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 $R_0/2$. Then, the final value of the kinetic energy of particle is

• A. $\frac{1}{2}m{v_0}^2$
• B. $m{v_0}^2$
• C. $\frac{1}{4}m{v_0}^2$
• D. $2m{v_0}^2$

Answer 1

The correct option is D $2m{v_0}^2$

Since no external torque is acting on the system, thus according to law of conservation of angular momentum,
Initial angular momentum = Final angular momentum
$m{v}_0{R}_0=mvr\Rightarrow m{v}_0{R}_0=\frac{mv{R}_0}{2}\left(\because \mathrm{New}\mathrm{radius},r=\frac{R_0}{2}\right)\Rightarrow v=2v_0\mathrm{New}\mathrm{kinetic}\mathrm{energy}=\frac{1}{2}m{v}^2=2m{v}_0^2$