕ˙ 2 cos ϕ2 ),
y y˙ c = (l/ l/2)( 2)(ϕ˙ 1 sin ϕ1
− ϕ˙ sin ϕ ) 2
2
and ΘC = m mll2 /12 follow the required energies l V = V = mg (cos ϕ1 + cos ϕ2 ) , 2 1 1 T = T = m(x˙ 2c + y˙ c2 ) + ΘC ϕ˙ 22 2 2 1 2 2 1 = ml [ϕ˙ 1 + ϕ˙ 22 + 2ϕ˙ 1 ϕ˙ 2 cos( cos(ϕ ϕ1 + ϕ2 )] + ml2 ϕ˙ 22 8 24
−
for the Lagrangean L = = T T
− V V .. Introducing the derivatives −
∂L ml2 1 = ϕ˙ 1 ϕ˙ 2 sin( sin(ϕ ϕ1 + ϕ2 ) mgl sin ϕ1 , ∂ϕ 1 4 2 ∂L ml2 1 = ϕ˙ 1 ϕ˙ 2 sin( sin(ϕ ϕ1 + ϕ2 ) mgl sin ϕ2 , ∂ϕ 2 4 2 d ∂L ml2 [ϕ ¨1 + ϕ ¨2 cos( cos(ϕ ϕ1 + ϕ2 ) ϕ˙ 2 (ϕ˙ 1 + ϕ˙ 2 ) sin in((ϕ1 + ϕ2 )] , = dt ∂ ϕ˙ 1 4 d ∂L ml2 = [4ϕ ¨2 + 3ϕ ¨1 cos( cos(ϕ ϕ1 + ϕ2 ) 3ϕ˙ 1 (ϕ˙ 1 + ϕ˙ 2 ) sin in((ϕ1 + ϕ2 )] dt ∂ ϕ˙ 2 12
−
−
−
−
−
−
−
into the Lagrange equations d ∂L ∂L d ∂L = 0 , dt ∂ ϕ˙ 1 ∂ ϕ1 dt ∂ ϕ˙ 2
−
∂L =0 ∂ ϕ2
yields the coupled equations of motion 2 2
ϕ ¨1 + ϕ ¨2 cos( cos(ϕ ϕ1 + ϕ2 )
sin(ϕ ϕ + ϕ ) + 2(g/l 2(g/l)sin )sin ϕ = 0 , − ϕ˙ sin( 4ϕ ¨ + 3ϕ ¨ cos( cos(ϕ ϕ + ϕ ) − 3ϕ˙ sin( sin(ϕ ϕ + ϕ ) + 6(g/l 6(g/l)sin )sin ϕ = 0 . 2
1
1
2
1
2 1
2
1
1
2
2
Formalism Problem 9.9 A
simple pendulum is attached to point 0 of a disk which rotates with constant angular velocity Ω about the vertical axis.
Ω
P9.9
r l
0
Derive the equation of motion using the Lag Lagran range ge for formal malism ism.. Dis Disreg regard ard the weight of the point mass.
Solution
The system is cons conserv ervati ative ve.. Since the weight is neglected (V (V = 0), only the kinetic T T energy is needed in the Lagrangean. To describe the motion, we introduce the angle ψ = Ωt, prescribed by the rotating disk, and the angle ϕ relative to the disk. With
y
m
m l ϕ r
ψ
x = = r r cos ψ + l cos( cos(ψ ψ + ϕ) , y = r sin ψ + l sin( sin(ψ ψ + ϕ) , x˙ = r Ωsin ψ l(Ω + ϕ˙ ) sin in((ψ + ϕ) , y˙ = r rΩcos Ωcos ψ + l(Ω + ϕ˙ )cos( )cos(ψ ψ + ϕ) , x˙ 2 + y˙ 2 = r 2 Ω2 + l2 (Ω + ϕ˙ )2 + 2rl 2rlΩ(Ω Ω(Ω + ϕ˙ )cos ϕ
−
223
0 x
−
follows
1 1 L = = T T = m(x˙ 2 + y˙ 2) = m r2 Ω2 + l2 (Ω+ ϕ˙ )2 + 2rl rlΩ(Ω+ Ω(Ω+ ϕ˙ )cos ϕ . 2 2 Introducing the derivatives
∂L m = 2l2 Ω + 2l 2 l2 ϕ˙ + 2rl 2rlΩcos Ωcos ϕ , ∂ ϕ˙ 2 d ∂L m = 2l2 ϕ ¨ 2rl rlΩ Ωϕ˙ sin ϕ , dt ∂ ϕ˙ 2
−
−
into the Lagrange equation d ∂L ∂L =0 dt ∂ ϕ˙ ∂ϕ
∂L = ∂ϕ
Ω(Ω + ϕ˙ )sin ϕ − m2 2rlrlΩ(Ω
leads to the equation of motion r ϕ ¨ + Ω2 sin ϕ = 0 . l Remark : For small angles (sin ϕ
vibrations with ω = Ω
≈ ϕ) this equation describes harmonic
r/l (see also Problem 8.9).
