Answer key for third exam version A
1
1. Say that the disk has mass
M
1
and radius
R
1
, while the sphere has mass
M
2
and
R
2
. At the top of the plane, the disk has potential energy
M
1
gh
, while the
sphere has potential energy
M
2
gh
, where
h
is the height of inclined plane. At
the bottom of the inclined plane the potential energies are both defined to be
zero, and the kinetic energies equal to the initial potential energies
K
1
=
M
1
gh
=
1
2
M
1
v
2
1
+
1
2
I
1
ω
2
=
1
2
M
1
v
2
1
+
1
4
M
1
R
2
1
v
2
1
R
2
1
=
3
4
M
1
v
2
1
=
⇒
v
1
=
v
u
u
t
4
gh
3
speed of c.m. of disk
K
2
=
M
2
gh
=
1
2
M
2
v
2
2
+
1
2
I
2
ω
2
=
1
2
M
2
v
2
2
+
1
5
M
2
R
2
2
v
2
2
R
2
2
=
7
10
M
2
v
2
2
=
⇒
v
2
=
v
u
u
t
10
gh
7
speed of c.m. of sphere
The sphere is going faster at the bottom of the inclined plane for any height
h
,
and this does not depend on the mass nor the radius of the sphere. We did a
class demonstration using a solid disk and a hoop to show the same qualitative
result: the object with the smaller moment of inertia will win the race to the
bottom because proportionately less kinetic energy is in rotational motion and
proportionately more kinetic energy is in translational motion.
Answer C
2. The mass
M
X
of planet
X
is not given, but it can be obtained from knowing
the radius
R
X
and the surface acceleration
g
X
(see Lecture 16, pages 1–2):
g
X
=
G
M
X
R
2
X
=
⇒
GM
X
=
g
X
R
2
X
For the weight of a mass
m
at a distance
h
above the surface of planet
X
to be
equal to the weight on the Earth’s surface we have
mg
E
=
GM
X
m
(
R
X
+
h
)
2
=
⇒
(
R
X
+
h
)
2
=
GM
x
g
E
=
g
X
R
2
X
g
E
h
=
R
X
v
u
u
t
g
X
g
E

1
= 5
.
16
×
10
7
v
u
u
t
13
.
2
9
.
8

1
= 8
.
30
×
10
6
meters
Answer E
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Answer key for third exam version A
2
3. We can solve this problem by using either conservation total energy, or by
conservation of angular momentum, between the minimum distance
r
1
and the
maximum distance
r
2
. The angular momentum method is easier
L
1
=
L
2
=
mv
1
r
1
=
L
2
=
mv
2
r
2
=
⇒
v
1
=
r
2
r
1
v
2
The speed
v
2
at the maximum distance is given as 908 m/s.
This gives the
maximum speed at the closest distance as
v
1
= 5
.
46
×
10
4
m/s.
Answer C
4.
The net torque is related to the angular acceleration by
τ
=
Iα
with
I
=
2
MR
2
/
5 in this case (
M
= 1
.
85 kg,
R
= 0
.
45
/
2 m).
We can compute the
angular acceleration from
ω
2
=
ω
2
0
+ 2
α
(
θ

θ
0
), where
ω
0
= 2
.
40
*
2
π
rad/sec,
and
θ

θ
0
= 18
.
2
*
2
π
radians. Finally, we will have
τ
=
2
5
(1
.
85)(0
.
225)
2
(2
.
40
×
2
π
)
2
2
×
18
.
2
×
2
π
= 0
.
0372 Nm
Answer = B
5.
This answer was actually given in the MasteringPhysics Assignment 12,
item 13 for the
Satellite in Orbit
.
The extra hint for this item
1
emphasized
that MasteringPhysics comment, and specifically mentioned the Space Shuttle
astronauts appearing “weightless” was continuous falling but never reaching the
Earth.
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 Spring '08
 Maguire
 Physics, Energy, Kinetic Energy, Mass, Potential Energy, exam version, Wblock, c.m. xcm, +120o

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