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

• A. $2m{v}_0^2$
• B. $\frac{1}{2}m{v}_0^2$
• C. $m{v}_0^2$
• D. $\frac{1}{4}m{v}_0^2$

Answer 1

The correct option is A $2m{v}_0^2$

Angular momentum L is conserved since there is no torque acting on the particle.
$\therefore m{v}_0{R}_0=m{v}^{\prime }\frac{R_0}{2}$
$\Rightarrow v^{\prime }=2V_0$
$K{E}_{final}=\frac{1}{2}m{v}^{\prime 2}=2m{V}_0^2$