Version 047/AACDD – midterm 04 CC10 – turner – (57340)
1
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printout
should
have
17
questions.
Multiplechoice questions may continue on
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before answering.
001
10.0 points
A uniform solid disk with radius R and
mass M is mounted on a fixed axle passing
through its center (which is not shown).
A
constant force F is applied as shown. At the
instant shown, the angular velocity of the disk
ω
is in the

ˆ
z
direction (where +ˆ
x
is to the
right, +ˆ
y
is up, and +ˆ
z
is out of the page,
toward you). The length of the string is d.
The angular momentum, torque and kinetic
energy of the disk are respectively:
1.

1
2
MR
2
ω
ˆ
z
,
RF
ˆ
z
,

1
4
MR
2
ω
2
2.
1
2
MR
2
ω
ˆ
z
,
RF
ˆ
z
,

1
4
MR
2
ω
2
3.
1
2
MR
2
ω
ˆ
z
,

RF
ˆ
z
,

1
4
MR
2
ω
2
4.
1
2
MR
2
ω
ˆ
z
,

RF
ˆ
z
,
1
4
MR
2
ω
2
5.

1
2
MR
2
ω
ˆ
z
,

RF
ˆ
z
,
1
4
MR
2
ω
2
correct
6.

1
2
MR
2
ω
ˆ
z
,
RF
ˆ
z
,
1
4
MR
2
ω
2
7.
1
2
MR
2
ω
ˆ
z
,
RF
ˆ
z
,
1
4
MR
2
ω
2
8.

1
2
MR
2
ω
ˆ
z
,

RF
ˆ
z
,

1
4
MR
2
ω
2
Explanation:
The direction of both L and thtat of torque
are along the

ˆ
z
direction.
The moment of inertia of the disk is given
by
I
=
1
2
M R
2
The magnitude of angular momentum is
L
=
I
ω
=
1
2
M R
2
ω
The magnitude of the torque is given by
τ
=
R
2
+
d
2
F sin
θ
and the value of
sin
θ
is given by
sin
θ
=
R
√
R
2
+
d
2
Thus, the value of torque is
τ
=
R F
The kinetic energy is given by
K
=
1
2
I
ω
2
=
1
4
M R
2
ω
2
002
10.0 points
Consider a ball with its mass 0.1 kg is head
on colliding with a leadsphere. Initially the
leadsphere is at rest and the ball is moving
along the positive x direction with a speed 6
m/s and the final velocity is along the negative
xdirection with a speed 4 m/s. Assume the
leadsphere is so massive that its final kinetic
energy is negligible compared to the kinetic
energies of ball, both before and after the
collision.
Consider following choices for the
change in the momentum vector of the lead
sphere
Ia.
Δ
p
lead

sphere
=
Δ
p
ball
Ib.
Δ
p
lead

sphere
= 0
Ic.
Δ
p
lead

sphere
=

Δ
p
ball
and the following choices for the change of the
internal energy
IIa.
Δ
E
int
= 1
J
IIb.
Δ
E
int
=

1
J
IIc.
Δ
E
int
= 0
1.
Ia, IIa
2.
Ia, IIb
3.
Ib, IIc
4.
Ic, IIc
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Version 047/AACDD – midterm 04 CC10 – turner – (57340)
2
5.
Ic, IIb
6.
Ib, IIb
7.
Ib, IIa
8.
Ia, IIc
9.
Ic, IIa
correct
Explanation:
Let :
m
= 0
.
1 kg
,
v
i
= 6 m
/
s
,
and
v
f
= 4 m
/
s
We take the ball plus the leadsphere as a
system. Neglecting any external impulse, the
momentum principle implies that
Δ
p
ball
+
Δ
p
lead

sphere
= 0
.
So Ic is the correct choice.
For the ball + leadsphere system, the en
ergy principle states that
Δ
E
tot
=
Δ
K
+
Δ
E
int
=
W
+
Q .
Since collision process occurs quickly, W and
Q are assumed to be negligible. Thus
Δ
K
+
Δ
E
int
= 0
.
This leads to
Δ
E
int
=

Δ
K
ball

Δ
K
lead

sphere
=

Δ
K
ball
=
K
i

K
f
=
m
2
v
2
i

m
2
v
2
f
=
(0
.
1 kg)((6 m
/
s)
2

(4 m
/
s)
2
)
2
= 1 J
.
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 Spring '08
 Turner
 Physics, Angular Momentum, Energy, Kinetic Energy, Momentum, IIA, vcm, Version 047/AACDD

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