Rotational Motion Question 83
Question: A mass m moves in a circle on a smooth horizontal plane with velocity $ v _0 $ at a radius $ R _0 $ . The mass is attached to a string which passes through a smooth hole in the plane as shown. The tension in the string is increased gradually and finally m moves in a circle of radius $ \frac{R _0}{2} $ . The final value of the kinetic energy is
Options:
A) $ mv_0^{2} $
B) $ \frac{1}{4}mv_0^{2} $
C) $ 2mv_0^{2} $
D) $ \frac{1}{2}mv_0^{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
-
Conserving angular momentum
$ L _{i}=L _{t} $
$ \Rightarrow $ $ mv _0R _0=mv’( \frac{R _0}{2} ) $
$ \Rightarrow $ $ v’=2v _0 $
So, final kinetic energy of the particle is
$ K _{f}=\frac{1}{2}mv{{’}^{2}}=\frac{1}{2}m{{(2v _0)}^{2}} $
$ =4\frac{1}{2}mv_0^{2}=2mv_0^{2} $
