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

Answer: (A)

According to law of conservation of angular momentum
$mvr=m{v}^{\prime }{r}^{\prime }$
$v_0{R}_0=v(\frac{R_0}{2});v=2v_0...(i)$
$\therefore \frac{K_0}{K}=\frac{\frac{1}{2}m{v}_0^2}{\frac{1}{2}m{v}^2}=(-\frac{v_0}{v}{)}^2$
or $\frac{K}{K_0}=(\frac{v}{v_0}{)}^2=(2{)}^2...$ (Using (i))
$K=4K_0=2m{v}_0^2$