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. $ 2 m 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

The correct answer is A:\(2mv_0^2\)
According to law of conservation of angular momentum 
\(mvr = mv'r'\)
\(v_0 R_0 = v (\frac{R_0}{2}) ; v = 2v_0 \, \, \, \, ...(i)\)
\(\therefore \frac{K_0}{K} = \frac{\frac{1}{2} mv_0^2}{\frac{1}{2}mv^2} = ( - \frac{v_0}{v}) ^2\) 
or \(\frac{K}{K_0} = (\frac{v}{v_0})^2 = (2)^2 \, \, \, . . .\) (Using (i)) 
\(K = 4K_0 = 2mv_0^2\)