Version 024 Exam 3 hoffmann (57870)

This print-out should have 15 questions.

Multiple-choice questions may continue on
the next column or page find all choices
before answering.
001 10.0 points
~ = (1.7 N) + (3.6 N) is applied
A force F
to an object that is pivoted about a fixed
axis aligned along the z coordinate axis. The
~ = (4.1 m) +
force is applied at the point R
(4.5 m) .
Find the z-component of the net torque.
1. 2.24
2. 6.7
3. 0.12
4. 0.57
5. 4.61
6. 7.11
7. 4.4
8. 3.62
9. 2.64
10. 0.72
Correct answer: 7.11 N m.
Explanation:
Basic Concept:

1. 28939.7 J correct
2. 162.9 J
3. 57879.3 J
4. 2.16 J
5. 1466.1 J
6. 733.05 J
7. 1023.53 J
Explanation:
Let :

I = 18.1 kg m2 ,
f = 9 rev/s , and
F = 3.6 N .

Applying the work-energy principle,

W = K =
=

1
1
I 2 = I (2 f )2
2
2

1
(18.1 kg m2 )
2
2

~
~ = ~r F
Solution: From the definition of torque
~ F
~
~ = R
= [Rx + Ry ] [Fx + Fy ]
= [(4.1 m) + (4.5 m) ] [(1.7 N) + (3.6 N) ]
= [(4.1 m) (3.6 N) (4.5 m) (1.7 N)]k
= (7.11 N m) k

In this case, the net torque only has a zcomponent.

002 10.0 points
A flywheel of radius 0.6 m and moment of
inertia of 18.1 kg m2 rotates initially at a
rate of 9 rev/s.
If a force of 3.6 N is applied tangentially
to the flywheel to slow it down, how much
work will be done by this force in bringing the
flywheel to a stop?

4 (9 rev/s)

2
rev

= 28939.7 J .
003 10.0 points
A rod of mass m and length L is hinged with a
frictionless hinge at one end. The moment of
1
inertia about the center of mass is
m L2 .
12
Attached to the end of the rod opposite to
the hinge there is a mass of magnitude 2 m.
The rod is released from rest in the horizontal
position.

m
2m
L
What is the speed of the mass 2 m when
the rod passes through the vertical position?

Version 024 Exam 3 hoffmann (57870)

2

2


Consider the mass at the end of the rod to be
1
L
1 1
2
2
mL + m
=
a point particle.
2 12
2
2
r
1
7gL
= m L2 2 .
1. k~v2m k =
6
15
r
1
Alternately, I rod,end = L2 , so
5gL
3
2. k~v2m k =
2


r
1 rod,end 2 1 1
3gL
2
rod
Kf = I
=
m L 2
3. k~v2m k =
2
2
3
2
p
1
4. k~v2m k = g L
= m L2 2 .
6
r
g
Since vm = L ,
5. k~v2m k =
L
Uirod + Ui2m = Ufrod + Kf2m + Kfrod
p
6. k~v2m k = 2 g L
1
3 m g L = m g L + m ( L)2
r
2
15 g L
1
correct
7. k~v2m k =
+ m L2 2
7
6
r
2gL
18
m
g
L
=
3
m
g L + 6 m (L )2
8. k~v2m k =
5
+ m L2 2
r
2
2gL
15 m g L = 7 m (L )2 = 7 m v2m
,
9. k~v2m k =
r
3
15
r
k~v2m k =
gL.
gL
7
10. k~v2m k =
2
004 10.0 points
Explanation:
Inside a hollow pipe of inner radius R, a hoop
Let us measure heights from the point at
of mass m and radius r R rolls without
the end of the rod when it is vertical. The inislipping.
tial energy is all potential, with a magnitude
Ei = Uirod + Ui2m
= m g L + (2 m) g L = 3 m g L .

~v

Energy is conserved. When the rod is vertical, the energy of the mass 2 m is all kinetic:
2m
E = Kmass
=

1
2
(2 m) v2m
= m ( L)2 .
2

The final potential energy of the rod is

1
mgL
2
and the kinetic energy is
Ufrod =

Kfrod = Kfrot + Kftr =

1
1
2
I 2 + m vrod
2
2

v~0

If the hoop has speed v0 at the bottom of

the pipe, what is its speed when it has rolled
halfway up the side of the pipe?
1. m g R
2. There is no way such a loop could roll
without slipping.

Version 024 Exam 3 hoffmann (57870)

q
v02 g r
q
4. v02 g R

3.

