boland (kbb544) – HW9 – mackie – (10611)
1
This
printout
should
have
14
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
A can of soup has a mass 120 g, height 17 cm,
and diameter 2
.
8 cm. It is placed at rest on
the top of an incline that is 1
.
4 m long at 28
◦
to the horizontal.
The acceleration of gravity is 9
.
8 m
/
s
2
.
Calculate the moment of inertia of the can
if it takes 0
.
88 s to reach the bottom of the
incline.
Correct answer: 6
.
40817
×
10
−
6
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 (120 g)
(227
.
273 rad
/
s)
2
×
bracketleftbigg
(9
.
8 m
/
s
2
)(1
.
4 m) sin(28
◦
)

1
2
(3
.
18182 m
/
s)
2
bracketrightbigg
= 6
.
40817
×
10
−
6
kg m
2
.
002
10.0 points
A solid steel sphere of density 7
.
5 g
/
cm
3
and
mass 0
.
6 kg spin on an axis through its center
with a period of 3
.
5 s.
Given
V
sphere
=
4
3
π R
3
, what is its angular
momentum?
Correct answer: 0
.
000307839 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
.
6 kg)
4
π
(7500 kg
/
m
3
)
bracketrightbigg
1
/
3
= 0
.
0267301 m
.
Using
ω
=
2
π
T
=
2
π
(3
.
5 s)
= 1
.
7952 s
−
1
and
I
=
2
5
M R
2
=
2
5
(0
.
6 kg)(0
.
0267301 m)
2
= 0
.
000171479 kg m
2
,
we have
L
≡
I ω
=
4
π M R
2
5
T
=
4
π
(0
.
6 kg)(0
.
0267301 m)
2
5 (3
.
5 s)
= 0
.
000307839 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.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
boland (kbb544) – HW9 – mackie – (10611)
2
v
9
.
8 m
/
s
2
4 m
9 kg
19
◦
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: 24
.
5694 kg
·
m
2
/
s.
Explanation:
Let :
ℓ
= 4 m
,
θ
= 19
◦
,
g
= 9
.
8 m
/
s
2
,
and
m
= 9 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
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '11
 MACKIE
 Physics, Angular Momentum, Mass, Rotation, Correct Answer, Boland

Click to edit the document details