224 P9.10
Lagrange
Problem 9.10
A homogeneous bar (length 2a, weight W = mg mg)) is suspended to the support A, wh whic ich h ca can n mo move ve hor horiz izon onta tall lly y and is held by a spring (spring constant k).
k A 2a
g
Find th Find thee eq equa uati tion onss of mo moti tion on us usin ing g th thee Langrange formalism.
m
Solution
The system system is conserv conservati ative ve and has two degrees of freedom. As generalized coordinates we choose the displacement w and angle ϕ, both measured from the equilibrium position. With
x a w
xC = w w + + a sin ϕ , yC = a cos ϕ , x˙ C = w˙ + aϕ ˙ cos ϕ , y˙ C = aϕ ˙ sin ϕ
ϕ
y
−
and ΘC = m ma a2/3 follow the energies V = mga cos ϕ + kw 2 /2 , 2 T = (m/ m/2) 2) x2C + yC + (Θ C /2)ϕ˙ 2
−
2
2
2
= (m/ m/2) 2) w˙ + a ϕ˙ + 2a 2aw˙ ϕ˙ cos ϕ + (ma (ma2 /6)ϕ˙ 2 . Herewith, the derivatives of the Lagrangean L = = T T
− − − − −
d dt d dt ∂L ∂ϕ
∂L = m w ¨ + aϕ ¨ cos ϕ aϕ ˙ 2 sin ϕ , ∂ w˙ ∂L 4 2 = m a ϕ ¨ + aw ¨ cos ϕ aw˙ ϕ ˙ sin ϕ , ∂ ϕ˙ 3
=
− V are − ∂L = −k w , ∂w
m aw˙ ϕ ˙ + ag sin ϕ ,
and the Lagrange equations d ∂L dt ∂ w˙
∂L = 0 , ∂w
d ∂L dt ∂ ϕ˙
yield the coupled equations of motion w ¨ + a ϕ¨ cos ϕ
2
− a ϕ˙
sin ϕ +
k w = 0 , m
4 a ϕ ¨+w ¨ cos ϕ + g sin ϕ = 0 . 3
∂L =0 ∂ϕ
C a
Formalism
225
Problem 9.11
The double pendulum consists of 4 bars of equal length l and mass m. The spring is unstrained in the vertical position of the hanging bars.
k
a) Determine the equations of motion by using the Lagrange equations. b) Linearize the equations for small amplitudes.
m, l
Solution
x
a) The position of the conservative system system is uniquely described by the angles α and β . Accordingly, it has two degrees of freedom. To fo form rmul ulate ate the pote potent ntia iall and kin kineti eticc energies, we introduce the x, y -coordinate system and determine first the coordinates and velocities of points ①, ② as well as the length change ∆ of the spring: 1 2
1 2
①
β
l 2
②
cos β
l 2
y l sin α
x˙ 1 = l α˙ cos α ,
˙ cos β ) , x˙ 2 = l l((α˙ cos α + 21 β cos β
y1 = l cos α ,
y2 = l l(cos (cos α + 21 cos β ) , ˙ sin β ) , y˙ 2 = l(α˙ sin α + 1 β β sin
−
l cos α
α
x2 = l l(sin (sin α +
−lα˙ sin α ,
g
k
x1 = l sin α ,
y˙ 1 =
+
P9.11
l 2
sin β
sin β ) ,
2
v12 = x˙ 21 + y˙ 12 = l 2 α˙ 2 (cos2 α + sin2 α) = l 2 α˙ 2 , ˙ cos α cos β + ˙ 2 cos2 β v22 = x˙ 22 + y˙ 22 = l 2 (α˙ 2 cos2 α + α˙ β β cos + 41 β ˙ sin α sin β + ˙ 2 sin2 β ) + α˙ 2 sin2 α + α˙ β β sin + 1 β 4
˙ cos(α = l 2 [α˙ 2 + α˙ β β cos( α ∆= 2
(l sin α +
∆ = 2l
2
l)2
+
− β ) + l2
cos2
1 ˙2 4 β ] ,
√
√ √ α − l 2 = l 2 1 + sin α − 1
√ 2 + sin α − 2 1 + sin α
.