5. Not enough information is provided.

6. r g
q
7. v02 g (R r) correct
q
8. v02 + g r
q
9. v02 + g (R r)
Explanation:
The moment of inertia of a solid ring is
simply Icm = M R2 . The hoop does not slip,
so vcm = v0 = r .
Applying conservation of energy with zero
gravitational potential energy at y = R r,

E0 = Ef
KE0 + P E0 = KEf + P Ef


 
1
1
1
1
2
2
2
2
m v0 + I 0 =
mv + m
2
2
2
2
+m g y
 v 2



 v 2
0
2
2
2
2
m v0 + m r
= mv + mr
r
r
+2 m g (R r)
2 v02 = 2 v 2 + 2 g (R r)
q
v = v02 g (R r) .

005

10.0 points

The pulley shown has a radius 0.232 m and

moment of inertia 0.847 kg m2 . One end of
the 0.359 kg mass is connected to a spring of
force constant 67 N/m, and the other end is
fastened to a cord wrapped around the pulley.
The incline is at an angle 23.3 .

m
k

If the pulley is wound clockwise so as to

stretch the spring a distance 0.109 m from its
equilibrium position and then released from
rest, find the angular speed of the pulley when
the spring is again unstretched (i.e., at the
springs equilibrium position). Assume the
pulley axle and the incline are frictionless.
The acceleration of gravity is 9.8 m/s2 .
1. 1.83059
2. 1.87215
3. 2.04001
4. 1.41782
5. 2.50354
6. 2.88281
7. 1.68467
8. 1.32265
9. 1.4006
10. 1.12652
Correct answer: 1.12652 rad/s.
Explanation:
g = 9.8 m/s2 ,
I = 0.847 kg m2 ,
R = 0.232 m ,
k = 67 N/m ,
m = 0.359 kg ,
d = 0.109 m , and
= 23.3 .
v = R and conservation of energy gives
Let :

W = U + K , so
m v2 I 2
k d2
0=
+
m g d sin
.
2
2
2
k d2
(I + m R2 ) 2
= m g d sin +
.
2
2

Version 024 Exam 3 hoffmann (57870)

Since
m g d sin = (0.359 kg) (9.8 m/s2 ) (0.109 m)
sin 23.3
= 0.151685 J , then

2 m g d sin + k d2
I + m R2

2 (0.151685 J) + (67 N/m)(0.109 m)2

0.847 kg m2 + (0.359 kg)(0.232 m)2

Given :

M = 13.9 kg ,
R = 1.85 m ,
i = 2.36 rad/s ,
m = 0.0714 kg , and
r = 0.740 m .

Solution: Angular momentum is conserved,

so

= 1.12652 rad/s .

Lcyl,f + Lputty,f = Lcyl,i + Lputty,i

Icyl f + Iputty f = Icyl i




1
1
2
2
2
M R + m r f =
M R i ,
2
2
so that

006 10.0 points

A solid, vertical cylinder with a mass of 13.9
kg and a radius of 1.85 m rotates with an
angular speed of 2.36 rad/s about a fixed
vertical axis through its center. A 0.0714 kg
piece of putty is dropped vertically at a point
0.740 m from the cylinders center of rotation
and sticks to the cylinder.
What is the final angular speed of the system?
1. 1.92948
2. 1.11164
3. 4.00719
4. 3.29174
5. 3.5044
6. 3.60896
7. 4.81561
8. 1.84354
9. 2.35613
10. 2.44066
Correct answer: 2.35613 rad/s.
Explanation:
Basic Concepts:
The cylinder is a solid disk and the putty
can be modeled as a point mass.
Lcyl = Icyl =

1
M R2
2

Lputty = Iputty = m r 2
Lf = Li

f =

1
M R 2 i
2

1
M R2 + m r 2
2
i
=
2 m r2
1+
M R2
(2.36 rad/s)
=
2 (0.0714 kg) (0.74 m)2
1+
(13.9 kg) (1.85 m)2
= 2.35613 rad/s .

007 10.0 points

Two blocks of masses 3 kg and 26 kg are
placed on a horizontal, frictionless surface. A
light spring is attached to one of them, and
the blocks are pushed together with the spring
between them. A cord holding them together
is burned, after which the block of mass 26 kg
moves to the right with a speed of 14 m/s.
What is the velocity of the other mass in
m/s?
1. -9.35
2. -35.36
3. -75.4286
4. -96.6
5. -11.408
6. -28.0
7. -69.0
8. -121.333
9. -120.0

Version 024 Exam 3 hoffmann (57870)

10. -10.2125
Correct answer: 121.333 m/s.
Explanation:

By momentum conservation
m1 v1 + m2 v2 = 0
so the velocity of m1 is
v1 =

m2 v2
m1

008 10.0 points

What is the kinetic energy of a satellite of
mass m that orbits the Earth of mass M in a
circular orbit of radius R?

