Kinematics
deals only with relationships acceleration, and time.
among position, velocity,
A body in motion can be considered a particle if rotation of the body is absent or insignificant. A particle does not possess rotational kinetic energy. All parts of a particle have the same instantaneous displacement, velocity, and acceleration. A rigid body does not deform when loaded and can be considered a combination of two or more particles that remain at a fixed, finite distance from each other. At any given instant, the parts (particles) of a rigid body can have different displacements, velocities, and accelerations if the body has rotational as well as translational motion.
a 9 r
v
acceleration gravitational acceleration position radius distance time velocity
ft/sec2
m/s2
ft/sec2 ft
m/s2 rn
ft
m
ft
m
sec ft/sec
s m/s
r
e w
dr dt dv
d2r
dt
dt2
V=-
a=-=-
[position]
14.1
[velocity]
14.2
[acceleration]
14.3
RECTILINEAR MOTION
Symbols 0:
If r is the position vector of a particle, the instantaneous velocity and acceleration are
angular acceleration angular position angular velocity
rad/sec'' rad tad/see
rad/s2 rad rad/s
Subscripts 0 initial final f n normal r radial t tangential transverse
e
A rectilinear system is one in which particles move only in straight lines. (Another name is linear system.) The relationships among position, velocity, and acceleration for a linear system are given by Eqs. 14.4 through 14.6. s(t)
=
J
v(t)dt
=
JJ
a(t)de
14.4
v(t) = d:;t) = ja(t)dt
14.5
_ d2s(t) dt2
14.6
( ) _ dv(t) a t dt
Rectangular Coordinates
INTROOOCTIONTo"KINEMATICS Dynamics is the study of moving objects. The subject is . divided into kinematics and kinetics. Kinematics is the study of a body's motion independent of the forces on the body. It is a study of the geometry of motion without consideration of the causes of motion. Kinematics
•
_______________________________________
The position of a particle is specified with reference to a coordinate. system. Three coordinates are necessary to identify the position in three-dimensional space; in two dimensions, two coordinates are necessary. A coordinate can represent a linear position, as in the rectangular coordinate 'system, or it can represent an angular position, as in the polar system. Professional Publuutions, lne.
Consider the particle shown in Fig. 14.1. Its position, as well as its velocity and acceleration, can be specified in three primary forms: vector form, rectangular coordinate form, and unit vector form.
out of the integrals in Eqs. 14.4 and 14.5. The initial distance from the origin is so; the initial velocity is a constant, va; and a constant acceleration is denoted ao.
Figure 14.1 Rectangular Coordinates y
• (x, y. z)
a(t)
=
«o
v(t)
=
ao
s(t) = ao
J JJ
14.10
dt = va
+ aot
2
dt
aot2
= So
v (t) 2
~ath
=
]4.1]
+ vat + -2-
v6 + 2ao(s
- so)
14.12 14.13
of particle
x k
CURVILINEAR MOTION z
o
The vector form of the particle's position is r, where the vector r has both magnitude and direction. The rectangular coordinate form is (x, y, z). The unit vector form is
Curvilinear motion describes the motion of a particle along a path that is not a straight line. Special examples of curvilinear motion include plane circular motion and projectile motion. For particles traveling along curvilinear paths, the position, velocity, and acceleration may be specified in rectangular coordinates as they were for rectilinear motion, or it may be more convenient to express the kinematic variables in terms of other coordinate systems (e.g., polar coordinates).
0\
Transverse and Radial Components r
= xi + yj + zk
14.7
The velocity and acceleration are the first two derivatives of the position vector, as shown in Eqs. 14.8 and 14.9.
v=-
dr dt
= xi + yj + z k dv
a=-=dt
14.8
d2r
In polar coordinates, the position of a particle is described by a radius, r, and an angle, B. The position may also be expressed as a vector of magnitude r and direction specified by unit vector er. Since the velocity of a particle is not usually directed radially out from the center of the coordinate system, it can be divided into two components, called radial and transverse, which are parallel and perpendicular, respectively, to the unit radial vector. Figure 14.2 illustrates the radial and transverse components of velocity ina polar coordinate system, and the unit radial and unit transverse vectors, er and ee, used in the vector forms of the motion equations.
0
0
dt2
=xi+jjj+ik
14.9 v = »;e;
[position]
14.14
[velocity)
]4. ]5
[acceleration]
/4./6
+ Veee
= fer + reee Constant Acceleration Acceleration is a constant in many cases, such as a free-
.falling body with constant acceleration g. If the acceleration is constant, the acceleration term can be taken
= (..r-r B02) er
+ (rB + 2rB)ee
J
-
Kinemati(s
Figure i4.2 Radial
;;d Tr~~5ve~5e Coordi~~te~
.
of a particle around a fixed circular path. The behavior of a rotating particle is defined by its angular position, B, angular velocity, w, and angular acceleration, a. These variables are analogous to the s, v, and a variables for linear systems .. Angular variables can be substituted one-for-one in place of linear variables in most equations.
y
r
path of particle
B
\8
eQ~ __e~r __ ~l
w=-
_
dB
dt
[angular position]
14.19
[angular velocity]
14.20
[angular acceleration]
14.21
dJ.JJ
x
a=-
dt d2e
dt2
Tangential and Normal Components
o
14~3
A particle moving in a curvilinear path will have instantaneous linear velocity and linear acceleration. These linear variables will be directed tangentially to the path, and, therefore, are known as tangential velocity, Vt, and tangential acceleration, at, respectively. The force that constrains the particle to the curved path will generally be directed toward the center of rotation, and the particle will experience an inward acceleration perpendicular to the tangential velocity and .acceleration, known as the normal acceleration, an. The resultant acceleration, a, is the vector sum of the tangential and normal accelerations. Normal and tangential components of acceleration are illustrated in Fig. 14.3. The vectors en and et are normal and tangential to the path, respectively. p is the principal radius of curvature.
RELATIONSHIPS BETWEEN fiNEAifAND ROTATIONAL VARIABLES ...
_-_
. .................•...........••.•.•.
__
s
= rB
Vt
= rw
14.22 14.23 dVt
at
14.24
= ra = dt
vr
an=-=rw r
2
14.25
14.17 14.18
PROJECTILE MOTION A projectile is placed into motion by an initial impulse. (Kinematics deals only with dynamics during the flight. The force acting on the projectile during the launch phase is covered in kinetics.) Neglecting air drag, once the projectile is in motion, it is acted upon only by the downward gravitational acceleration (i.e., its own weight). Thus, projectile motion is a special case of motion under constant acceleration.
Figure 14.3 Tangential and Normal Coordinates instantaneous center of rotation
y
Consider a general projectile set into motion at an angle of B from the horizontal plane, and initial velocity Vo, as shown in Fig. 14.4. In the absence of air drag, the following rules apply to the case of travel over a horizontal plane. x
• The trajectory is parabolic. • The impact velocity is equal to initial velocity, Vo· • The range is maximum when B
Plane Cirtular Motion Plane circular motion (also known as rotational particle motion, ·angular motion, or circular motion) is motion _____________________________________
=
45° .
• _The time for the projectile to travel from the launch point to the apex is equal to the time to travel from apex to impact point. Professional Publications, Inc.
14-4
FEReview Manual
• The time for the of its flight path tially stationary down from that
1
-
projectile to travel from the apex to impact is the same time an iniobject would take to fall straight height.
2. What = O?
is the acceleration
of the particle
at tiuie
t
(A) 2 mfs2 (B) 3 mfs2
(0) 5 mfs2
Figure 14.4 Proiectile Motion
(D) 8 mfs2 Th
#17 394 .
y v{t)
. is ;
~----
path of projectile
ch<
Solution:
ize
(D tiv
At t
Fa
= 0,
nO"
a = 8 mffp
x
Answer is D. The following solutions to most common projectile problems are derived from the laws of uniform acceleration and conservation of energy. ax = 0
14.26 14.27
ay = -g
Vx = Vxo = Vocosf
14.28 14.29 14.30
Vy = VyO- gt = Vosine - gt x
= vxot = votcose
Y
= Vyot
1 - 29t
2
= vot sin e -
1 29t
2
14.31
3. What ticle? (A) (B) (0) (D)
21.8 27.9 34.6 48.0
speed reached by the par-
mfs mjs mfs mfs #78394
>
Solution: The maximum of the velocity function is found by equating the derivative of the velocity function to zero and solving for t.
sAMpLE·PROBLEMs··
v
Problems 1-3 refer to a particle whose curvilinear motion is represented by the equation s = 20t + 4t2 - 3t3. 1. What is the particle's
dv
-
dt
=
20
+ 8t - 9t2
-18t
= 8
= 0
8 s = 0.444 s 18
t = -
initial velocity?
20 mjs 25 mjs 30 mfs
(A) (B) (C) (D)
is the maximum
Vmax =
=
32 mfs
20
+ (8)(0.444
s) - (9)(0.444
S)2
21.8 mfs
Answer is A. #76394
Solution:
4. Choose the equation that best represents body or particle under constant acceleration. ds
v = -d
t
At t
=
= 20 + 8t - .9t
2
0, v
= 20 + (8) (0) = 20 mfs
- (9) (0)2
(A) a
(B)
(0) v (D)
= 9.81
v = Vo
=
vo
mjs2
a rigid
+ voft
+ aot +
lot a(t)dt
a = vVr
Answer is A. Professional Publications, Inc.
---
••• ..
14-6
FE Review Manual
._
Solution: gt2
e- 2 votsine + y = 0
Y = vot sin
t2
!J...
-
2 y
= -1500
m since it is below the launch plane.
(9" m) .~1 2"
.
2s
t2
m -
(1000 -; ) t sin 30° - 1500 m
-b
m
m
S2
S
(B) 1.1 mjs t -
1500 m
=
0
(C) 10 m/s (D) 11 m/s
± ...jb2 - 4ac
= .
