Rotational Motion Question 54
Question: A round disc of moment of inertia $ I _2 $ about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia $ I _1 $ rotating with an angular velocity $ \omega $ about the same axis. The final angular velocity of the combination of discs is:
[AIPMT (S) 2004]
Options:
A) $ \frac{I _2\omega }{I _1+I _2} $
B) $ \omega $
C) $ \frac{I _1\omega }{I _1+I _2} $
D) $ \frac{(I _1+I _2)\omega }{I _1} $
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Answer:
Correct Answer: A
Solution:
- Key Idea: When no external torque acts on a system of particles, then the total angular momentum of the system remains always a constant.
The angular momentum of a disc of moment of inertia $ I _1 $ and rotating about its axis with angular velocity $ \omega $ is
$ L _1=I _1\omega $
When a round disc of moment of inertia $ I _2 $ is placed on first disc, then angular momentum of the combination is
$ L _2=(I _1+I _2)\omega $
In the absence of any external torque, angular momentum remains conserved i.e.,
$ L _1=L _2 $
$ I _1\omega =(I _1+I _2)\omega $
$ \Rightarrow $ $ \omega ‘=\frac{I _2\omega }{I+I _2} $
