Rotational Motion Question 36
Question: A ring of mass M and radius R is rotating about its axis with angular velocity $ \omega $ . Two identical bodies each of mass m are now gently attached at the two ends of a diameter of the ring. Because of this, the kinetic energy loss will be:
[JEE ONLINE 25-04-2013]
Options:
A) $ \frac{m(M+2m)}{M}{{\omega }^{2}}R^{2} $
B) $ \frac{Mm}{(M+m)}{{\omega }^{2}}R^{2} $
C) $ \frac{Mm}{(M+2m)}{{\omega }^{2}}R^{2} $
D) $ \frac{(M+m )M}{(M+2m)}{{\omega }^{2}}R^{2} $
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Answer:
Correct Answer: C
Solution:
- [c] Kinetic energy Kinetic energy Solving we get loss in K.E.
By using angular momentum conservation
$MR^2×ω=(MR^2+MR^2+MR^2)ω′$
$ω′=\frac{Mω}{M+2m}$
BETA