#84394
[quadratic formula]
2a
500 ±
x
0 (A) 0 m/s
4.905 - 2t - 500 t
=
V( -500)2
- (4)(4.905)(-1500) (2) (4.905)
= +104.85 s, -2.9166 = votcos8
4. A golfer on level ground attempts to drive a golf ball across a 50 m wide pond, hitting the ball so that it travels initially at 25 isi]«. The ball travels at an initial angle of 45° to the horizontal plane. How far will the golf ball travel, and does it clear the pond?
s
=
(1000 :)
(104.85 s) cos 30°
=
90803 m
(90800 m)
32 m; 45 m; 58 m; (D) 64 m;
(A) (B) (C)
Answer is D.
the the the the
ball ball ball ball
does not clear the pond does not clear the pond clears the pond clears the pond #85994
FE-STYLE EXAM PROBLEMS Problems 1 and 2 refer to a particle for which the position is defined by
set)
= 2 sin ti + 4 cas tj [t in radians]
5. Rigid link AB is 12 m long. It rotates counterclockwise about point A at 12 rev/min. A thin disk with radius 1.75 m is pinned at its center to the link at point B. The disk rotates counterclockwise at 60 rev /min with respect to point B. What is the maximum tangential velocity seen by any point on the disk?
1. What is the magnitude of the particle's velocity at t = 4 rad?
(A) 2.61 (B) 2.75 (C)
3.30
(D) 4.12 #82689
2. What is the magnitude of the particle's acceleration at t = 7r?
(A) (B) (C) (D)
2.00 2.56 3.14 4.00 #83689
3. For the reciprocating pump shown, the radius of the crank 'is 'T' = 0.3 m,and the rotational speed is n = 350 rpm. What is the tangential velocity of point A on the crank corresponding to an angle of e = 35° from the horizontal?
Professional Publications, Inc.
IIIIiIII
(A) (B) (C) (D)
6 m/s 26 m/s 33 m/s 45 mjs
.;..
__
.;..
--
•••
-.~
II I
-
Kinematics
For the following pr-ob lerns use the NCEES book as your only reference.
Hand-
6. A particle has a tangential acceleration of at (represented by the equation given) when it moves around a point in a curve with instantaneous radius of 1 m. What is the instantaneous angular velocity (in rad/s) of the particle? at
=
2t - sin t
(A) t2 + cost
+ 3 In
+ 3 cot t
Problems 10 and 11 refer to the following information. The position (in radians) of a car traveling around a curve is described by the following function of time (in seconds).
8(t)
= t3 -
2e -
4t
+ 10
[in m/s2] 10. What is the angular velocity at t = 3 s? (A)
[csc r]
-16 rad/s
(B) -4 rad/s
t2-cost+3lnlcsctl
(B) (C) t2 - cost + 3 In [sin t] (D) t2+cost+3Inlsintl
(C) (D)
11 rad/s 15 rad/s #2886687
#3555794
7. A stone sound of the
14-7
stone is dropped down a well. 2.47 s after the is released, a splash is heard. If the velocity of in air is 342 tis]«, find the distance to the surface water in the well.
(A) 2.4 m (B). 7.2 m
11. What is the angular acceleration at t
=
5 s? .
(A) 4 rad/s2 (B) 6 rad/s2 (C) 26 rad/s2 (D) 30 rad/s2 . #2886 687
(C) 28 m (D) 30 m #2878687
Problems 8 and 9 refer to the following situation. A motorist is traveling at light in an intersection light's red cycle is 15 s. the intersection without the light turns green.
70 km/h when he sees a traffic 250 m ahead turn red. The The motorist wants to enter stopping his vehicle, just as
12. The rotor of a steam turbine is rotating at 7200 rev /min when the steam supply is suddenly cut off. The rotor decelerates at a constant rate and comes to rest after 5 min. What was the angular deceleration of the rotor? (A) (B)
0.40 rad/s? 2.5 rad/s2 (C) 5.8 rad/s2 (D) 16 rad/s2 #2887687
What. uniform deceleration of the vehicle will just put the motorist in the intersection when the light turns green? . 8.
0.18 m/s2 0.25 m/s2 0.37m/s2 (D) 1.3 m/s2
(A) (B) (C)
(A) 18000 rev #2884687
9. If the vehicle decelerates at a constant rate of 0.5 m/s2, what will be its speed when the light turns green? (A) 43 km/h (B) 52 km/h (C) 59 km/h (D) 63 km/h
(B) 22000 rev (C) 72 000 rev (D) 390000 rev #2888687
Problems 14-i6 refer to the following situation.
#2885687
____________________________________
13. A flywheel rotates at 7200 rev /min when the power is suddenly cut off. The flywheel decelerates at a constant rate of 2.1 rad/s" and comes to rest 6 min later. How many revolutions does the flywheel make before coming to rest?
A projectile has an initial velocity of 110 m/s and a launch angle of 20° from the horizontal. The surroundin bIT terrain is level , and air friction is to be disregarded.
Professional Publications; Inc.
14-8
FEReview Manual
-----------!!IIIIII----------I-------- __ ._
(A) (B) (C) (D)
Use the relationship tional variables.
3.8 s 7.7 s 8.9 s 12 s
Vt =
#OOOOMM 298
15. What projectile?
Soluti
Solution 3:
14. What is the flight time of the projectile?
is the horizontal
-
between the tangential
and rota-
The spec'
rw
w = angular velocity of the crank
distance traveled by the = (350 :i:)
(21l' ::~)
(60 ~) mm
(A) 80 m (B) 400 m (C) 800 m (D) 1200 m
= Vt =
(A) (B)
72 m 140 m (C) 350 m (D) 620 m
(0.3
rad) m) ( 36.65 -s-
The the
= 11.0 m/s
#OOOOMM 298
16. What is the maximum elevation achieved by the projectile?
36.65 rad/s
This value is the same for any point on the crank at r = 0.3 m. Answer is D.
#OOOOMM 298
Th<
Solution 4:
con
The elevation of the ball above the ground is
SOLUTIONS TO FE-STYLE EXAM PR()BLEMS
per
of Solution 1:
Y = Vyot -
. ds(t)
= --;It =
v ()t At t
=
gt2
.
= vat sin
"'2
e-
1
gt2 "'2
2. . cos zi -- 4sintJ When the ball hits the ground, y = 0, and
4 rad, v(4)
Iv(4)1
=
2cos (4 rad)! - 4 sin (4 rad)j
=
-1.3li - (-3.03)j
=
V(-1.31)2
=
votsine
gt2 "'2
Solving for t, the time to impact is
+ (3.03)2
= 3.30
2va sin ()
t= -'--9
Answer is C.
So
Substitute the time of impact into the expression for x to obtain an expression for the range.
Solution 2: From Prob. 1, v( t)
=
x 2 cos ti - 4 sin tj dv(t)
at( ) = -= -2Sintldt a(rr) = -2sin1l'i-4cos1l'j
+ 4.0j \a(1l') \ = V(O)2 + (4.0)2 =
=
Answer is D.
Oi
4.0
.
4costJ
.
= vatcose 2v2 = _0
sin
9
2vosine)
= vc (
mr
9.812' s
Answer is D.
cose
e cos ()
(2) (25 ---'----;m~s~ = 63.7 m
9
sin 45° cos45°
Solution 5:
Solution 7:
The maximum tangential velocity of point B with respect to point A is
The elapsed time is the sum of the time for the stone to fall and the time for the sound to return to the listener. The distance, x, traveled by the stone under the influence of a constant gravitational acceleration is
= rw = r(27rf)
Vt;BIA
(12 m) ( 27r -rad) rev 60~
( 12 -.rev) nun
min
1 2 + vot + 2,gt
X = Xo
Xo
and Vo are both zero.
= 15.08 m/s
1 x = _gt2 2
The maximum tangential velocity of the periphery of the disk with respect to point B is
Solving for t, the time for the stone to drop is
= rw =r(27r f)
Vt,disklB
(60 re.v) min
(1.75 m) (27r rad)
rev 60 _s_ min
=
The time for the sound (traveling at velocity c) to return to the listener is
11.00 sss]«
The velocities combine when the two velocity vectors coincide in direction. The maximum velocity of the periphery of the disk with respect to point A is the sum of the magnitudes of the two velocities.
The total time taken is
+
h
f!
-- x+ -x = 2.47 c
9
Vt,disklB
= Vt,BIA + Vt,disklB m
15.08 -
=
26.08 m/s
s
m
+ 11.00 -
=
t2 = 2.47 s
s
Substitute values for 9 and c.
s
R
x
9.81 -
342-
S2
Answer is B.
x
+ ------m
m
= 2.47 s
s
By trial and error with the answer choices given (or by solving the quadratic equation), x =28 m
Solution 6:
Answer is C. dVt
at=:" Vt =
dt
J
Solution 8:
at dt
The initial speed of the vehicle is
=J(2t~sint+3
cott) dt
= t2 + cos t + 3 In [sin tl
t2
Vt
W
+ cost + 3 In
(70 ~)
[in m/s) [sin r]
vo
rim
=
t2
+ cas t + 3 In
Answer is D.
___________________________________
=
[sin tl
[in rad/s)
3600 ~
=
= - = ------------~--~
(1000 ~)
19.44
mls
The distance traveled under a constant deceleration is x = Xo
+ vot
-
2,1 at 2 Professional Publications, Inc.
14-10
FEReview Manual
Letting Xo = 0, the acceleration required to travel a distance x in time t starting with velocity vo is
W
= Wo
(2)(vot - x)
t2
- at
(1200 _r~v) (271 _rad) mm rev (s ---6-0-~s~---'---- a( 5 min) 60 ~
o= (15 S)2
ex
0.37 m/s2
=
.~. .Ans'"
2
2.51 rad/s
Answer is B.
Answer is C.