,
Herewith, the potential energy is obtained as
− y2 − mgy − mgy + 21 k∆ √ 1 =−mgl 3cos α + cos β + kl 2 + sin α − 2 1 + sin α 2
V = V = 2mg
1
1
2
2
2
.
When determining the kinetic energy, we consider that the motion of the upper bars is a pure pure rotat rotation ion (angular (angular velocit velocity y α˙ ) about the pins, the motion of the horizontal bar is a pure translation (velocity v1 ) and the
226
Lagrange
motion of the lower bar is a combination of translation (velocity v (velocity v 2 ) plus ˙ ). rotation (angular velocity β ). With the moment of inertia Θ1 = ml 2 /3 of an upper bar with respect to the pin and the moment of inertia Θ2 = ml2 /12 of the lower bar with respect to its center of mass ②, it follows
1 1 1 1 ˙ 2 T = 2 Θ1 ˙α2 + mv12 + mv22 + Θ2 β 2 2 2 2 4 1 ˙ 2 1 ˙ = ml 2 α˙ 2 + β + α˙ β β cos( cos(α α β ) . 3 6 2
−
With L = = T T
and the derivatives − V V and −
∂L ∂α ∂L ∂β d dt d dt
ml2 ˙ = α˙ β β sin( sin(α α β ) 3mgl sin α 2 ml2 ˙ 1 = α˙ β β sin( sin(α α β ) mgl sin β , 2 2 ∂L 8 1¨ = ml 2 ¨ α + β β cos( cos(α α β ) ∂ ˙ α˙ 3 2 ∂L 1¨ 1 = ml 2 β + ¨ + α cos( cos(α α β ) ˙ 6 2 ∂ β
−
− −
− kl
2
cos α 1
− −
−
− √ − −
1 , 1 + sin α
) sin(α − − 21 β ˙ (α˙ − β ˙ )sin(α
β ) ,
n(α α − − 21 α˙ (α˙ − β ˙ ) sisin(
β ) ,
we obtain from the Lagrange equations d ∂L dt ∂ ˙α˙
−
∂L = 0 , ∂α
d ∂L ˙ dt ∂ β
∂L =0 ∂ β
the equations of motion: 8 1¨ ¨ α + β β cos( cos(α α 3 2
1 ˙2 β ) + β sin(α sin( α 2
−
1¨ 1 β + + ¨ α cos( cos(α α 6 2
− β ) − 21 α˙
2
sin(α sin( α
b) For small amplitudes α amplitudes α linearizatio linear izations ns apply: sin α
− β ) + 21 gl sin β = 0.
1 1 + sin α
≈
≈ √
− ≈ − ≈ − − ≈
− ≈
Herewith, the equations of motion simplify to 3 ¨ α ¨+ β + + 16
= 0,
β 1,1, β 1 and α˙ 1, β ˙ 1 the following
cos α 1 , sin(α sin(α β ) (α β ) , cos(α cos(α β ) 1 , 1 1 ˙ 2 sin(α 1 α , β sin(α β ) 0 , α˙ 2 sin( sin(α α β ) 1 + α 1 + sin α
√
≈ α,
−
− √
g k β ) + 3 sin α + cos α 1 l m
9g 3k + α = 0 , 8l 8m
g ¨ + β + 3 α ¨ + 3 β = 0 . l
− ≈ 0.
Chapter 10
Hydrodynamics
10
228 Hydrodynamics
The velocity field v(x(t), t) describes the motion of a fluid. The vector x assignes to each location in the fluid a velocity v at time t. The velocity field is stationary for ∂ v/∂ /∂tt = 0, otherwise instationary .
Pathline: trajectory that a material point of the fluid (fluid element) follows over a time period. The pathline x(t) yields from the solution of the differential equation
(t1 )
x
z
d x(t) = v (x(t), t) . dt
(t2 )
x
y
Streamlines: fami family ly of cur curve vess wh whos osee tan tan-gents coincide in each point x with the direction of the local velocity vector. The streamline field follows from the differential equation
x
z
d x(s) = v (x(s), t) , ds
x
y
where s is the arclength of the streamline. Notice: pathlines and streamlines are identical for a stationary velocity velocity
field.