Find the magnitude of the net gravitational

force exerted by the two larger masses on the
20.3 kg mass. The value of the universal gravitational constant is 6.672 1011 N m2 /kg2 .
1. 4.25415e-05
2. 0.0
3. 5.76876e-05
4. 5.6135e-05
5. 2.76116e-05
6. 5.26626e-05
7. 3.84771e-05
8. 4.67056e-05
9. 1.71929e-05
10. 7.44172e-05
Correct answer: 1.71929 105 N.
Explanation:

GM m
1. K =
R2
2. K = 0
1 GM m
4
R
1 GM m
4. K =
2 R2
1 GM m
5. K =
correct
2
R
Explanation:
The gravitational force on the satellite provides the centripetal force needed to keep it
in circular orbit:
GM m
v2
= FG = Fc = m
R2
R
GM m
2
, so
mv =
R
3. K =

K=

1
1 GM m
m v2 =
.
2
2
R

009 (part 1 of 2) 10.0 points

Objects with masses of 201 kg and 528 kg
are separated by 0.321 m. A 20.3 kg mass is
placed midway between them.
528 kg
201 kg
20.3 kg
b

0.321 m

Let : d = 0.1605 m ,
m = 20.3 kg ,
M1 = 201 kg ,
M2 = 528 kg , and
G = 6.672 1011 N m2 /kg2 .
At the midpoint between the two masses, the
forces exerted by the masses M1 and M2 are
oppositely directed and since
F =G

mM
,
r2

the net force on the mass m is

m (M1 M2 )
d2


= 6.672 1011 N m2 /kg2
(20.3 kg) (201 kg 528 kg)

(0.1605 m)2
= 1.71929 105 N

F =G

which has a magnitude of 1.71929 105 N .

010 (part 2 of 2) 10.0 points
Leaving the distance between the 201 kg and
the 528 kg masses fixed, at what distance from
the 528 kg mass (other than infinitely remote

Version 024 Exam 3 hoffmann (57870)

ones) does the 20.3 kg mass experience a net
force of zero?
1. 0.247019
2. 0.230709
3. 0.178549
4. 0.227976
5. 0.283791
6. 0.175071
7. 0.198516
8. 0.257437
9. 0.224745
10. 0.237614
Correct answer: 0.198516 m.
Explanation:
Let :

d = 0.321 m .

5. r0 F0 k 2 2 .



6. r0 F0 k .



7. r0 F0 k 2 .
8. Zero.

9. r0 F0

= 0.198516 m .

011  10.0 points


A force F = F0 + + k acts on a rigid
body at a point r = r0 ( ) away from the
axis of rotation.
What is the resulting torque on the body?



1. r0 F0 2 k .



2. r0 F0 2 k .



3. r0 F0 + + k .



4. r0 F0 k 2 .

2 k . correct

Explanation:
~ = r F

= F0 r0 + k k


= F0 r0 k + k



= F0 r0 2 k .

Let R be the distance from M2 ; then d r

is its distance from M2 .
G m M1
G m M2
=
2
(d r)
r2
r 2 M1 = (d r)2 M1
p
p
r M1 = (d r) M2
p
p 
p
M1 + M2 = d M2 and
r

d M2

r=
M1 + M2

(0.321 m) 528 kg

=
201 kg + 528 kg

012 10.0 points

Assume: A bullet of mass m and cube of
mass M undergo an inelastic collision, where
m M.
Note: The moment of inertia of this cube
(with edges of length 2 a and mass M ) about
8 M a2
an axis along one of its edges is
.
3
A solid cube is resting on a horizontal surface. The cube is constrained to rotate about
an axis at its bottom right edge (due to a
small obstacle on the table). A bullet with
speed vmin is shot at the left-hand face at a
4
height of a. The bullet gets embedded in
3
the cube.

2
a
m~vmin
4
a
3

Mg
Find the minimum value of vmin required to
tip the cube so that it falls on its right-hand
face.
r


m
3ga
51
1. vmin =
M

Version 024 Exam 3 hoffmann (57870)

r


m
2. vmin =
5ga
21
M
r


m
3. vmin =
2ga
21
M
r


M
3ga
2 1 correct
4. vmin =
m
r


m
3ga
31
5. vmin =
M
Explanation:
Basic Concepts:
X
~ = const
L
U + K = 0

For the cube to tip over the center of mass

(CM) must rise so that it is over the axis of
rotation AB. To do
this the CM must be
raised a distance of a
21 .
From conservation of energy
 1

(1)
M ga
2 1 = Icube 2 .
2
From conservation of angular momentum


8 M a2
4a

mv =
3
3
 mv 
=
.
(2)
2M a
Thus, substituting Eq. 2 into 1, we have

 2 2 


m v
1 8 M a2
=
M
g
a
2

1
2
3
4 M 2 a2
Solving for v yields
r


M
3ga
21 .
vmin =
m

013 10.0 points

A 160 g basketball has a 35.1 cm diameter
and may be approximated as a thin spherical
shell.
2. 1