Solution 13: 1 B = Bo + wot - 2at
Solution 9: v
=
0
= va + at krn = 70 _
+ (1200 x
( -0.5
~)
(158) (3600 ~)
+ .:....:. __ -'-----=-'-_~'-
=
",-8
h
= 43
)
min
(15 s) - 250 m]
(2) [( 19.44 ~)
••••
Solution 12:
a = ~--'----".,,----'-
=
-
•
1000 k~
e=
r:2d)
(271)(6 ruin) - ~ (2.1
(60 rr:n)]
2
103 rad
X
135.4 x 103 rad .
km/h
::i:)
[(6 min)
135.4
2
271
= 21.5
X
103 rev
Answer is A. Alternative solution: The average rotational ing deceleration is
speed dur- .
rev . 7200 -. -0
Solution 10:
mm
W
dB = 3t 2 -
(t) =. -dt
w(3)
=
..4t - 4
=
(3)(3)2 - (4)(3) - 4
=
11 rad/s
2 3600 rev /min
e = wt
rev)
= ( 3600 min
2
(6 min)
= 21,600 rev
Answer is B.
Answer is C.
Solution 14: The vertical component Vy =
Solution 11:
0=
a(t)
=
dw(t)
dt = 6t - 4
a(5)
Answer is C.
Professional Publications, Inc.
=
(6)(5) - 4
=
26 rad/s2
of velocity is zero at the apex.
va sine - gt
(no 7) sin 20° -
(9.81 ~)
t
t = 3.84 s
The projectile takes an equal amount of time to return to the ground from the apex. The total flight time is
T = (2)(3.84 s) = 7.68 s (7.7 s) Answer is B.
=
-
Kinematics
• solution 15:
Solution 16:
Calculate the range from the horizontal component of velocity. x = vxt =Vo coset
The elevation at time t is
=
(110
7)
cos200(7.68
=794m
y = vot sin
s) = (110
Answer is C.
e ~ (~)
7)
- (~) =72
14-11
2
gt
(3.84 s) sin 20° (9.81 ~)
(3.848)2
m
Answer is A.
______________________________________
Professional Publications, Inc.
Kinetics
Subscripts 0
f k n r R s t
e
initial friction dynamic normal radial resultant static tangential transverse
INTRODUCTION TO KINETICS Nomendature a f F 9
9c k
m N
p r
r
R s t
T v
W
acceleration linear frequency force gravitational acceleration gravitational constant (32.2) spring constant mass normal force linear momentum position radius resultant force distance time period velocity weight
ft/sec2 Hz lbf
m/s2 Hz N
ft/sec2
m/s2
S () j.L
¢
w
-
angular acceleration deflection angular position coefficient of friction angle natural frequency
lbm- ft /lbf-sec2 lbf/ft lbm lbf lbf-sec
N/m kg N N-s
ft
m
ft lbf ft sec sec ft/sec lbf
m
rad/sec?
rad/s2 m rad
ft rad
.............................................
MOMENTUM The vector linear momentum (momentum) is defined by Eq. 15.1. It has the same direction as the velocity vector. Momentum has. units of force x time (e.g., lbfsec or N·s). p=mv
N p=-
m s s
m/s N
Symbols Q
Kinetics is the study of motion and the forces that cause motion. Kinetics includes an analysis of the relationship between the force and mass for translational motion and between torque and moment of inertia for rotational motion. Newton's laws form the basis of the governing theory in the subject of kinetics.
mv
gc
[81]
15.10
[U.S.]
15.1b
Momentum is conserved when no external forces act on a particle. If no forces act on the particle, the velocity and direction of the particle are unchanged. The law of conservation of momentum states that the linear momentum is unchanged if no unbalanced forces act on the particle. This does not prohibit the mass and velocity from changing, however. Only the product of mass and velocity is constant.
NEWrON/S·FiRSTAND·SECOND·iAWS·OF
MOTIO~........
.r:
deg
deg
rad/sec
rad/s
Newton's first law of motion states that a particle will remain in a state of rest or will continue to move with
Professional publications, Inc.
15-2
FEReview Manual
-
•
constant velocity unless an unbalanced acts on it.
external force
This law can also be stated in terms of conservation of momentum: If the resultant external force acting on a particle is zero, then the linear momentum of the particle is constant. Newton's second law of motion states that the acceleration of a particle is directly proportional to the force acting on it and is inversely proportional to the particle mass. The direction of acceleration is the same as the direction of force. This law can be stated in terms of the force vector required to cause a change in momentum: The resultant force is equal to the rate of change of linear momentum.
F= dp
15.2
dt For a fixed mass,
_
The magnitude of the frictional force depends on the normal force, N, and the coefficient of friction, IL, between the body and the contacting surface.
15.5 The static coefficient of friction is usually denoted with the subscript s, while the dynamic coefficient of friction is denoted with the subscript k. ILk is often assumed to be 75 percent of the value oflLs. These coefficients are complex functions of surface properties. Experimentally determined values for various contacting conditions can be found in handbooks. For a body resting on a horizontal surface, the normal force is the weight of the body. If the body rests on an inclined surface, the normal force is calculated as the component of weight normal to that surface, as illustrated in Fig. 15.1. N
=
mgcos¢
N
=
mgcos¢
15.6b
dv
Figure 15./ Frictional and Normal Forces
dt
=ma
F=
[US.]
---;It
dt
=m-
15.6a
9c
F _ dp _ d(mv) -
[SI]
ma
gc
[SI]
15.3a
[US.]
15.3b
s
impending
motion
WEIGHT The weight, W, of an object is the force the object exerts due to its position in a gravitational field, g. W=mg
[SI]
15.4a
W= mg gc
[US.]
15.4b
gc is the gravitational constant, approximately 32.2 lbmft/lbf-sec2.
Friction is a force that always resists motion or impending motion. It always acts parallel to the contacting surfaces. If the body is moving, the friction is known as dynamic friction. If the body is stationary, friction is known as static friction.
Professional Publications, Inc.
The frictional force acts only in response to a disturbing force, and it increases as the disturbing force increases. The motion of a stationary body is impending when the disturbing force reaches the maximum frictional force, ILsN. Figure 15.1 shows the condition of impending motion for a block on a plane. Just before motion starts, the resultant, R, of the frictional force and normal force'~ equals the weight of the block. The angle at which motion is just impending can be calculated from the coefficient of static friction. ¢ = tan
-lJ.Ls
15.1
Once motion begins, the coefficient of friction droPS slightly, and a lower frictional force opposes movement. This is illustrated in Fig. 15.2.
----
••
Kinetics
"Figure 15.2 "Frictional Force ~ersus DisturbingForce " " " ... " .. " .
no rnotron
I(
~
Tangential and Normal Components For a particle moving along a circular path, the tangential and normal components of force, acceleration, and velocity are related.
impending motion
I
~e~~~b~~~~~~e~
I
LFn
~sN
Newton's second law can be applied separately to any direction in which forces are resolved into components. The law can be expressed in rectangular coordinate form (i.e., in terms of z- and y-component forces), in polar coordinate form (i.e., in tangential and normal components), or in radial and transverse component form.
Rectangular Coordinates Equation 15.8 is Newton's second law in rectangular coordinate form and refers to motion in the x-direction. Similar equations can be written for the y-direction or any other coordinate direction. [81]
15.8
In general, Fx may be a function of time, displacement, and/ or velocity. If Fx is a function of time only, then the motion equations are = Vxo
;,
=
+
J
Xo + vxot
"(Fx(t)) ---:;:n-
+
J
dt
[81]
[81]
15.15
=mae
[81]
15.16
[81]
15.17
.................
FREEVIBRATION
.
::, .
Vibration is an oscillatory motion about an equilibrium point. If the motion is the result of a disturbing force that is applied once and then removed, the motion is known as natural (or free) vibration. If a continuous " force or single impulse is applied repeatedly to a system, the motion is known as forced vibration. A simple application of free vibration is a mass suspended from a vertical spring, as shown in Fig. 15.3. After the mass is displaced and released, it will oscillate up and down. If there is no friction (i.e., the vibration is undamped), the oscillations will continue forever.
Figure 15.3 Simple Mass-Spring System
15.9 '
If Fx is constant (i.e., is independent of time, displacement, or velocity), then the motion equations become
[81]
e
vx(t)
= v-o + (~)
15.11
t 15.12
x(t) = Xo
+ vxot +
F.
t2
2~ t2
= Xo
•
(d~t)
=m
15.10
vx(t)dt
e h
s
15.14
LFe
KiNETICSOF"A"PARTiCLE""
x(t)
[81]
For a particle moving along a circular path, the radial and transverse components of force are
disturbing force
ri e
e :,
=
m(vI)
Radial and Transverse Components
4~
g
= man =
t mat
LF
~motion
vx(t)
15-3
ax
+ vxot + --2
__________________________________
The system shown in Fig. 15.3 is initially at rest. The mass is hanging on the spring, and the equilibrium position is the static deflection, 8st. This is the deflection due to the gravitational force alone. mg = k8st
15.13
[81]
15.180
[u.s.]
IS.18b
Professional Publications, Inc.
15-4
FEReview Manual
•
The system is then disturbed by a downward force (i.e., the mass is pulled downward from its static deflection and released). After the initial disturbing force is removed, the mass will be acted upon by the restoring force (- kx) and the inertial force ( -mg) . .Both of these forces are proportional to the displacement from the . equilibrium point, and they are opposite in sign from the displacement. The equation of motion is
F=ma
+ Ost) = m/i: + Ost) = mii: mx+kx =0
mg - k(x kOst - k(x
[81J
15.19
[81]
15.20
[81J
15.21 15.22
[81]
sAMp(fpROBIEM"S
_
.
....................
-
1. For which of the following situations is the net force acting on a particle necessarily equal to zero? (A) The particle is traveling around a circle.
a~ constant
velocity .
(B) The particle has constant linear momentum. (C) The particle has constant kinetic energy. (D) The particle has constant angular momentum. #86691
A
Solution: This is a restatement of Newton's first law of motion which says that if the resultant external force acting on a particle is zero, then the linear momentum of the particle is constant.