Stream Filament Theory: In what follows we restrict ourselves to the stationary motion motion of an incompressible fluid fluid in a streamtube, where the flow behavior is characterized by the behavior at a median streamline. This one-dimensional theory is described by the following basic equations: a) Continuity equation A1 v1 = A 2 v2
or
A1
Q = = Av Av = = const
where Q = = A A v is the volume flow.
v1
A2 v2 outflow
inflow
b) BERNOULLI’s theorem For an inviscid fluid fluid holds 1 v 2 + g z + p + p = const 2
or
v2 p + z + + = H H = = const . 2g g
229
Hydrodynamics
where v 2 /2 = dynamic pressure (specific kinetic energy), g z = geodetic pressure (specific potential energy), p = static pressure (pressure energy), H = hydraulic (total) head,
p0
2
v /2g = velocity head,
z 1
v1
z = elevation head,
h
p/ g = pressure head.
v2
z 2
Example: outflow from a reservoir: 1 1 v12 + g z 1 + p0 = v22 + g z 2 + p0 . 2 2 In the special case v1 = 0 (h =const) follows TORRICELLI’s law (outflow from big fluid tanks). v2 =
2gh .
For viscous fluids (flow with energy losses) the generalized BERNOULLI’s theorem 1 2 1 v1 + g z 1 + p1 = v22 + g z 2 + p2 + ∆ pv 2 2 is valid, where 1 ∆ pv = ζ v12 = pressure loss, 2
c) Balance of Momentum F
= Q (v2
−v ) 1
or
F x = m ˙ (v2x
1x ) ,
F y = m ˙ (v2y
1y ) ,
F z = m ˙ (v2z
1z ) ,
where F
ζ = pressure loss coefficient.
−v −v − v
= resulting force exerted on the closed fluid volume within the streamtube (control volume),
Q = m = ˙ mass flow, Q v1 = inflowing momentum, Q v2 = outflowing momentum.
230 P10.1
Velocity Field
Problem 10.1 v (x, t)
A flow is described by the plane velocity field
= 2ax ex
− 2ay e
y
Determine the equation for the streamlines and sketch the profile for the specific streamline through the point A with coordinates x = 0.5 m; y = 4 m. Solution
The differential equation for the streamlines reads in com-
ponents dx = v x = 2ax , ds
dy = v y = ds
−2ay .
Dividing the 1st by the 2nd equation yields dx vx = = dy vy
− xy
and by separation of variables it follows dx dy + = 0 . x y Integration leads to ln x + ln y = ln xy = C =: ln c
ln xy = ln c .
Accordingly, the streamlines are given by the hyperbola y=
c . x
y [m] 5
For a streamline through point A, the integration constant c is calculated with the given data as 2
4
A
3 2
c = 0.5 m 4 m = 2 m .
·
1
Having c, the profile of the streamline can be sketched. 1
2
3
4
x [m]
Remarks:
•
For x , the y-component of the velocity vector vanishes: 2ax ex . v
•
Because the flow is stationary (∂ v /∂t = 0), the streamlines and the pathlines coincide.
→
→∞
Pathlines and Streamlines
231
Problem 10.2
A planar flow is described by the velocity field −t v (x, t) = ax e + b e e , x
y
where a and b are given constants. a) Determine the pathline of the particle, which at time t = 0 is at point P = (1, 1). b) Determine the streamline, which at time t = 0 passes the point P = (1, 1).
Solution Because
the flow is instationary, the pathlines and streamlines
do not coincide! a) The components of the pathline are determined from dx = ax dt dy = b e−t dt
dx = x
y=
a dt t
−b e−
ln
x = at C 1
x = C 1 eat ,
+ C 2 .
With the initial conditions x (t = 0) = 1, y (t =0) = 1, we obtain x(t) = eat ,
y(t) = b (1
t
− e− ) + 1 .
b) For t = 0, the differential equations of the streamlines are given by dx = ax , ds
dy = b ds
dx a = x. dy b
Separation of variables and integration lead to b a
dx = x
dy
y=
b ln x + C 3 . a
The boundary condition yields y(x = 1) = 1
C 3 = 1 ,
and thus, it follows y(x) =
b ln x + 1 . a
Remark: From the parameter representation of the pathline in a), by
1 ln x), we can obtain the representation y(x) = eliminating t (t = a b (1 x−1/a ).
−
P10.2
232 P10.3
Velocity field
Problem 10.3
The velocity field of a planar, instationary flow is des-
cribed by v (x,y,t)
= a xy ex + b t ey
where a, b are given constants. As initial conditions, x = x 0 , y = y0 at t = 0 are prescribed. a) Determine the pathlines and streamlines. b) Where has a fluid particle been at time t = 0, which was detected at time t1 = 1 s at point (x1 , y1 ) = (1, 0)m?
