Pre-Quiz-04 Worksheet/Practice; Allowed Equations (below); Topic: forces a1 a2 a (conserve string)
F ma m
tan
L y L x
Fy Fx
d 2x dt
2
; sin
m Ly L
dv dt
; K
F y F
F Fr , k F N
; cos
; ar
Lx L
F x F
v2 r
; a
; L
F F ; 1
m1 m2
2
RPM
L x 2 Ly 2 [length ]; F
rev min
2 rad 60 s
;
Fx 2 F y 2 [force] ;
Problem 1: A small weight of mass m hangs from a string in an automobile which accelerates at rate
A . The acceleration of gravity is g . What is the static angle of the string from the vertical, and what is its tension?
Problem 2: If you have the nerve and a tight grip, you can yank a tablecloth out from under the dishes on a table. An expensive crystal glass of mass m and distance d from the edge in gravitational field
g comes to rest before falling off the table. Assume that the coefficient of friction of the glass sliding on either the tablecloth or sliding on the tabletop is K . Question (a): What is the longest time t max in which the cloth can be pulled out and the glass come to
rest (i.e., the total time for the process)?
in which the tablecloth alone may be pulled out? Question (b): What is the maximum allowed time t max
(For the trick to be effective the cloth should be pulled out so rapidly that the glass does not move appreciably.) Problem 3: An automobile of mass m enters a turn whose radius is R . The road is banked at angle , and the coefficient of friction between wheels and road is .
(1.1) Find the maximum and minimum speeds for the car to stay on the road without skidding sideways.
Problem 4: In a concrete mixer, cement, gravel, and water are mixed by tumbling action in a slowly rotating drum. If the drum spins too fast the ingredients stick to the drum wall instead of mixing (on planet earth, gravity is g ). Assume that the drum of a mixer has radius R and that it is mounted with
its axle horizontal. Question (a): What is the fastest the drum can rotate without the ingredients sticking to the wall all the time?
Question (b): Compute the fictitious force due to the centripetal acc eleration and the normal force, each at the bottom of the mixer and at the top of the mixer.
Problem 5: A particle of mass m slides without friction on the inside of a cone. The axis of the cone
is vertical, and gravity is directed downward. The apex half-angle of the cone is , as shown. The path of the particle happens to be a circle in a horizontal plane. The speed of the particle is v0 .
(1.2)
Draw a force diagram and find the radius of the circular path in terms of v0 , g , and .
Pre-Quiz-04 Worksheet/Practice; Allowed Equations (below); Topic: forces a1 a2 a (conserve string)
F ma m
tan
L y L x
Fy
d 2x dt
2
; sin
Fx
m Ly L
dv dt
; K
F y F
F Fr , k F N
; cos
; ar
Lx L
F x F
v2 r
; a
; L
F F ; 1
2
m1 m2
RPM
L x 2 Ly 2 [length]; F
rev min
2 rad 60 s
;
Fx 2 F y 2 [force] ;
Problem 1: A small weight of mass m hangs from a string in an automobile which accelerates at rate
A . The acceleration of gravity is g . What is the static angle of the string from the vertical, and what is its tension? We shall analyze the problem both in an inertial frame and in a frame accelerating with the car.
(1.3) Let’s consider the inertial system first,
F
x
mA Tx T sin ;
st
F
y
0 Ty mg T cos mg ;
(1.4)
nd
Dividing the 1 equation of (1.4) by the 2 , we obtain an angle and tension of,
tan
sin cos
mA / T mg / T
A m/T g m/T
A
; T Tx 2 Ty 2 ( mA) 2 ( mg ) 2 m A2 g 2 ; g
(1.5)
Now, let’s consider the system accelerating with the automobile, in which a fictitious force F fict appears,
F
x
0 Tx Ffict T sin mA T sin mA;
F
y
0 Ty mg T cos mg ;
(1.6)
From (1.6), we obtain the results (1.5) quite immediately. Problem 2: If you have the nerve and a tight grip, you can yank a tablecloth out from under the dishes on a table. An expensive crystal glass of mass m and distance d from the edge in gravitational field
g comes to rest before falling off the table. Assume that the coefficient of friction of the glass sliding on either the tablecloth or sliding on the tabletop is K . Question (a): What is the longest time t max in which the cloth can be pulled out and the glass come to
rest (i.e., the total time for the process)? There are two legs of the journey: (1) the glass is accelerated by the sliding friction while it is on the tablecloth, and over a distance x1 d / 2 to a maximum velocity of v0 (2) the glass is de-celerated over the remaining
nd distance x2 x1 d / 2 by the sliding friction to a velocity of v 0 . Newton’s 2 Law for each leg of the
journey, in which the normal-force is always F N mg , is,
F ma 1
1
K
FN K mg a1 K g ;
F
2
ma2 K FN K mg a2 K g a1; (1.7)
The constraint is that the glass must undergo accel eration a1 over a distance x1 , and a2 a1 over x2 , used v f 2 vi 2 2 ax x
used v f 2 vi 2 2 ax x
used v f vi ax t
v0 02 2a1x1 2 K g ( d / 2)
used v f vi ax t
K gd 0 a1t1 ; v 0 v0 2 2a2x 2 v0 a 2t 2 ;
(1.8)
The expressions (1.8) yield a value for tmax t1 t 2 as,
tmax t1 t 2
v0 a1
v v0 a2
v0 a1
0 v0
a1
2
v0 a1
2
K gd K g
2
d K g
;
(1.9)
in which the tablecloth alone may be pulled out? Question (b): What is the maximum allowed time t max
Since t1
12 t max t 2 , we have tmax
d / ( K g ) .