The solution to this second-order differential equation is Answer is B. x(t)
=
C1 coswt
+ C2 sinwt
15.23
C1 and C2 are constants of integration that depend on the initial displacement and velocity of the mass. w is known as the natural frequency of vibration or angular frequency. It has units of radians per second. It is not the same as the linear frequency, i, which has units of hertz. The period of oscillation, T, is the reciprocal of the linear frequency.
2. One newton is the force required to (A) (B) (C) (D)
give a 1 g accelerate accelerate accelerate
mass an acceleration of 1 m/s2. a 10 kg mass at a rate of 0.10 mjs2. a 1 kg mass at a rate of 1.00 em/s2. a 1 kg mass at a rate of 9.81 m/s2. #87689
Solution:
w={£ w= )kf=~=~· T T= ~= f w gc m
[81]
15.240
[U.8.]
15.24b
15.25
211"
211"
=
xocoswt
+ (~)
sinwt
Professional Publications, Inc.
= xocos
wt
3. A 550 kg mass initially at rest is acted upon by a force of 50et N. What are the acceleration, speed, and displacement of the mass at t = 4 s? (A) 4.96 (B) 4.96 (C) 4.96 (D) 4.96
m/s2, m/s2, m/s2, m/s2,
4.87 mis, 19.5 m 4.96 mis, 19.8 m 135.5 mis, 1466 m 271 us]«, 3900 in
15.27
For the special case where the initial displacement is Xo and the initial velocity is zero, the solution of the equation of motion is
x( t)
Answer is B.
15.26
For the general case where the initial displacement is Xo and the initial velocity is va, the solution of the equation of motion is
x(t)
Newton's second law can be expressed in the form of F = ma. The unit of force in S1 units is the newton, which has fundamental units of kg·m/s2. A newton is the force required to accelerate a 1 kg mass at a rate of 1 m/s2 or a 10 kg mass at a rate of 0.10 m/s2.
15.28
#88691
5.
Solution: 50e4 N 550 kg
a=-=--F m
=
4.96 rn/s2
---.
•
Kinetics·
r
4
v
s(t) s
=
50et N 550 kg dt
)0
et 4 1110
=
= 4.87
m/s
=
50e N --k- de a 550 gas
ti it t a
= ~87tl~ =
(4.87
=
7)
1
= 4.96
it
Solution:
- 11
(4.87
m)
Choose a coordinate system so that the x-direction . parallel to the inclined plane.
L
dt
Fx
=
= mgx
max
max = mgsin45°
(4 s) - 0
19.48 rn
=
15-5
p,=
mgsin45°
is
- Ff
- p,mgcos45° -:-max
gsin45°
mgcos45°
- ax
9 cos 45°
(9.81 ~) sin 45° -1.5 ~ =
Answer is A.
(9.81 Problems
=
4 and 5 refer to the following situation.
• A 5 kg block begins from rest and slides down an inclined plane . • After 4 s, the block has a velocity of 6 m/s.
4. If the angle of inclination block traveled after 4 s?
(A) 1.5
is 45°, how far has the
:Z) cos 45°
0.78
Answer is C. 6. A constant force of 750 N is applied through a pulley system to lift a mass of 50 kg as shown. Neglecting the mass and friction of the pulley system, what is the acceleration of the 50 kg mass?
m
(B) 3 m (C) 6 m (D) 12 m #89689
F= 750 N
Solution: v(t)
=
va
+ aot m
ao=
v(t) -va
(A) 5.20 m/s2 (B) 8.72 m/s2
6 -;-0
= _o?...-_
t
4 s
(C) (D)
= 1.5 m/s2
sCt)
=
So + vat
=0+0+ =
1
+ 2aot (~)
16.2 m/s2 20.2 m/s2 #91 694
2
Solution: (1.5 ~)
(4S)2
Apply Newton's second law to the mass and to the two frictionless, massless pulleys. Refer to the following free-body diagrams.
12 m
Answer is D. T,
5. What is the coefficient of friction between the plane - and the block? (A) 0.15 (B) 0.22 (C) 0.78 (D) 0.85
1. mass A
#90689
F T, pulley B
pulley C
15·6
FEReview Manual -----------------------
a=
_
mass A:
Tl - mg = ma
pulley B:
2T2 - T, = 0
3
pulley C:
T2
=F =
2
2T2 -
Tl - mg m
750 N
2F -
mg
m
problE In sta - and b initial . pulley
F(N)
mg 2
m
4
t (5)
(2) (750 N) - (50 kg) (9.81 ~) (A) (B)
0 m/s 0.075 m/s (C) 0.15 m/s (D) 0.30 ta]»
50 kg
=
20.2 m/s2
Answer is D.
#94691
7. A mass of 10 kg is suspended from a vertical spring with a spring constant of 10 N/m. What is the period of vibration?
Problems 3 and 4 refer to the following situation.
(A) (B)
0.30 s 0.60 s (C) 0.90 s (D) 6.3 s #92394
• The 52 kg block shown starts from rest at position A and slides down the inclined plane to position B.
F
• When the block reaches position B, a 383 N horizontal force is applied.
C C
• The block comes to a complete stop at position C.
(
• The coefficient of friction between the block and the plane is f.-L = 0.15.
Solution: T
1m
= 27rV k = 27r~
~
C-
F . are rl
10 ~
( (
= 6.3 s
(
( Answer is D. , For'
bool
FE-STYLE EXAM PROBLEMS
7. \
c
1. If the sum of the forces on a particle is not equal to zero, the particle is
(A) moving with constant velocity in the direction of the resultant force. (B) accelerating in a direction opposite to the resultant force. (C) accelerating in the same direction as the resultant force. (D) moving with a constant velocity opposite to the direction of the resultant force. #93689
2. A varying force acts on a 40 kg weight as shown in the following force versus time diagram. What is the object's velocity at t = 4 s if the object starts from rest?
Professional Publications, Inc.
3. Find the velocity at position B. (A) 2.41 m/s (B) 4.12 m/s (C) 6.95 m/s (D) 9.83 m/s
8. . #95681
4.
Find the distance between positions Band
C.
1
of a no C spriJ
(A) 3.23 m (B) 4.78 rn (C) 7.78 m (D) 10.1 m #96681
.-
--
-
-
Kineti(s
problems 5 and 6 refer to the following pulley system. In standard gravity, block A exerts a force of 10000 N and block B exerts a force of 7500 N. Both blocks are initially held stationary. There is no friction and the pulleys have no mass. ,~
15· 7
9. A spring has a constant of 50 N/m. The spring is hung vertically, and a mass is attached to its end. The spring end displaces 30 em from its equilibrium position. The same mass is removed from the first spring and attached to the end of a second (different) spring, and the displacement is 25 em. What is the spring constant of the second spring?
(A) 46 N/m (B) 56 N/m (C) 60 N/m
(D) 63 N/m #39351294
10000 N
5. Find the acceleration of block A after the blocks are released. (A) 0 m/s2 (B) 1.4 m/s2 (C) 2.5 m/s2 (D) 5.6 m/s2
o m/s
(B)
3.5 m/s
(B)
0.59
m
#3942295
6. Find the velocity of block A 2.5 s after the blocks are released.
(A)
(A) 0.35 m (C) 0.77 m (D) 0.92 m
#97994
\
10. A cannonball of mass 10 kg is fired from a cannon of mass 250 kg. The initial velocity of the cannonball is 1000 km/h. All of the cannon's recoil is absorbed by a spring with a spring constant of 520 N I cm. What is the maximum recoil distance of the cannon?
(C) 4.4 m/s (D) 4.9 m/s #98994
11. A child keeps a 1 kg toy airplane flying horizontally in a circle by holding onto a 1.5 m long string attached to its wing tip. The string is always in the plane of the circular flight path. If the plane flies at 10 us]«, what is the tension in the string? (A) 7 N (B) 15 N (C) 28 N (D) 67 N
For the following problems use the NCEES Handbook as your only reference. 7. What is the period of a pendulum center point 20 times a minute? (A) (B) (C) (D)
that passes the
0.2 s 0.3 s 3s 6 s
#3471 193
12. A car with a mass of 1530 kg tows a trailer (mass of 200 kg) at 100 km/h. What is the total momentum of the car-trailer combination? (A) (B) (C) (D)
4600 N-s 22 000 N-s 37000N·s 48 000 N·s
#2142689
e
8. A variable force of (40 N)cos is attached to the end of a spring whose spring constant is 50 N[ux. There is no deflection when = 90°. At what angle, will the spring deflect 20 em from its equilibrium position?
e
e,
#1584 689
13. A car is pulling a trailer at 100 km/h. A 5 kg cat riding on the roof of the car jumps from the car to the trailer. What is the change in the eat's momentum? (A) -25 N-s(loss) (B) a N-s (C) 25 N·s (gain) {D) 1300 N-s (gain)
(A) -14° (B) 25°
(C) 64° (D) 76° #39331294
#4.169 795
15-8
FEReview Manual
-
1
_
14. A 3500 kg car accelerates from rest. The constant forward tractive force of the car is 1000 N, and the con-: stant drag force is 150 N. What distance will the car travel in 3 s?
(A) 0.19 m (B) 1.1 m (C)
1.3 m
(D) 15
m #2108
689
An:
LFx
SOLUTIONS TO FE-STYLE EXAM PROBLEMS
x
= ma
Wx - flN
Solu'
= max
Ref,
mg sin e - flmg cas e = ma;
Solution 1: Newton's second law, F = ma, can be applied separately to any direction in which forces are resolved into components, including the resultant direction.
ax
2
Since force and acceleration are both vectors, and mass is a scalar, the direction of acceleration is the same as the resultant force.
= 9
sin
fl9 cas
e
5 [1 3 - (0.15)
=
(9.81 ~)
=
2.415 m/s2
v
= v~ + 2ao(s - so)
va
=
2
v
G~)]
So = 0
= 2aos =
Ap
Answer is C. v =
(20 m)
(2) (2.415 ~)
= 96.6
Solution 2:
e-
J
m2/s2
96.6
~2
Bu
= 9.83 iis]»
Answer is D.
