A mass moves in a circle on a smooth horizontal plane with velocity at a radius . 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 moves in a circle

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Solution

Q.
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 [2015]

1 $2 m v_{0}^{2}$

2 $\frac{1}{2} m v_{0}^{2}$

3 $m v_{0}^{2}$

4 $\frac{1}{4} m v_{0}^{2}$

Ans.
(1)

According to law of conservation of angular momentum

   $m v r = m v' r'$

   $v_{0} R_{0} = v (\frac{R_{0}}{2});$ $v = 2 v_{0}$ ...(i)

$∴$    $\frac{K_{0}}{K} = \frac{\frac{1}{2} m v_{0}^{2}}{\frac{1}{2} m v^{2}} = {(\frac{v_{0}}{v})}^{2}$

or    $\frac{K}{K_{0}} = {(\frac{v}{v_{0}})}^{2} = {(2)}^{2} (Using (i))$

   $K = 4 K_{0} = 2 m v_{0}^{2}$

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