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Thevalingam, Donald – Homework 30 – Due: Apr 30 2007, midnight – Inst: Eslami
1
This
printout
should
have
20
questions.
Multiplechoice
questions
may
continue
on
the next column or page – fnd all choices
beFore answering.
The due time is Central
time.
001
(part 1 oF 1) 10 points
The center oF mass oF a pitched baseball or
radius 5
.
36 cm moves at 18
.
3 m
/
s.
The ball
spins about an axis through its center oF mass
with an angular speed oF 50
.
2 rad
/
s.
Calculate the ratio oF the rotational energy
to the translational kinetic energy. Treat the
ball as a uniForm sphere.
Correct answer: 0
.
00864758
.
Explanation:
The wanted ratio is
ratio
=
Iω
2
2
mv
2
2
=
µ
2
mr
2
5
¶
ω
2
mv
2
=
2
r
2
ω
2
5
v
2
=
2(0
.
0536 m
/
cm)
2
(50
.
2 rad
/
s)
2
5(18
.
3 m
/
s)
2
= 0
.
00864758
keywords:
002
(part 1 oF 1) 10 points
A circular disk with a mass
m
and radius
R
is mounted at its center, about which it can
rotate Freely. The disk has moment oF inertia
I
=
1
2
m R
2
.
A light cord wrapped around it supports a
weight
m g
.
R
I
ω
m
T
g
±ind the total kinetic energy oF the system,
when the weight is moving at a speed
v
.
1.
K
=
1
2
m v
2
2.
K
=
3
4
m v
2
correct
3.
K
=
m v
2
4.
K
=
4
5
m v
2
5.
K
=
1
3
m v
2
6.
K
=
5
2
m v
2
7.
K
=
3
2
m v
2
8.
K
=
2
3
m v
2
9.
K
=
5
4
m v
2
Explanation:
The total kinetic energy is the sum oF the
kinetic energy oF the hanging weight,
K
m
=
1
2
m v
2
, plus the kinetic energy oF rotation oF
the disk,
K
d
=
1
2
I ω
2
=
1
4
m v
2
.
ThereFore,
K
=
K
g
+
K
d
=
1
2
m v
2
+
1
4
m v
2
=
3
4
m v
2
.
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View Full DocumentThevalingam, Donald – Homework 30 – Due: Apr 30 2007, midnight – Inst: Eslami
2
keywords:
003
(part 1 of 3) 10 points
For any given rotational axis, the radius of
gyration,
K
, of a rigid body is de±ned by the
expression
K
2
=
I
M
, where
M
is the total
mass of the body and
I
is the moment of
inertia about the given axis.
In other words,
the radius of gyration is the distance between
an imaginary point mass
M
, and the axis of
rotation with
I
for the point mass about that
axis is the same as for the rigid body.
Find the radius of gyration of a solid disk of
radius 3
.
93 m rotating about a central axis.
Correct answer: 2
.
77893
m.
Explanation:
In all part the axis is the central axis, as
stated in the problem. The moment of inertia
of a solid disk with radius
R
and mass
M
is
I
d
=
M R
2
2
.
Hence,
K
d
=
r
M R
2
2
M
=
R
√
2
= (3
.
93 m)
√
2
= 2
.
77893 m
.
004
(part 2 of 3) 10 points
Find the radius of gyration of a uniform rod of
length 7
.
73 m rotating about a central axis.
Correct answer: 2
.
23146
m.
Explanation:
The moment of inertia of a uniform rod of
length
L
and mass
M
is
I
r
=
M L
2
12
Hence,
K
r
=
r
M L
2
12
M
=
L
6
√
3
=
(7
.
73 m)
6
√
3
= 2
.
23146 m
.
005
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 Physics, Work

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