Use the impulse-momentum principle. The impulse is the area under the F-t curve. There are two right triangles. F=
Solution 4:
LFx
dp
=ma
dt FD..t
= m!:::.v FD..t
D..v
= --
(2) [(~)
(3N)(2S)]
mg sin e - pcose - fl(mgcose
= --=-~'--:,------=-
tri
+ Psine) = ma
40 kg
= 0.15 m/s
a
= (~)
[mgsine - pcose
- fl(mgcose
+ P sin e)]
Cc TI:
P = g sin e - flg cos e - - (cos e + fl sin e) m Answer is C.
Solution 3:
Professional Publications, Inc.
(9.81 ~)
52 kg
Choose a coordinate system parallel and perpendicular
to the plane, as shown.
[1 3_ (383 N) 5
=
=
(0.15TG~)]
[12 13
+ (0.15)
(~)] 13
-4.809 m/s2
---
-
Kinetin 2
'2
=
9.83 m/s
v
=
30
-vo 3
[from Prob. 3]
=0
2
_
= 2ao = (2) = 10.05
(9.83
But F
m)2 S
(-4.809
=
ma, so
;) Since the tension in the rope is the same everywhere,
m :
Answer is D. 10000 N - T 10000 N
Solution 5: Refer to the following free-body diagrams. T
T - 7500 N 7500 N
T = 8571 N
T
From block A,
g
T)
g(WA -
WA (9.81 ~) =
Apply Newton's second law to the free body of mass A. \
=
But m
= Wig,
(10000:) 9.812'
(10000 N - 8571 N) 10000 N
1.4 m/s2
Answer is B.
so
(aA)
=
Solution 6:
10000 N - TA
s
VA = Vo
Apply Newton's second law to the free body of mass B.
+ aAt
=
0 + (1.~ ;)
(2.5 s)
= 3.5 ui]»
Answer is B.
Solution 7: Combine the equations and solve for aA, setting TA TB, and aA = -aB' WA - mAaA
=
aA=
WB - mBaB WA - WB mB+mA
(
m)
= 9.81 s2
=
=
=
= WE + mBaA
A pendulum will pass the center point two times during each complete cycle. Therefore, 10 cycles are completed in 60 s. elapsed time T = -"------,-no. of cycles
g(WA - WB) WB+WA
(10000 N -7500 N) 7500 N + 10000 N
1.4 m/s2
___________________________________
- __'J1I I
Alternate Solution:
v =vo+2aO(S-30) vo
15-9-
-
60 s 10
=
6 s
_ Answer is D.
.~h~oo~~~~m~l~
15·10
FEReview Manual
_
Solution 8: From Hooke's law, the relationship between and deflection, x, or a linear spring is
force,' F,
(6.5)(250 kg) [(40~)
(l~OO ~)
J
h
3600
r
F=kx
= (40 N)cose
(20 cm) cm 100 m
(0.5) (520 c:)
(100 :)
x2
(50 :) =
case
= 0.25
Answer is C.
e = cos-1(0.25) -;; 75.5°
:r; = 0.77 m
(76°)
Solution 11: The normal acceleration the airplane) is
Answer is D. Solution 9:
an
(perpendicular
vZ = ..l.
The gravitational force on the mass is the same for both springs. From Hooke'~ law, . F
=
kz =
k1Xl
=
to the path of
T
(10 kZX2
~)2
1.5 m
k1Xl
=
66.7 m/s2
X2
(50 ~)
The tension force.
(30 cm)
in the string
is equal to the centripetal
25 cm = 60 N/m
=
(1 kg) (66.7 ~)
= 66.7 N
Answer is C. Answer is D.
Solution 10: Use the conservation of momentum equation to determine the velocity of the cannon after the ball is fired. Initially, the cannon and cannonball are both at rest. Since the cannon recoils, its velocity direction will be opposite (i.e., negative) to the direction of the cannonball.
Solution 12: (100 k:) v
3600 ~
= 27.78 (10 kg) (0)+(250 =
=
+ (250 kg) (v.)
(1530 kg + 200 kg) (27.78
= 48060 Vc
= -40
7)
N-s
km/h
Use the conservation of energy principle to determine the compression of the spring. The kinetic energy of the cannon will be equal to the elastic potential energy stored in the spring.
KE=PE
Professional Publicalions, Inc.
sii]»
P=mv
kg) (0) (10 kg) (1000 ~)
(1000 :)
= --'-----"---.",.----
Answer is D. Solution 13: The law of conservation of momentum states that the linear momentum is unchanged if no unbalanced forces act on an object. This does not prohibit the mass and velocity from changing; only the product .of mass and
m-
"Tfr1 I '
"\1: I." .
Kinetics
15-11
j
!
velocity is constant. In this case, both the total mass and the velocity are constant. Thus, there is no change.
;1
II Answer is B.
Solution 14: F
=
1000 N - 150 N
=
850 N
F
a=-
m 850 N 3500 kg =
0.243 m/s2 1
S
= vot + zat
+ (~)
=
0
=
1.09 m
2
(0.243 ~)
(3 s)2
Answer is B.
i
i ,
I i
l
I
Kinetics of Rotational Motion
Subscripts
o c
f n
o S
t
initial centroidal friction normal or natural origin or center static tangential or torsional
Nomenclature a
A d F 9
. ge G h I
J kt
m M r r R t
v W
acceleration area distance force gravitational acceleration gravitational constant (32.2) shear modulus angular momentum mass moment of inertia area polar moment of inertia torsional spring constant length mass moment radius radius of gyration moment arm time velocity weight
ft/sec2 ft2 ft lbf
m/s2 m2
ft/sec2
m/s2
lbm-ft.Zlbf-sec'' lbf/ft2 ft-lbf-sec
Pa N·m·s
Ibm-ft2
kg.m2
i: =
(y2
+ z2)dm
16.1
ft4
m4
t;
(x
2
+ z2)dm
16.2
ft-lbf/rad ft lbm ft-lbf ft ft ft sec ft/sec lbf
N-m/rad rn kg N-m rn m m s m/s N
(x2
+ y2)dm
16.3
rad/sec'' rad deg
rad/s2 rad deg
Ibm/ft3 rad/sec rad/sec
kg/m3 radj's rad/s
m N
MASS MOMENT OF INERTIA .........
The mass moment of inertia measures a solid object's resistance to changes in rotational speed about a specific axis. Ix, Iy, and Iz are the mass moments of inertia with respect to the x-, y-, and z-axes, respectively. They are not components of a resultant value.
t,
The centroidal mass moment of inertia, Ie, is obtained when the origin of the axes coincides with the object's center of gravity. Once the centroidal mass moment of inertia is known, the parallel axis theorem is used to find the mass moment of inertia about any parallel axis. In Eq. 16.4, d is the distance from the center of mass to the parallel axis.
Symbols Q
o ()
fJ, p
w to
•
angular acceleration angular position superelevation angle coefficient of friction density angular velocity natural frequency
J =J =J
. Iany parallel
axis
=
Ie
+ md
2
16.4
For a composite object, the parallel axis theorem must be applied for each of the constituent objects.
I
-2
.
2
= Ie,l + m]:d1 + Ic,2 + m2d2 + ...
16.5
Professiol!ol Publications, Inc.
16-2
FEReview Manual
_
The radius of gyration, r , of a solid object represents the distance from the rotational axis at which the object's entire mass could be located without changing the mass moment of inertia.
r=ff
]6.6
1= r2m
]6.7
Table 16.1 (at the end of this chapter) lists the mass moments of inertia and radii of gyration for some standard shapes.
For a rigid body rotating about an axis passing through its center of gravity located at point 0, the scalar value of angular momentum is given by Eq. 16.9.
ho =Iw Iw ho=gc
[SI]
]6.90
[U.S.]
]6.9b
]6. ]0
M=-
General rigid body plane motion, such as rolling wheels, gear sets, and linkages, can be represented in two dimensions (i.e., the plane of motion). Plane motion can be considered as the sum of a translational component and a rotation about a fixed axis, as illustrated in Fig. 16.1.
gra tie~
Although Newton's laws do not specifically deal with rotation, there is an analogous relationship between applied moment and change in angular momentum. For a rotating body, the moment (torque), M, required to change the angular momentum is
dho dt
PLA"t~fMotioNOF A RIGID BODY
•••
The rotation of a rigid body will be about the center of gravity unless the body is constrained otherwise. If the moment of inertia is constant, the scalar form of Eq. 16.10 is [SI]
]6.110
[U.S.]
]6.11b
Figure] 6.] Components of Plane Motion p
Velocity and position in terms of rotational variables can be determined by integrating the expression for acceleration.
plane motion
]6.12
p
16.]3 translation
]6.14
rotation
Instantaneous Center of Rotation Rotation About a Fixed Axis Rotation about a fixed axis describes a motion in which all particles within the body move in concentric circles about the axis of rotation.
°
The angular momentum taken about a point is the moment of the linear momentum vector. Angular momentum has units of distance x force x time (e.g., ft-lbf-sec or Ncn-s). It has the same direction as the rotation vector and can be determined from the vectors by use of the right-hand rule (cross product). ho = r
X
mv mv
ho=rx.
Professional Publications, Inc.
9c
[SI]
]6.80
[U.S.]
16.8b
Analysis of the rotational component of a rigid body's plane motion can sometimes be simplified if the location of the body's instantaneous center is known. Using the instantaneous center reduces many relative motion problems to simple geometry. The instantaneous center (also known as the instant center and Ie) is a point at which the body could be fixed (pinned) without changing the instantaneous angular velocities of any point on the body. Thus, for angular velocities, the body seems to rotate about a fixed instantaneous center. The instantaneous center is located by finding two points for which the absolute velocity directions are known. Lines drawn perpendicular to these two velocities will intersect at the instantaneous center. (This
--
ac in, se:
N! th at tT'"
be tr av ce in
•
_.------------------------------_
Kinetics of Rotational Motion
crraphic procedure is slightly different if the two velocities are parallel, as Fig. 16.2 shows.) For a rolling wheel, the instantaneous center is the point of contact with the supporting surface. o
.. ,...
