zhou (cz3574) – HW9 – mackie – (10611)
1
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001
10.0 points
A can of soup has a mass 570 g, height 14 cm,
and diameter 6
.
6 cm. It is placed at rest on
the top of an incline that is 4
.
4 m long at 24
◦
to the horizontal.
The acceleration of gravity is 9
.
8 m
/
s
2
.
Calculate the moment of inertia of the can
if it takes 1
.
7 s to reach the bottom of the
incline.
Correct answer: 0
.
000191834 kg m
2
.
Explanation:
The speed of the center of mass of the can
at the bottom of the incline is
v
f
= 2¯
v
=
2
L
t
.
Thus the angular speed at the bottom of the
incline is
ω
f
=
v
f
R
.
Using energy conservation, we have
m g L
sin
θ
=
1
2
m v
2
f
+
1
2
I ω
2
f
.
Solving for
I
,
I
=
2
m
ω
2
f
bracketleftbigg
g L
sin
θ

1
2
v
2
f
bracketrightbigg
=
2 (570 g)
(156
.
863 rad
/
s)
2
×
bracketleftbigg
(9
.
8 m
/
s
2
)(4
.
4 m) sin(24
◦
)

1
2
(5
.
17647 m
/
s)
2
bracketrightbigg
= 0
.
000191834 kg m
2
.
002
10.0 points
A solid steel sphere of density 7
.
68 g
/
cm
3
and
mass 0
.
9 kg spin on an axis through its center
with a period of 3
.
8 s.
Given
V
sphere
=
4
3
π R
3
, what is its angular
momentum?
Correct answer: 0
.
000548564 kg m
2
/
s.
Explanation:
The definition of density is
ρ
≡
M
V
=
M
4
3
π R
3
,
Therefore
R
=
bracketleftbigg
3
M
4
π ρ
bracketrightbigg
1
/
3
=
bracketleftbigg
3 (0
.
9 kg)
4
π
(7680 kg
/
m
3
)
bracketrightbigg
1
/
3
= 0
.
0303574 m
.
Using
ω
=
2
π
T
=
2
π
(3
.
8 s)
= 1
.
65347 s
−
1
and
I
=
2
5
M R
2
=
2
5
(0
.
9 kg)(0
.
0303574 m)
2
= 0
.
000331765 kg m
2
,
we have
L
≡
I ω
=
4
π M R
2
5
T
=
4
π
(0
.
9 kg)(0
.
0303574 m)
2
5 (3
.
8 s)
= 0
.
000548564 kg m
2
/
s
.
003
10.0 points
A small metallic bob is suspended from the
ceiling by a thread of negligible mass.
The
ball is then set in motion in a horizontal circle
so that the thread describes a cone.
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zhou (cz3574) – HW9 – mackie – (10611)
2
v
9
.
8 m
/
s
2
1
.
7 m
7 kg
33
◦
Calculate the magnitude of the angular mo
mentum of the bob about a vertical axis
through the supporting point.
The acceler
ation of gravity is 9
.
8 m
/
s
2
.
Correct answer: 15
.
7328 kg
·
m
2
/
s.
Explanation:
Let :
ℓ
= 1
.
7 m
,
θ
= 33
◦
,
g
= 9
.
8 m
/
s
2
,
and
m
= 7 kg
.
Consider the free body diagram.
T
m g
θ
Newton’s second law in the vertical and
horizontal projections, respectively, gives
T
cos
θ

m g
= 0
T
cos
θ
=
m g
and
T
sin
θ

m ω
2
ℓ
sin
θ
= 0
T
=
m ω
2
ℓ ,
where the radius of the orbit is
R
=
ℓ
sin
θ
.
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 Spring '10
 Hiyishi
 Angular Momentum, Mass, Rigid Body, Rotation, Correct Answer, Zhou

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