homework 10 – – Due: Apr 4 2007, 4:00 am
1
Question 1, chap 12, sect 5.
part 1 of 1
6 points
The rigid object shown is rotated about an
axis perpendicular to the paper and through
center point
O
.
O
ω
34 kg
34 kg
68 kg
68 kg
16 m
16 m
8 m
8 m
What is the moment of inertia of the ob-
ject?
Neglect the mass of the connecting rods and
treat the masses as point masses.
Correct answer: 26112 kg m
2
(tolerance
±
1
%).
Explanation:
The moment of inertia of a rigid body whose
mass is concentrated in several points is given
by
I
=
s
i
m
i
r
2
i
.
For the body at hand, the mass is concen-
trated in the four small balls. Two of them
have masses
m
1
,
2
= 34 kg and are located at
distances
r
1
,
2
= 16 m from the rotation axis
O
; the other two balls have
m
3
,
4
= 68 kg each
and are located at distances
r
3
,
4
= 8 m from
the axis. Therefore,
I
=
4
s
i
=1
m
i
r
2
i
= 2
×
(34 kg)(16 m)
2
+ 2
×
(68 kg)(8 m)
2
= 26112 kg m
2
.
Question 2, chap 12, sect 5.
part 1 of 3
6 points
Four particles with masses 5 kg, 5 kg, 5 kg,
and 5 kg are connected by rigid rods of negli-
gible mass at the corners of a rectangle. The
origin is at the center of the ±gure.
x
y
O
ω
6 m
8 m
5 kg
5 kg
5 kg
5 kg
What is the moment of inertia about the
y
-axis?
Correct answer: 320 kg m
2
(tolerance
±
1 %).
Explanation:
Let :
m
1
= 5 kg
,
top right
m
2
= 5 kg
,
top left
m
3
= 5 kg
,
bottom left
m
4
= 5 kg
,
bottom right
w
= 8 m
,
ℓ
= 6 m
,
x
0
= 4 m
,
and
y
0
= 3 m
.
The distance
r
i
in the expression for the
moment of inertia is simply the distance of
each particle to the
y
axis.
i.e.
r
i
=
x
0
.
Hence
I
y
=
s
m
i
r
2
i
=
x
2
0
s
m
i
= (4 m)
2
(20 kg)
= 320 kg m
2
.
Question 3, chap 12, sect 5.
part 2 of 3
6 points
What is the moment of inertia about the
z
-axis?
Correct answer: 500 kg m
2
(tolerance
±
1 %).
Explanation: