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 plane as shown. The tension in the string is increased gradually and finally m moves in a circle of radius R 0 2 . The final value of…

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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 plane as shown. The tension in the string is increased gradually and finally m moves in a circle of radius R 0 2 . The final value of the kinetic energy is:

Aicharya · Accepted Answer

From conservation of angular momentum. $\begin{aligned} & m v_0 R_0=m v^{\prime}\left(\frac{R_0}{2}\right) \ & \Rightarrow v^{\prime}=2 v_0 \end{aligned}$ Hence, final $K E=\frac{1}{2} m v^{\prime 2}=\frac{1}{2} m\left(2 v_0\right)^2$ Final $KE=2 m v_0^2$ ~

A mass m moves in a circle on a smooth horizontal plane with velocity v 0 at a radius R 0 . The mass is…

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