Equation 16.16 gives the centrifugal force on a body of mass m with distance r from the center of rotation to the center of mass.
Fe
.
Figure 16.2. Graphic Method of Finding the Instantaneous Center
16-3
mvi = man = -= mrw 2 r
_ man _ mv 2 t Fe--------ge ger
_ mrw
ge
[SI]
16.160
[u.s.]
16.16b
2
BANKING OF CURVES. IC
IC
The absolute velocity of any point, P, on a wheel rolling (Fig. 16.3) with translational velocity, vo, can be found by geometry. Assume that the. wheel is pinned at point C and rotates with its actual angular velocity, =w= voir. The direction of the point's velocity will be perpendicular to the line of length I between the instantaneous center and the point.
e
v=lw=
lvo -
r
16.15
If a vehicle travels in a circular path on a fiat plane with instantaneous radius r and tangential velocity Vt, it will experience an apparent centrifugal force. The centrifugal force is resisted by a combination of roadway banking (superelevation) and sideways friction. The vehicle weight, W, corresponds to the normal force. For small banking angles, the maximum frictional force is Ff =J.lsN
16.17
= J.lsW
For large banking angles, the centrifugal force contributes to the normal force. If the roadway is banked so that friction is not required to resist the centrifugal force, the superelevation angle, can be calculated from Eq, 16.18.
e,
tan 8
v2 = -.i.
16.18
gr
Figure 16.3 Instantaneous Center of a Rolling Wheel
TORSIONAL FREE VIBRATION The torsional pendulum in Fig. 16.4 can be analyzed in a manner similar to the spring-mass combination. Disregarding the mass and moment of inertia of the shaft, the differential equation is
CENtRIFUGACFORCf
t
16.19
.
Newton's second law states that there is a force for every acceleration that a body experiences. For a body moving around a curved path, the total acceleration can be separated into tangential and normal components, By Newton's second law, there are corresponding forces in the tangential and normal directions. The force associated with the normal acceleration is known as the centripetal force. The centripetal force is a real force on the body toward the center of .rotation. The so-called centrifugal force is an apparent force on the body directed away from the center of rotation. The centripetal and centrifugal forces are equal in magnitude but opposite in sign.
Figure 16.4 Torsional Pendulum
J
G
L
)9
16-4
FEReview Manual
-
•
For the torsional pendulum, stant kt can be written
the torsional spring con-
16.20
_
, ;:
.....
(A) 2.25 rad/s (clockwise) (B) 3.25 rad/s (counterclockwise) (e) 5.50 rad/s (clockwise) (D) 12.5 rad/s (clockwise) #100694
The solution to Eq. 16.20 is directly analogous to the solution for the spring-mass system.
B(t)
=
Bocoswnt
+ (::)
sinwnt
16.21
SAMpLE' PROBLEMs
Solution:
Solut
Find the instantaneous center of rotation. The absolute velocity directions at points Band e are known. The instantaneous center is located by drawing perpendiculars to these velocities as shown. The angular velocity of any point on rigid body link Be is the same at this instant.
The asse fron be 3 ity (
1. Why does a spinning ice skater's angular velocity increase as she brings her arms in toward her body? (A) Her mass moment of inertia is reduced. (B) Her angular momentum is constant. (e) Her radius of gyration is reduced. (D) all of the above
B
VB
= 50
t
4m II ------------1,-
m/s I/IC
1° I
I 13m
#99689
I I I
Solution:
4
As the skater brings her arms in, her radius of gyration and mass moment of inertia decrease. However, in the absence of friction, her angular momentum, h, is constant. From Eq. 16.9,
vc-
VB
h I
I
C
= ABwAB
= (5 m) ( 10 rad) -s- = 50 m/s
w=-
Since angular velocity, w, is inversely proportional to the mass moment of inertia, the angular velocity increases when the mass moment of inertia decreases.
wBC
=
VB
OB
50 m
= 4 ~ = 12.5 rad/s
[clockwise]
Answer is D.
Answer is D. 2. Link AB of the linkage mechanism shown in the illustration rotates with an instantaneous counterclockwise angular velocity of 10 rad/s, What is the instantaneous angular velocity of link Be when link AB is horizontal and link CD is vertical?
3. Two 2 kg blocks are linked as shown. Assuming that the surfaces are frictionless, what is the velocity of block B if block A is moving at a speed of 3 m/s?
4. ba fri, ca: ill,
",WAB = 10 rad/s (counterclockwise) ~/~5m --5-m--'I
~
So
I
4 C . -
T fo
5m
T D._
~ Professional Publicalions, Inc.
--
•
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FEReview Manual
_
1. A 2 kg mass swings in a vertical plane at the end of a 2 m cord. When = 30°, the magnitude of the tangential velocity of the mass is 1 ui]«. What is the tension in the cord at this position?
o
e
f----'--'------j
(A)
10 m/s ~
10.0 m/s
(B) 15.0 m/s (C) (D)
16.2 m/s 18.5 m/s #106193
4. A car travels around an unbanked 50 m radius curve without skidding. The coefficient of friction between the tires and road is 0.3. What is the car's maximum speed? (A) (B) (C) (D)
14 25 44 54
km/h km/h km/h km/h #107394
5. Traffic travels at 100 km/h around a banked highway curve with a radius of 1000 m. What banking angle is necessary such that friction will not be required to resist the centrifugal force?
(A) 1.4°
(A) 18.0 N
(B) 2.8° (C) 4.5° (D) 46°
(B) 19.6 N (C) 24.5 N (D) 29.4 N
#2107689 #104
694
Forthe following problems use the NCEES Handbook as your only reference. 2. of in of
A 2 kg mass swings in the horizontal plane of a circle radius 1.5 m and is held by a taut cord. The tension the cord is 100 N. What is the angular momentum the mass? (A) 5.77 N·m·s (B) 26.0 N-m· s (C) 113Nun-s (D) 150 N-m·s #105691
6. The center of gravity of a roller coaster car is 0.5 ill above the rails. The rails are 1 m apart. What is the maximum speed that the car can travel around an unbanked curve of radius 15 m without the inner wheel losing contact with the top of the rail? (A) 8.58 m/s (B) 12.1 m/s (C) 17.2 m/s (D) 24.2 m/s #2133689
3. A disk rolls along a fiat surface at a constant speed of 10 m/s. Its diameter is 0.5 m. At a particular instant, point P on the edge of the disk is 45° from the horizontal. What is the velocity of point P at that in-
stant? Professional Publkcfions, Inc.
7. A 50 kg cylinder has a height of 3 m and a radius of 50 em. The cylinder sits on the z-axis and is oriented with its major axis parallel to the y-axis. "What is the mass moment of inertia about the z-axis?
---
(A) (B) (C) (D)
4.1 kg·m2 16 kg·m2 41 kg·m2 150 kg·m2
0.47 kg·m2 0.56 kg·m2 0.87 kgrrr' 3.7 kg·m2
(A) (B) (C) (D) #3553794
#2095689
8. A uniform thin disk has a radius of 30 em and a mass of 2 kg. A constant force of 10 N is applied tangentially at a varying, but unknown, distance from the center of the disk. The disk accelerates about its axis at 3t rad/s2. What is the distance from the center of the disk at which the force is applied at t = 12 s?
11. A 1 kg uniform rod 1 m long is suspended from the ceiling by a frictionless hinge. The rod is free to pivot. What is the product of inertia of the rod about the pivot point?
(A) 0 kg·m2 Cl,W
(B)
~
0.045 kg-rrr'
(C) 0.13 kg.m2 (D) 0.33 kg·m2 R = 30 em
lever
#2096689
12. A wheel with a radius of and accelerates clockwise. The rad/s'') of the wheel is defined the resultant linear acceleration rim at t = 2 s?
(A) 32.4 em (B) (C)
(D)
3.6.0 em 54.0 em 108 em
0.75 m starts from rest angular acceleration (in by a = 6t - 4. What is of a point on the wheel
(A) 6 m/s2 (B) 12 m/s2 (C)
#3975395
13 m/s2
(D) 18 m/s2 9. A torsional pendulum consists of a 5 kg uniform disk with a diameter of 50 em attached at its center to a rod 1.5 m in length. The torsional spring constant is 0.625 Nrri/rad. Disregarding the mass of the rod, what is the natural frequency of the torsional pendulum? (A) 1.0 rad/s (B) 1.2 radys (C) 1.4 rad/s (D) 2.0 rad/s #4170
#2893 687
13. A uniform rod (AB) of length L and weight W is pinned at point C and restrained by cable OA. The cable is suddenly cut. The rod starts to rotate about point C, with point A moving down and point B moving up. What is the instantaneous linear acceleration of point B?
795
10. A 3 kg disk with a diameter of 0.6 m is rigidly attached at point B to a 1 kg rod 1 m in length. The rod-disk combination rotates around point A. What is the mass moment of inertia about point A for the combination?
~
a
0.6 m
I__
'-
..;. •••
---------------------Cl~" L
Professional Publications, Inc.
16-8
FEReview Manual
(A) (B)
(C)
(D)
1
3g 16
(A) 8° (B) 21°
g
(C) 36°
4
(D) 78°
_
3g
#1582689
7
3g 4 #2911
687
14. A uniform rod (AB) of length L and weight W is pinned at point C. The rod is accelerating with an instantaneous angular acceleration (in rad/s'') of 0: = 12g /7 L. What is the instantaneous reaction at point C?
