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
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