Rotational Motion Question 39
Question: A thin circular ring of mass M and radius r is rotating about its axis with a constant angular velocity $ \omega $ . Two objects each of mass m are attached gently to the opposite ends of diameter of the ring. The ring will now rotate with an angular velocity:
[AIPMT 1998]
Options:
A) $ \frac{\omega (M-2m)}{(M+2m)} $
B) $ \frac{\omega M}{(M+2m)} $
C) $ \frac{\omega M}{(M+m)} $
D) $ \frac{\omega (M+2m)}{M} $
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Answer:
Correct Answer: B
Solution:
Angular momentum remains conserved in the universe.
According to conservation of angular momentum
$ L= $ constant
or $ l\omega =constant $
$ \therefore $ $ l _1{\omega_1}=l _2{\omega_2} $
Initial moment of inertia
$ l _1=MR^{2} $
and angular velocity
$ {\omega_1}=\omega $
Hence, $ l _1{\omega_1}=MR^{2}\omega $
When two objects of mass m are attached to opposite ends of a diameter, the final readings are
$ l _2=MR^{2}+mR^{2}+mR^{2} $
$ =(M+2m)R^{2} $
So, $ l _2{\omega_2}=(M+2m)R^{2}{\omega_2} $ ….(iii)
$ \therefore $ From Eqs.(i), (ii) and (iii)
$ MR^{2}\omega =(M+2m)R^{2}{\omega_2} $
$ \Rightarrow $ $ {\omega_2}=\frac{\omega M}{M+2m} $