17. A wheel with a 0.'75 m radius has a mass of 200 kg. The wheel is pinned at its center and has a radius of gyration of 0.25 m. A rope is wrapped around the wheel and supports a hanging 100 kg block. When the wheel is released, the rope begins to unwind. What is the angular acceleration of the wheel? (A) 5.9 rad/s2 (B) 6.5 rad/s2 (C) 11 rad/s2 (D) 14 rad/s2 #2907687
B
A
• 1. 4
Solution 1:
L
-
(B)
-
(D)
Use tangential and normal components.
W
(A)
(C)
SOLUTIONS TO FE-STYLE EXAM PROBLEMS
4
W 3 4W 7
7W
12
#2912687
15. A 1530 kg car is towing a 300 kg trailer. The coefficient of friction between all tires and the road is 0.80. How fast can the car and trailer travel around an unbanked curve of radius 200 m without either the car or trailer skidding? (A) 40.0 km/h (B) 75.2 km/h (C) 108.1 km/h (D) 143 km/h
an
r
Sum forces in the normal direction.
= man = T
LPn
= =
man
- Wsin60°
+ mg sin 60°
(2 kg) (0.5 ~) + (2 kg) (9.81 ~) (sin 60°)
= 18.0 N
#1579689
Professional Publications, Inc.
2 m
= 0.5 m/s2
T
16. A 1530 kg car is towing a 300 kg trailer. The coefficient of friction between all tires and the road is 0.80. The car and trailer are traveling at 100 km/h around a banked curve of radius 200 m. What is the necessary banking angle such that tire friction will not be necessary to prevent skidding?
2 (1 7)2
v = -...1. = -'---"'-.:....-
Answer is A.
Solution 2: tension
= centripetal
force
---
------------------------------
lKinetics of Rotational Motion
mv2t T= __ r
mv2
--
v
= rmv = r mw
ho
=
V
(0.3) (9.81 ~) (50 m) = 12.13
m/s
(12.13 ~) (3600 ~) v=
5.77 rad/s 2
= vf1gr =
100 N (2 kg)(1.5 m) =
=f1mg
r
r
16-9
1000 = (1.5 m)2 (2 kg) (5.77 r:d)
r:
= 43.67 km/h
26.0 N·m·s Answer is C.
Answer is B. Solution 5: Since there is no friction force, the superelevation B, can be determined directly.
Solution 3: Use the instantaneous center of rotation to solve this problem. Assume the wheel is pinned at point A.
angle,
v2 tanB = ~ gr
e = tan "
(;!) (100 ~) (
(1000 ~))
2
3600 ~
=tan-l~--~--~~------~~
(9.81 ~) (1000 m) Z2
= (2)(0.25
=
m)2 - (2)(0.25 m)2(cos 135°)
4.50°
[lawof cosines] Answer
= 0.2134 m2 I
= VO.2134 m2 Zvo
vp
= ---;;:= =
= 0.462 m
Solution 6:
(0.462 m) (10 ~) 0.25 m
The wheel will lose contact with the top of the rail when the reaction on the rail is zero. Refer to the following illustration. Wheel A is the inner wheel.
18.5 m/s
The velocity of point P is perpendicular
is C.
to the line AP.
Answer is D.
YCG
= 0.5
m
Solution 4: The car uses friction to resist the centrifugal force.
______________________________
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16-10
FEReview Manual
1
The forces acting on the car are the centrifugal force, Fe, its weight, and the reaction at the outer wheel. (The reaction at the inner wheel is zero.) Take moments about rail B.
L MB = 0 = W XCG -
Solution 9: The radius of the disk is
R= .~D 2
FeYCG
50 em
- -(2) (100 ::)
mv;YcG r
=mgxCG-
_
_...::..::.....:....:c.
=
The mass moment
v« = JgXCG'r YCG (~)
of inertia
of the disk is
= ~MR2
I
(9.81 ~)
TJ
0.25 m
2
(15 m) = (~) (5 kg)(0.25
m)2
0.5 m
=
= 0.15625 kg-rrr'
12.1 m/s
Natural frequency can be determined for the torsional spring constant.
Answer is B.
kt
Solution 7: Find the formula for Ix in the Mass Moments of Inertia table. m(3R2 + 4h2) Ix = ---''------"12 (50 kg)[(3)(0.5
m)2
+ (4)(3
153.1 kg-rrr'
= w2I
w=/¥
~----,--,-0.625 N·m rad 0.15625 kg·m2
m)2]
12 =
from the equation
2 rad/s
=
(150 kgrn") Answer is D.
Answer is D. Solution 10:
Solution 8:
The mass moment of inertia of the rod about its end is
The centroidal
moment
of inertia is
ML2 = -3-
Irod,A
I = ~mR2 2
=
= 0.09 The acceleration a
at
t =
= 3t = =
(1 kg)(1 m)2
(0.5)(2 kg)(0.3 m)2
3
kg·m2
=
12 s is
(3
r:d)
The mass moment center is
(12 s)
of inertia
MR2
(3 kg)
Ia r=F
(Tr 2
=
2 (0.09 kg·m ) (36 ~)
The distance
10 N 0.324 m
0.135 kg·m2
T
AC is AC
=
tl
J AB2 + BC
2
(32.4 em) (1 m)2
Answer is A.
Professional Publications, Inc.
of the disk about its own
Idisk,c = -2-
36 rad/s2
Mo=Fr=Ia
=
0.33 kg.m2
..
=
1.04
+ (O.~
ill
r
rn
._--------_ _---------------_ .._
.
•
Using the parallel axis theorem, the mass moment of inertia of the disk about point A is Idisk,A
The resultant
acceleration is
a Va; + a~ =
+ mdiskAC2 = 0.135 kg·m2 + (3 kg) (1.04 m)2 = Idisk,e
(6 ~)
= 3.38 kgrrr'
=
2
+ (12
~)
2
13.4 m/s2
The total moment of inertia of the rod and disk is IA = =
+
Irod,A
Answer
Idisk,A
+ 3.38
0.33kg·m2
kgm"
= 3.71 kg.m2 Answer
is C.
Solution 13: Point C is L/4 from the center of gravity of the rod. The mass moment of inertia about point C is
is D.
Ie
=
Solution 11: =
The product of inertia for the rod is zero because the pivot point lies on an axis of symmetry.
= Answer is A. =
leG
+ md2
(112) mL
2
+m (~
mL2 C~+ 116) (:8)mL
r
2
Solution 12: The angular acceleration
at t
=
2 s is
The sum of moments on the rod is
a(2 s) = (6)(2 s) - 4 = 8 rad/s2 The equation for the angular velocity is w =
=
J
a(t)dt
WL mgL
J(6t - 4)dt 2
= 3t
-
4
4
4t +wo The angular acceleration is
However, the wheel starts from rest, so Wo = O. At t
=2
LMe
s, the angular velocity is w(2 s)
=
Ie
mgL 4 (:8) mL2
(3)(2 S)2 - (4)(2 s)
= 4
The tangential
a=---
rad/s
acceleration of the point is
= (0.75 m) =
rad) (8 ~ The tangential
6 m/s2
The normal acceleration the wheel) is
12g 7L acceleration of point B is
(directed toward the center of
= (~) = (0.75 m) ( 4 =
12 m/s2
(~1)
3g
rad)2
S
7
Answer
is C.
,\
I
II
16-12
FEReview Manual -----------------------------
_
Solution 14:
Solution 16:
The mass moment of inertia of the rod about its center of gravity is
The velocity is
(100 k:)
(1000 :)
v - --'-----'---;;---Take moments about the center of gravity of the rod. All moments due to gravitational forces will cancel. The only unbalanced force action on the rod will be the reaction force, Rc, at point C.
LMCG
L
-
3600 ~
=
27.78 m/s
The necessary superelevation
angle is
=Rc (~)
MCG =
ICGacG
Rc= 4W 7 Answer is C.
Answer is B.
Solution 15:
Solution 17:
To keep the vehicle and trailer from skidding, the centripetal force must be less than or equal to the frictional force. At the limit,
The mass moment of inertia of the wheel is
Fe =Ff The unbalanced
= --
m
=
= mblockR(g =p,g
The acceleration
is given by
(0.8) (9.81 ~) M=Ia mblockR(g
The normal acceleration can be calculated from the tangential velocity.
= =
m/s
(39.6 ~)
==
mwheelr2a
(100 kg)(0.75 m) (9.81 ~)
- (200 kg)(0.25 m)2
(7.848 ~) (200 m)
= 39.6
- Ra)
mblockRg
_
Janr
J
- a)
- Ra)
= 7.848 m/s2
Vt
= mR(g
M = F R = (mg - ma)R
p,N
an=-
m p,mg
torque (moment) on the wheel is
=
+ (100 kg)(0.75
m)2
2
10.7 rad/s
Answer is C.
(3600 ~)
1000 : = 143 km/h Answer is D.
Professional Publications, Inc.
----
Energy and Work
Wark, W, is the act of. changing the energy of a mass. Work is a signed, scalar quantity. Work is positive when
a force acts in the direction of motion and moves a mass from one location to another. Work is negative when a force acts to oppose motion. (Friction, for example, always opposes the direction of motion and can only do negative work.) The net work done on a mass by more than one force can be found by superposition. The work performed by a force is calculated as a dot product of the force acting through a displacement.
Nomenclature a e E F g \
\gc h I Imp k
m p
r
t v W X
acceleration coefficient of restitution energy force gravitational acceleration gravitational constant (32.2) height above datum mass moment of inertia impulse spring constant mass linear momentum distance time velocity work displacement
w= ft-lbf lbf
J N
Ibm-ftj'lbf-sec-'
angular velocity
17.1
Kinetic Energy Kinetic energy is a form of mechanical energy associated with a moving or rotating body. The linear kinetic energy of a body moving with instantaneous linear velocity v is
1
2
ft
m
KE = Zmv
lbm-ft2 Ibf-sec lbf/ft lbm lbm-ft/sec ft sec ft/sec ft-lbf ft
kg·m2 N·s
KE= mv 2gc
rad/sec
rad/s
[SI]
17.20
[U.S.]