Solution
a) The pathlines are determined from
dx = a xy , dt
dy = b t . dt
With y(t = 0) = y 0 the second equation yields y(t) = y 0 +
1 2 bt . 2
Introducing the result into the first equation, after separation of variables, integration and using x(t = 0) = x 0 we obtain
dx 1 = a y0 + b t2 dt x 2
x(t) = x 0 ea (y
0
t+bt3 /6)
.
The streamlines are calculated from dx = a xy , ds
dy = b t . ds
The second equation in conjunction with y(s = 0) = y 0 yields y(s) = b ts + y0 . Again, introducing the result into the first equation, applying separation of variables and using x(s = 0) = x 0 leads to dx = a (b ts + y0 ) ds x
x(s) = x 0 ea (y
0
s+bt2 /2)
.
b) Introducing the conditions x = x1 = 1 m, y = y1 = 0 for t = 1 s into the pathline yields y0 =
−b/2 ,
x0 = eab/3 .
Bernoulli’s
Problem 10.4
From a big reservoir water is taken via a pipe. To increase the mass flow Q, a diffuser is attached at the end of the pipe. Because cavitation danger is to be avoided, the pressure must not drop below pmin at any location of the pipe. a) Determine the maximum allowable diffuser cross-section ADmax .
p0
Theorem
233 P10.4
①
A
h
②
AD
b) Calculate the temporal mass flow Q for this case. c) Determine the height h∗ , the highest point of the pipe can be lifted, if the diffuser is not present.
a) Considering v0 = 0 (big reservoir) and applying Bernoulli’s theorem for the points and ① as well as for and ② of a streamline, the corresponding velocities can be determined: 2 2 2 1 1 v + p + gh = v + p + gh v = ( p0 p1 ) , 0 1 1 0 1 2 2 Solution
2 1 v 0 2
+ p0 + gh = 21 v22 + p0
v2 =
−
2gh .
Herewith follows from the continuity equation the cross section of the diffuser: AD v2 = A v1
AD = A
v1 = A v2
−
p0 p1 . gh
It is a maximum, when we insert for p 1 the minimum allowable pressure pmin : p0 pmin ADmax = A . gh
−
b) The temporal mass flow Q in this case is gicen by Q = v 2 AD =
2gh ADmax .
c) In the same way as in a) we obtain for a streamline between and ① as well as between and ② after lifting point ① to the height h∗ p0 + gh =
2 1 v 1 2
+ pmin + gh ∗ ,
v2 =
2gh .
With AD = A, the continuity equation yields v1 = v2 . By introducing this and solving for h∗ we obtain p0 h∗ =
− p
g
min
.
234 P10.5
Bernoulli’s
Theorem ②
Problem 10.5
The lower trapezoidal tank (constant depth f ) is filled via a pipe (cross section Ap ) from a big reservoir located above.
A p
p0
a
①
a) Determine the maximum height a = amax , the pipe may protrude the fluid level of the reservoir, such that the pressure in the pipe does not drop below pD . b) When is the height h(t) = H/2 reached in the lower tank? c) When h(t) = H/2 is reached, the valve of the lower tank is opened. Determine the cross section AV of the valve, such that the fluid level keeps constant.
H
③
p0 A(h) ④
α
AV
h(t) ⑤
e Solution
a) According to Torricelli’s law (big reservoir), the velocity at point ③ is given by v3 =
2gH .
Thus, from the continuity equation follows at point ② Ap v2 = A p v3
v2 = v 3 =
2gH .
Using Bernoulli’s theorem for a streamline between the points and ② p0 + 0 + 0 = p D + g amax +
①
1 v22 , 2
and introducing v2 , the maximum height amax can be determined: amax
1 = p0 g
− p − D
1 p0 pD (2gH ) = 2 g
−
−H .
b) The relation between the filling height h and time t follows from the continuity equation between point ③ and point ④ AR v3 = A(h) v4
Bernoulli’s
Theorem
235
in conjunction with the rise velocity of the fluid v4 =
dh dt
and the cross section of the tank A(h) = (e + 2h tan α)f as dh Ap = v 3 . dt A(h) Separation of variables and integration leads to
t
Ap
h
2gH dt =
t0 =0
(e + 2h tan α)f dh .
h0 =0
This yields Ap
2gH t = (e h + h2 tan α)f .
From this result the required time tr , to reach the filling height H/2, is found by introducing h = H/2: tr =
e
2
H H + tan α 2 4
AR
√ f 2gH .
c) From Bernoulli’s theorem, applied between the points p0 + g
④
and
⑤,
H 1 1 + v42 = p 0 + 0 + v52 , 2 2 2
the velocity at the valve can be calculated. It leads with the requirement of a constant fluid level, i.e. v4 = 0, to v5 =
gH .
