DATE:
November 6, 2011
DUE:
November 11
→
14 (before Noon), 2011
ENG102, Solution to homework set #7.
1.
Look at Problem 3/354 on page 266 in the book.
a. Solve the problem with the additional information that each mass is a solid
sphere with radius
r
= 0
.
1
m
.
Both momentum and angular momentum are conserved. The change from the
original problem is that the moment of intertial around
G
is now:
I
= 2
mR
2
+
4
5
mr
2
, where
R
= 0
.
4
m
. Thus:
H
G
=
m
(
v
A
+
v
B
)0
.
4
m
=
I
˙
θ
′
(1)
⇒
˙
θ
′
=
H
G
I
=
v
A
+
v
B
2
·
0
.
4
m
+
4
5
r
2
0
.
4
m
= 9
.
756
s
−
1
(2)
p
=
m
(
v
A

v
B
) = 2
mv
′
(3)
⇒
v
′
=
v
A

v
B
2
=

1
m/s
(4)
b. What is the loss in kinetic energy?
Initial and Final kinetic energy are:
E
k
=
1
2
mv
2
A
+
1
2
mv
2
B
(5)
E
′
k
=
1
2
I
(
˙
θ
′
)
2
+
m
(
v
′
)
2
=
1
2
H
2
G
I
+
m
4
(
v
A

v
B
)
2
(6)
Δ
E
k
=
1
2
H
2
G
I

1
4
m
(
v
2
A
+
v
2
B
+ 2
v
A
v
B
) = =
1
2
H
2
G
I

1
4
m
(
v
A
+
v
B
)
2
(7)
=
1
4
m
(
v
A
+
v
B
)
2
bracketleftBigg
1
1 +
2
5
r
2
R
2

1
bracketrightBigg
=

0
.
390
m
(m
/
s)
2
(8)
Notice that
r
→
0
gives
Δ
E
k
= 0
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Winter '08
 Eke
 Angular Momentum, Kinetic Energy, Rigid Body, Rotation, rod, ek

Click to edit the document details