17.2b
2
N/m kg kg-m.'s m s m/s J m
Symbols w
l=
ENERGYANDWORIC The energy of a mass represents the capacity of the mass to do work. Such energy can be stored and released. There are many forms that the stored energy can take, including mechanical, thermal, electrical, and magnetic energies. Energy is a positive, scalar quantity, although
The rotational kinetic energy of a body moving with instantaneous angular velocity w is KE = ~Iw2
[S1]
17.30
KE = Iw2 2gc
[U.S.]
/7.3b
For general plane motion in which there are translational and rotational components, the kinetic energy is the sum of the translational and rotational forms. The change in kinetic energy is calculated from thedifference of squares of velocity, not from the square of the velocity difference. .I
17.40
17.4b
the change in energy can be either positive or negative .
•
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17-2
FEReview Manual
1
Potential Energy Potential energy (also known as gravitational potential energy) is a form of mechanical energy possessed by a mass due to its relative position in a gravitational field. Potential energy is lost when the elevation of a mass decreases. The lost potential energy usually is converted to kinetic energy or heat. PE
=
mgh
PE= mgh
[81]
[u.s.]
17.5a
_
the work done by the constant gravitational force to the change in kinetic energy.
LINEAR IMPULSE Impulse is a vector quantity equal to the change in momentum. Units of linear impulse are the same as those for linear momentum: Ibf-sec or N-s. Figure 17.1 illustrates that impulse is represented by the area under the force-time curve.
11.5b
t2
Imp =
i
17.9
Fdt
h
Elastic Potential Energy Figure 17.1 Impulse
17.6
If the applied force is constant, impulse is easily calculated.
ENERGY CONSERVATION ......................... PRINCIPLE
11.10
According to the energy conservation principle, energy cannot be created or destroyed. However, energy can be transformed into different forms. Therefore, the sum of all energy forms of a system is constant.
'E,E = constant
The change in momentum is equal to the impulse. This is known as the impulse-momentum principle. For a linear system with constant force and mass, Imp
17.7
For many problems, the total energy of the mass is equal to the sum of the potential (gravitational and elastic) and kinetic energies.
F(t2 - h) Fdt
Conversion of one form of energy into another does not violate the conservation of energy law. Most problems involving conversion of energy are really special cases. For example, consider a falling body that is acted upon by a gravitational force. The conversion of potential energy into kinetic energy can be interpreted as equating
= 6.(mv) = mdv
F=
17.8 Generally, the principle of conservation of energy is applied to mechanical energy problems (i.e., conversion of work into kinetic or potential energy).
17.11
= 6.p
Rewriting this equation for a constant force and mass moving in any direction demonstrates that the impulsemomentum principle follows directly from Newton's second law.
Because energy can neither be created nor destroyed, external work performed on a conservative system must go into changing the system's total energy. This is known as the work-energy principle.
-
00 VI
Tl: to
gc
A spring is an energy storage device because a compressed spring has the ability to perform work. In a perfect spring, the amount of energy stored is equal to the work required to compress the spring initially. The stored spring energy does not depend on the mass of the spring. Given a spring with spring constant (stiffness) k, the spring's elastic potential energy is
1
mdv -=ma dt
[81]
11.12
[81]
11.13
[81]
11.14
IMPACTS According to Newton's second law, momentum is conserved unless a body is acted upon by an external force such as gravity or friction. In an impact or collision, contact is very brief, and the effect of external forces is insignificant. Therefore, momentum is conserved, even though energy may be lost through heat generation and deforming the bodies.
Figl
_---------------
1_
Consider two particles, initially moving with velocities and V2 on a collision path, as shown in Fig. 17.2. The conservation of momentum equation can be used to find the velocities after impact, v~ and ~.
Energy and Work
17-3
Solution:
VI
1 2mv2
KE= d(KE) dt
=
(2) (~)
(mv) (dV) dt
2
=mva This combination of variables (kg·m2/s3) corresponds to a watt (i.e., power).
Figure 11.2 Dired Centrallmpad
.~
Answer is D.
m1
V1
Problems 2 and 3 refer to the following illustration. A 50 kg block is released down a curved, frictionless surface. The radius of the curve is 5 m.
The impact is said to be an inelastic impact if kinetic energy is lost. The impact is said to be perfectly inelastic or perfectly plastic if the two particles stick together and move on with the same final velocity. The impact is said to be an elastic impact only if kinetic energy is conserved.
A simple way of determining whether the impact is elastic or inelastic is by calculating the coefficient of restitution, e. The coefficient of restitution is the ratio of relative velocity differences along a mutual straight line. The collision is inelastic if e < La, perfectly inelastic if e = a, and elastic if e = 1.0. (When both impact velocities are not directed along the same straight line, the coefficient of restitution should be calculated separately for each velocity component.)
5m
What point A?
2.
is the tangential
(A) 6.52
velocity of the block at
m/s
(B) 9.22 sss]« (C) 42.5 mls (D) 85.0
mls #109
689
Solution:
relative separation velocity e= . relative approach velocity
The vertical distance between points 0 and A is h
11.17
=
(5 m) sin 60°
=
4.33 m
This drop in elevation corresponds to a decrease in potential energy and an increase in kinetic energy.
t.PE = t.KE mgt.h
1. The first derivative of kinetic energy with respect to time is (A) force. (B) momentum. (C) work. (D) power.
v
#108
691
=
V2gf:lh
=
9.22
Answer is B.
=
mls
=
V (2)
mv2 -2-
(9.81 ~)
(4.33 m)
=
FEReview ~anual --
17-4
What is the instantaneous acceleration in the direction of travel of the block at point B? 3.
(A) 0 m/s2 (B)
2.45
m/s2
Two balls, both of mass 2 kg, collide. head on. The velocity of each ball at the time of the collision is 2 m/s. The coefficient of restitution is 0.5.
(A) 1 mls and -1 tsi]» (B) 2 mls and -2 mls (C) 3 iii]« and -3 sxs]» (D) 4 mls and -4 mls
#110689
Solution:
#112691
At point B, all of the energy of the mass is kinetic, and the velocity is maximum. Acceleration is the rate of change of velocity. Since the velocity is maximum at point B, the acceleration is zero. There is centripetal acceleration directed away from point C, but this is not in the direction of travel.
. Solution: From the definition of coefficient of restitution, e vi - v~
4. A 1 kg disk with a diameter of 10 em and a width of 4 em is placed on edge at the top of an inclined ramp 1 m high. The ramp is inclined at 10°. At the bottom of the ramp is a spring whose spring constant is 2000 N 1m. The disk rolls down the ramp and compresses the spring while coming to a complete stop. What is the maximum compression of the spring?
Vi -Vi =1
2 V2 -
Answer is A.
=
VI
e(v2 - VI)
A
m-2S m). = (0.5) ( -2 S = -2
mls
[I]
#111
193
Since
Solution:
VI
= 2 mls VI
At the top of the ramp, all of the energy is gravitational potential energy; at the bottom, the energy is spring potential energy. Neglecting the small toss of gravitational potential energy in deflecting the spring, the energy balance is mgh
=
and
V2
=
-2
mis,
+ V2 = 2 ~ + (-2
~)=
a
So,
Solve Eqs. I and II simultaneously
1 "2kX2
vi
=
by adding them.
-1 mls
v; = 1 mls X=
J2~gh'
Answer is A. (2) (1 kg) (9.81 ~) (1 m) 2000 N m =
0.099 m
(9.9 cin)
is A.
Problems 5 and 6 refer to the following situation. Professional Publications, Inc.
6.
7. 5 fi t<
From the conservation of momentum,
(A) 9.9 cm (B) 11.4 em (C) 11.7 cm (D) 14.1 em
Answer
Sol
5. What are the final velocities of the balls?
(C) 4.91 m/s2 (D) 19.6 m/s2
I
-
What is the loss of energy in the collision?
(A) 1.4 N-m (B} 2.3 N-m (C) 6.0 N·m (D) 8.6 N-m #113691
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Energy and Work
Solution:
B
Each ball possesses kinetic energy before and after the collision. The velocity of each ball is reduced from 12m/s] to 11 m./s]. ~KE
17-5
=
(KE)i - (KE) f
=
(2)
-A
c
2 mv2) ( m;i --t
~ (2) [(2 kg) ~2
~)2 _ (2 kg) ~'
~)21
=6N-m compressed position
Answer is C.
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7. A 2 kg ball of clay moving at 40 m/s collides with a 5 kg ball of clay moving at 10 m/s directly toward the first ball. What is the final velocity if both balls stick together after the collision? 1. What is the initial spring compression?
(A) 4.29 m/s (B) 23.0 m/s
(A) 0.96 m (B) 1.3 m
(C)
30.0 m/s (D) 42.9 m/s
(C) lAm (D) 1.8 m
#114 689
#115 689
Solution:
2. What is the kinetic energy of the mass at point A?
vI
Since the balls stick together, = V2 and e = O. Thus, the collision is perfectly inelastic. Only momentum is conserved.
(2 kg) (40
7)
+ (5 kg) (-10
7) = (2 v'
=
(A) 19.8 J (B) 219 J (C) 392 J (D) 2350 J #116689
kg+5
kg)v'
4.29 m/s
Answer is A.
3.
What is the velocity of the mass at point A?
(A) (B) (C) (D)
3.13 m/s 4.43 m/s 9.80 m/s 19.6 m/s #117 689
FE-STYLE EXAM PROBLEMS Problems 1-4 refer to the following situation. • The mass m in the following illustration is guided by the frictionless rail and has a mass of 40 kg. • The spring constant, k. is 3000 N/m. • The spring is compressed sufficiently and released, such that the mass barely reaches point B.
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4. What is the energy stored in the spring if the spring is compressed 0.5 m? (A) 375 J (B) 750 J .' (C) 1500 J (D) 2100 J #118689
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