Finally, using the continuity equation AV v5 = A p v3 , the cross section of the valve is obtained as AV =
√
2 Ap .
236 P10.6
Bernoulli’s
Theorem
Problem 10.6
From a big reservoir, an ideal fluid (density ) flows out through a pipe (cross section A2 ) with a local smooth contraction (cross section A1 ). a) Determine the pressure p1 in the cross section A1 . b) The pipe will now be spot drilled at A1 . Calculate h2 , such that no fluid leaks from the drilled hole. c) Now a vertical standpipe is connected to the drilled hole, which dips into a lower fluid tank. Find the necessary cross section ratio A2 /A1 , such that fluid is sucked from the tank.
h1
po
h2
A1
A2
h1
po
h3
h2
A1
A2
po
a) First, from Torricelli’s law follows the outflow velocity at the point ② Solution
v2 =
2gh 2
and from that, using the continuity equation, the velocity in the cross section A1 :
h1
po
h2
A1
A1 v1 = A2 v2
A2 A2 v1 = v2 = A1 A1
①
2gh 2 .
A2
②
Introducing v1 into Bernoulli’s theorem for a streamline between the points and ① yields the pressure p1 : 1 2 v1 + 0 + p1 = 0 + gh 1 + p0 2
−
p1 = p 0 + g h1
A2 A1
2
h2 .
Bernoulli’s
Theorem
237
b) No fluid leaks from the drilled hole, if the pressure p1 is below the ambient pressure p0 : p1 < p0
− − A2 A1
p0 + g h1
h1
A2 A1
2
2
h2 < p 0
h2 < 0 .
This leads to the condition h2 >
A1 A2
2
h1 .
c) With the pressures p 4 = p 1 and p 3 = p 0 at the locations ④ and ③, we obtain from Bernoulli’s theorem for a streamline between the points ④ and ③ 1 2 v4 + gh3 + p1 = 0 + 0 + p0 . 2 Introducing p 1 and using the condition v42 > 0 leads to p0
− p − gh A h − A 1
3
> 0
2
2
1
1
h1
A1
h2 + h3 < 0 h3
and finally to A2 > A1
po
h2
④
po
A2 ③
h1 + h3 . h2
Remark: At the location
but not the velocity v1 !
④ of
the standpipe prevails the pressure p1
238 P10.7
Bernoulli’s
Theorem
Problem 10.7
A tank is filled by a pump through an opening. At the same time, fluid flows out through a leak at the bottom. a) Which stationary fluid level H will be reached? b) Determine for this case the loss of volume flow through the leak. c) Now the pump is shut down and the inflow is closed. How long does it take until the tank is empty?
p0 A0 H
①
A2
②
pump
A1 , p1
Solution
a) With the Bernoulli theorem applied between the points and ① as well as ① and ②, 1 2 1 v0 + p0 + gH = v12 + p1 , 2 2 1 2 v + p1 2 1
1 2 v + p0 , 2 2
=
and the continuity equation A1 v1 = A 2 v2 in conjunction with the stationarity condition v0 = 0, we have three equations for the three unknowns v1 , v2 and H . Solving the equations yields the velocities A2 v1 = v 2 = A1
2( p1 p0 ) A 22 (A21 A22 )
−
−
and the stationary fluid height H =
p1
− p
g
A21
0
A21
−A
2 2
.
It can be seen that a stationary state is only possible for A2 < A1 .
b) The loss of volume flow QV is determined by using the continuity equation QV = A 2 v2 = A 1 v1 .
Bernoulli’s
Theorem
239
Introducing v1 and v2 , respectively, yields QV = A 1 A2
2( p1 (A21
− p ) . −A ) 0 2 2
Alternatively, the loss of volume flow can be calculated by using v2 = 2gH (Torricelli) and A2 .
√
c) Due to the leak in the tank, the fluid level changes continuously. For the velocity of level decrease we have v(z ) =
, − dz dt
where z is the actual fluid level in the tank. Thus, Bernoulli’s theorem for a streamline between a point on the fluid surface and point ② reads 1 1 v(z )2 + p0 + gz = v22 + p0 . 2 2 Using the continuity equation A0 v(z ) = A 2 v2 we obtain for the decrease velocity of the fluid surface v(z ) =
−
dz = dt
2gzA22 . A20 A22
−
The time T , required to empty the tank, can be determined by separation of variables and integration:
− √ 0
2gA 22 A20 A22
dz = z
−
H
T
dt
0
T = 2
−
A20 A22 2gA 22
√
Here, for H , the result from a) can finally be introduced: T =
√ 2 g
− p1
p0 A21 (A20 ρ A22 (A21
2 2 2 2
−A ) . −A )
H.