(For the trick to be effective the cloth should be pulled out so rapidly that the glass does not move appreciably.) Problem 3: An automobile of mass m enters a turn whose radius is R . The road is banked at angle , and the coefficient of friction between wheels and road is .
(1.10) Find the maximum and minimum speeds for the car to stay on the road without skidding sideways. Let’s choose the x- and y-axis to be horizontal and vertical respectively (i.e., do not use “incline-planecoordinates”). Let Case (1) be the car having maximum velocity, and Case (2) be the car having minimum rd st velocity. Then, the friction-force is in the 3 quadrant for Case (1) and in the 1 quadrant for Case (2), and the nd nd normal force is in the 2 quadrant. Correspondingly, Newton’s 2 Law in the x-direction is,
F
x
m(aR ) FN , x F fr , x FN sin FN cos m
v 2 R
case 1 ; ( ) ( case 2 );
(1.11)
nd
The normal force F N is unknown in (1.11), and we require Newton’s 2 Law in the y-direction to get it,
N F y 0 FN , y F fr , y Fg FN cos FN sin mg F N
solve for F
mg cos sin
;
(1.12)
Solving (1.11) for v (i.e., v is that of case-1, and v is that of case-2, so obviously v v ), which we take to be positive, and using (1.12) to eliminate F N , we have,
v
R
F N m
(sin
cos )
R
mg cos sin
m
(sin
cos ) Rg
tan
; 1 tan
(1.13)
Writing (1.13) as an inequality, we have, v v v
Rg
tan
tan ; v Rg 1 tan 1 tan
(1.14)
Problem 4: In a concrete mixer, cement, gravel, and water are mixed by tumbling action in a slowly rotating drum. If the drum spins too fast the ingredients stick to the drum wall instead of mixing (on
planet earth, gravity is g ). Assume that the drum of a mixer has radius R and that it is mounted with its axle horizontal. Question (a): What is the fastest the drum can rotate without the ingredients sticking to the wall all the time?
Consider an ingredient of mass m (our results should be independent of m , however). The mass m must “fall” at some point along the rotation, so this is satisfied on asserting the normal force to be zero at the apex of rotation,
F
y
may m( aR ) m
v 2 R
N 0
solve for v N mg 0 mg v
solve for Rg R
g R
; (1.15)
Indeed, our result is independent of m . Question (b): Compute the fictitious force due to the centripetal acc eleration and the normal force, each at the bottom of the mixer and at the top of the mixer. Bottom of mixer Fictitious radial force Normal force
F fic maR m
Top of mixer
v
2
R
m
Rg R
mg
N F g Ffic mg mg 2mg ;
F fic maR m
v2 R
mg
N F g Ffic mg mg 0 ;
Problem 5: A particle of mass m slides without friction on the inside of a cone. The axis of the cone
is vertical, and gravity is directed downward. The apex half-angle of the cone is , as shown. The path of the particle happens to be a circle in a horizontal plane. The speed of the particle is v0 .
(1.16)
Draw a force diagram and find the radius of the circular path in terms of v0 , g , and . The free body diagram is as follows, in which appears a fictitious force. The angle can be written in terms of the reaction-forces,
F N
FN , xi FN , y j , as, ˆ
ˆ
F N , y
F N , x
ma
F fict
a a Ri ˆ
mg
v0 2 R
i ; F fict ma R m ˆ
v0 2 R
; tan
F N , y F N , x
;
(1.17)
nd
Newton’s 2 Law appears as,
F
x
m(aR ) FN , x FN , x
mv0 2 R
;
nd
F
y
0 FN , y mg FN , x tan mg F N , x
mg tan
; (1.18)
st
Using the 2 equation of (1.18) to eliminate F N , x in the 1 equation of (1.18), we get, mg tan
mv0 2 R
R solve for R
v0 2 g
tan
(1.19)
