This preview shows pages 1–2. Sign up to view the full content.
DATE:
October 26, 2011
DUE:
November 4, 2011
ENG102, Solution to homework set #6.
1.
Look at Problem 3/354 on page 266 in the book.
a. Solve the problem.
Both momentum and angular momentum are conserved. Thus:
H
G
=
m
(
v
A
+
v
B
)0
.
4
m
= 2
m
(0
.
4
m
)
2
˙
θ
′
(1)
⇒
˙
θ
′
=
v
A
+
v
B
2
·
0
.
4
m
= 10
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 kinetic energy:
E
k
=
1
2
mv
2
A
+
1
2
mv
2
B
.
Final kinetic energy:
E
′
k
=
m
(0
.
4
m
)
2
(
˙
θ
′
)
2
+
m
(
v
′
)
2
=
1
2
m
(
v
2
A
+
v
2
B
)
. Thus,
the kinetic energy is conserved, and there is no energy loss in the process.
c. Assume that the point
G
is held fixed (by a nail, or something) such that the
object can rotate freely around
G
. Solve the problem (1a) with this change.
Now momentum is not conserved. Angular momentum around the axis of the
nail (G) is conserved. Thus
H
G
=
m
(
v
A
+
v
B
)0
.
4
m
= 2
m
(0
.
4
m
)
2
˙
θ
′
(5)
⇒
˙
θ
′
=
v
A
+
v
B
2
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.
 Fall '08
 Eke

Click to edit the document details