240 P10.8
Bernoulli’s
Theorem
Problem 10.8 From
a big reservoir, fluid flows out through a pipe with a smoothly changing cross section. Determine the fluid levels z 1 and z 2 in the standpipes
po H
z 1
①
② ③
A1 Solution
z 2
The outflow velocity at point
③ follows
A2
from Torricelli’s
law as v3 =
2gH .
The pressures at the locations ① and in the standpipes are given by p1 = p 0 + gz 1 ,
② with
the fluid levels z 1 and z 2
p2 = p 0 + gz 2 .
Thus, applying Bernoulli’s theorem for a stremline between the points ① and ③, 2 1 2 v1
+ 0 + p1 = 21 v32 + 0 + p0 ,
and using the continuity equation A1 v1 = A 2 v3
v1 =
A2 v3 , A1
we first obtain
− √ −
1 p1 = p 0 + v32 1 2 Introducing v3 = p1 z 1 =
− p
ρg
0
A2 A1
2
.
2gH yields the fluid level in the standpipe:
= H 1
A2 A1
2
.
In the same way, the pressure p 2 is calculated by applying Bernoulli’s theorem for a streamline between ② and ③, 2 1 v 2 2
+ 0 + p2 = 21 v32 + 0 + p0 ,
and using the continuity equation v2 = v3 . Thus, it follows for the pressure p2 = p 0 and for the fluid level z 2 = 0 .
Momentum Balance Problem 10.9
From the drainpipe of a big reservoir, a water-jet hits a pivoted plate of weight W .
241 P10.9
①
p0
Determine a) the pressure in the pipe ②③ as a function of the coordinate z , b) the angle ϕ, the plate is rotated, if the jet flows off in plate direction.
h
AB
W b
ϕ z
②
h/2
a
④
③
AD = A C = 12 AB Solution
a) The cross section of the drainpipe
A(z ) = 2(A2
− A ) z/h + A 3
3
②③ is
given by
.
The continuity equation for an arbitrary point in the range ②③ and the point ③ reads A(z ) v(z ) = A 3 v3 . The velocity at point ③ is v3 = v4 = 3gh. Thus, the velocity in the range ②③ follows as hA3 v(z ) = 3gh . 2(A2 A3 )z + hA3
√
−
Bernoulli’s
theorem
2 2 1 3 1 v + p + gh = v(z ) 0 A 2 2 2
+ p(z ) + gz
between point ① and a point in the range ②③ leads with vA = 0 (big reservoir) to the pressure in the drainpipe: 3 h2 p(z ) = p 0 + gh 1 gz . 2 (2z + h)2
−
−
b) Using the sketched control volume, the momentum balance in the direction of the normal force N exerted to the plate yields Q(0
B
− v cos ϕ) = −N .
ϕ b
4
With the volume flow Q = A 4 v4 follows
W
N = A4 v42 cos ϕ .
N
Finally, from the equilibrium condition
B:
N
a cos ϕ
QvD
− W 2b sin ϕ = 0
the angle ϕ is determined:
sin ϕ =
QvD
6 gA 4ah . Wb
a
242 P10.10
Momentum
Problem 10.10
A trapezoidal tank of rectangular cross section (b1 b/2) is filled by a pump via a pipe. The pump produces a constant volume flow Q and the pipe has the cross section b2 /10.
×
a) Determine the rise velocity of the fluid in the tank. b) Calculate for z = h/2 the resulting force exerted on the pipe.
Solution
Q p0 ③
h
①
a
②
a
b1
z
b
a) With the varying width
b1 (z ) =
b (h h
− z )
of the trapezoidal tank the fluid surface is given by b b2 A1 (z ) = b 1 (z ) = (h 2 2h
− z ) .