2m

160 g
0
=

. 49

24

Starting from rest, how long will it take

a basketball to roll without slipping 2.12 m
down an incline that makes an angle of 24
with the horizontal? The moment of inertia of
a thin spherical shell of radius R and mass m
2
is I = m R2 , the acceleration due to gravity
3
is 9.8 m/s2 , and the coefficient of friction is
0.49 .
1. 1.19119
2. 1.52141
3. 1.35946
4. 1.40588
5. 1.60075
6. 1.15064
7. 1.57153
8. 1.27915
9. 1.96785
10. 1.33149
Correct answer: 1.33149 s.
Explanation:

Let : = 2.12 m ,
= 24 , and
R = 0.1755 m .

sin
g
m

mg
The moment of inertia of the ball about
the point of contact between the ball and the
inclined plane is
IP = Icm + m d2
2
= m R2 + m R2
3
5
= m R2 .
3

Version 024 Exam 3 hoffmann (57870)

The net torque about the point of contact
between the ball and the inclined plane is
5
m R2
3
3
m g R2 sin
= g sin .
a=
5
5
m R2
3

m g R sin = IP =

Because the sphere starts from rest, its center of mass moves a distance

Correct answer: 2.08093 1011 N.

Explanation:
Let:
mo = 2 kg ,
ma = 7 kg ,
mb = 4 kg ,

(xo , yo) = (0 m, 0 m) ,
(xa , ya ) = (7 m, 0 m) ,
(xb , yb ) = (0 m, 8 m) .

and

mo ma
(xa xo )2 + (ya yo )2
= (6.6726 1011 N m2 /kg2 )
(2 kg) (7 kg)

(7 m)2 + (0 m)2
= 1.90646 1011 N

Fao = G

= 1.33149 s .
014 10.0 points
Three masses are arranged in the (x, y)
plane as shown.
y (m)
9

acts along the x axis.

Applying the gravitational law for mo and
mb ,
mo mb
(xb xo )2 + (yb yo )2
= (6.6726 1011 N m2 /kg2 )
(2 kg) (4 kg)

(0 m)2 + (8 m)2
= 8.34075 1012 N

Fbo = G

4 kg
7
5
3
2 kg
1

9. 1.91293e-10
10. 5.10115e-11

Applying Newtons universal gravitational

law for mo and ma ,

1 2
at
2
s
r
2
2 (2.12 m)
=
t=
a
2.39161 m/s2

7 kg
3

x
(m)

What is the magnitude of the resulting

force on the 2 kg mass at the origin? The
value of the universal gravitational constant
is 6.6726 1011 N m2 /kg2 .
1. 2.21168e-11
2. 1.07846e-10
3. 3.80511e-11
4. 9.91613e-11
5. 2.08093e-11
6. 9.53979e-11
7. 4.62592e-11
8. 2.53207e-11

acts along the y axis.

The magnitude of the resultant force is
q
2 + F2
F = Fao
bo

= (1.90646 1011 N)2
1/2
+ (8.34075 1012 N)2
= 2.08093 1011 N .

015 10.0 points

Two masses of 25 kg and 17 kg are suspended
by a pulley that has a radius of 11 cm and
a mass of 6 kg. The cord has a negligible
mass and causes the pulley to rotate without

Version 024 Exam 3 hoffmann (57870)

slipping. The pulley rotates without friction.
The masses start from rest 4 m apart.
v=

11 cm

6 kg

2 (m1 m2 ) g h
2 (m1 + m2 ) + M

2 (25 kg 17 kg) (9.8 m/s2 )

2 (25 kg + 17 kg) + 6 kg

4m

= 2.63987 m/s .

25 kg
4m

keywords:
17 kg

Determine the speeds of the two masses as

they pass each other. Treat the pulley as a
uniform disk. The acceleration of gravity is
9.8 m/s2 .
1. 3.1305
2. 2.7633
3. 2.63987
4. 2.43384
5. 2.02073
6. 3.36242
7. 2.58061
8. 2.89879
9. 3.46482
10. 1.77701
Correct answer: 2.63987 m/s.
Explanation:
Let : M = 6 kg ,
R = 11 cm ,
m1 = 25 kg ,
m2 = 17 kg , and
h = 4 m.
From conservation of energy
K1 + K2 + Kdisk = U
m1
m2 v 2 M v 2
h
+
+
= (m1 m2 ) g
2
2
4
2
(2 m1 + 2 m2 + M ) v 2 = 2 (m1 m2 ) g h ,
v2

h
where is the height. Taking no slipping into
2
account,