Rotational Motion Question 50
Question: A thin circular ring of mass M and radius r is rotating about its axis with a constant angular velocity $ \omega $ . Four objects each of mass m, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be:
[AIPMT 2003]
Options:
A) $ \frac{(M+4m)\omega }{M} $
B) $ \frac{(M-4m)\omega }{M+4m} $
C) $ \frac{M\omega }{4m} $
D) $ \frac{M\omega }{M+4m} $
Show Answer
Answer:
Correct Answer: D
Solution:
-
Key Idea: If external torque acting on the system is zero, hence angular momentum remains conserved.
$ {\tau _{ext}}=0 $
or $ \frac{dL}{dt}=0 $
or $ L= $ constant
or $ I\omega = $ constant
$ \therefore $ $ I _1{\omega_1}=I _2{\omega_2} $ (i)
Here, $ I _1=Mr^{2},{\omega_1}=\omega ,I _2=Mr^{2}+4mr^{2} $
Hence, Eq.(i) can be written as $ Mr^{2}\omega =(Mr^{2}+4mr^{2}){\omega_2} $
$ \therefore $ $ {\omega_2}=\frac{M\omega }{M+4m} $
