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AMU Engineering 2017 solved paper Solved Paper
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© examsnet.com
Question : 8
Total: 150
A thin circular ring of mass
M
and radius
r
is rotating about its axis at angular velocity
ω
. Two particles, each of mass
m
are attached gently to the ring at points which are at opposite ends of diameter of the ring. New angular velocity of the ring is
M
ω
M
+
2
m
ω
(
M
+
2
m
)
M
ω
ω
m
(
M
+
2
m
)
Validate
Solution:
Since the two objects are attached gently to opposite ends of diameter the ring. So no external torque is applied to the system. Therefore, angular momentum remains constant.
i.e.,
I
1
ω
1
=
I
z
ω
2
∴
ω
2
=
I
1
ω
1
I
2
=
M
r
2
ω
(
M
r
2
+
2
m
r
2
)
=
M
ω
(
M
+
2
m
)
© examsnet.com
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