Thus, due to the constant volume flow Q, the velocity in v1 (z ) =
① follows
as
Q 2 Qh = 2 . A1 (z ) b (h z )
−
b) The force exerted by the fluid on the pipe can be determined from the momentum balance. For this purpose, we first calculate for z = h/2 the pressures and velocities at the points ② and ③ by using Bernoulli’s theorem and the continuity equation. For point ② follows from Bernoulli’s theorem between ① and ② 1 2 h 1 v1 + p0 + g = v22 + p2 + 0 , 2 2 2 and from the volume flow b2 Q = v 2 A2 = v 2 10 follows the velocity v2 =
10 Q . b2
Balance
243
Thus, we obtain for the pressure h p2 = p 0 + g 2
−
42 Q2 . b4
From the continuity equation Q = v 3 A3 = v 2 A2 between points v3 = v 2 =
②
and
③ follows
10 Q . b2
Thus, Bernoulli’s theorem v32 p3 v22 p2 + + a = + +0 2g g 2g g between points
②
p3 = p 0 + g
and
③ yields
− − h 2
a
the pressure at point ③ as
42 Q2 . b4
As control volume for the balance of momentum, we now choose the fluid within the pipe. It is loaded by its weight, by the pressure forces at ② and ③ and by the forces Rx , Ry exerted from the pipe-wall. The opposite forces are exerted from the fluid to the pipe-wall. Thus, the balance of momentum reads in components
→:
Q(v2
↑:
Q(0
− 0)
= p2 A2 + Rx ,
−
p3 A3
− (−v )) = − p A + R − gV . 3
3
3
y
Introducing the pressures and velocities yields with the fluid volume V = πab 2 /20 in the pipe the force components exerted on the pipe-wall 2
Qv3
b h Rx = p0 + g 10 2 b2 h Ry = p0 + g 10 2
Ry Rx gV
29 Q2 + , 5 b2
− ga
29 Q2 πab 2 + + g . 5 b2 20
p2 A2
Qv2
244 P10.11
Momentum
Problem 10.11
In a horizontal plane, a fluid jet hits with the velocity v0 a wall under the angle α. The jet depth in vertical direction is constant: h = const. a) Determine the velocities v1 and v2 of the two off-flowing jets. b) Calculate the width b1 and b2 of the off-flowing jets. c) Determine the normal force exerted on the wall.
b0 v0
②
①
b1
v2
v1
α
b2
a) The velocities v1 and v2 can be determined from Bernoulli’s theorem applied to streamlines between the points and ① and between the points and ②: Solution
1 2 1 v0 + p0 = v12 + p0 2 2
v1 = v 0 ,
1 2 1 v0 + p0 = v22 + p0 2 2
v2 = v 0 .
b) From the continuity equation follows with v1 and v2 the relation between the jet-widths: v0 b0 h = v1 b1 h + v2 b2 h
b0 = b 1 + b2 .
The balance of momentum in wall direction
→:
Q2 v 2
Q0 v0
− Q v − Q v cos α = 0 1 1
0 0
yields with Q0 = b 0 hv0 , Q1 = b 1 hv1 , Q2 = b 2 hv2 b0 cos α = b 2
−b
1
α
.
Q 1 v1
This provides the jet-widths b1 =
1 b0 (1 2
− cos α) ,
N
Q 2 v2
N
b2 =
1 b0 (1 + cos α) . 2
c) The normal force exerted on the wall is directly obtained from the balance of momentum perpendicular to the wall:
↑:
Q0 v0 sin α = N
N = v02 b 0 h sin α .
Balance
245
Problem 10.12
A horizontally placed bend of a pressure pipe is held by a concrete block B.
P10.12
A1
B
v1
Determine the horizontal and vertical force component exerted from the bend to the concrete block.
x
α
p1
② ①
v2 p2
A2
y
Given: v1 , p1 , A1 , A2 , α
Solution Using
the continuity equation, the outflow velocity v2 can be
determined: Q = A 1 v1 = A 2 v2
v2 =
A1 v1 . A2
The pressure at location ② follows from Bernoulli’s theorem applied to a streamline between points ① and ②: 1 2 1 v1 + p1 = v22 + p2 2 2
−
p2 = p 1 + v12 1 2
The fluid within the bend (control volume) is loaded by the forces Rx , Ry (exerted by the bend-wall) and the pressure forces at ① and ②. Thus, the balace of momentum F = Q(v2 v 1 ) reads in components
α
p1 A1
Ry
Q(v2 cos α
− v ) = p A − p A cos α + R
x
↓:
Q(v2 sin α
− 0)
y
=
1
2
2
Ry = A 1
2 vA 2
2+
2
A1 A2 + A2 A1
A1 A2 + A2 A1
Qv2
,
2
− − − 2 v 2 1
Rx
− p A sin α + R .
This leads to the forces Rx = A 1
.
p2 A2
→:
1
2
Qv1
−
1
A1 A2
+ p1
cos α
A2 A1
p1 1
A2 cos α A1
,
sin α .
Because of the equilibrium conditions these forces must be carried by the concrete block.
