homework 17 – RAHMAN, TARIQUE – Due: Apr 2 2008, 3:00 am
1
Question 1, chap 12, sect 5.
part 1 of 1
10 points
A circular disk, a ring, and a square, have
the same mass
M
and width 2
r
.
disk
2
r
ring
2
r
square
2
r
For the moment of inertia about their cen
ter of mass about an axis perpendicular to
the plane of the paper, which of the follow
ing statements concerning their moments of
inertia is true?
1.
I
square
> I
disk
> I
ring
2.
I
disk
> I
square
> I
ring
3.
I
ring
> I
disk
> I
square
4.
I
square
> I
ring
> I
disk
correct
5.
I
ring
> I
square
> I
disk
6.
I
disk
> I
ring
> I
square
Explanation:
Basic Concept:
Moment of inertia, rota
tional kinetic energy.
Solution:
In the ring,
the same mass of the disk is concentrated at
the maximum distance from the axis. There
fore
I
ring
> I
disk
In the square, the same mass of the ring lies
at distances which are
at least
the radius of
the ring. Therefore
I
square
> I
ring
thus
I
square
> I
ring
> I
disk
AlternativePart 1:
Using moments of iner
tia as found in the book and the parallel axis
theorem,
I
disk
=
1
2
m r
2
=
1
2
M r
2
I
ring
=
m r
2
=
M r
2
I
rod
cm
=
1
12
m ℓ
2
I
square
= 4
parenleftbigg
1
12
m ℓ
2
+
m d
2
parenrightbigg
m
=
M
4
ℓ
= 2
r
d
=
r
I
square
= 4
bracketleftbigg
1
12
parenleftbigg
M
4
parenrightbigg
(2
r
)
2
+
parenleftbigg
M
4
parenrightbigg
r
2
bracketrightbigg
=
4
3
M r
2
,
since
4
3
M r
2
> M r
2
>
1
2
M r
2
then
I
square
> I
ring
> I
disk
.
Question 2, chap 12, sect 5.
part 1 of 1
10 points
A solid sphere rolls along a horizontal,
smooth surface at a constant linear speed
without slipping.
What is the ratio between the rotational
kinetic energy about the center of the sphere
and the sphere’s total kinetic energy?
1.
None of these
2.
3
7
3.
5
3
4.
7
2
5.
2
7
correct
6.
3
5
7.
2
5
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homework 17 – RAHMAN, TARIQUE – Due: Apr 2 2008, 3:00 am
2
Explanation:
Basic Concepts:
KE
trans
=
1
2
mv
2
KE
rot
=
1
2
Iω
2
=
1
2
parenleftbigg
2
5
mr
2
parenrightbigg
ω
2
v
=
rω
Solution:
KE
rot
=
1
2
parenleftbigg
2
5
mr
2
parenrightbigg
parenleftBig
v
r
parenrightBig
2
=
1
5
mv
2
and
KE
tot
=
KE
trans
+
KE
rot
=
1
2
mv
2
+
1
5
mv
2
=
7
10
mv
2
Thus
KE
rot
KE
tot
=
1
5
mv
2
7
10
mv
2
=
parenleftbigg
1
5
parenrightbigg parenleftbigg
10
7
parenrightbigg
=
2
7
Question 3, chap 12, sect 5.
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 Fall '08
 Turner
 Physics, Angular Momentum, Kinetic Energy, Mass, Work, Moment Of Inertia, Rigid Body, Rotation

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