A mass $m$ moves in a circle on a smooth horizontal plane with velocity $v_0$ at a radius $R_0$. The mass is a

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 that passes through a smooth hole in the plane, as shown in the figure. 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:

Accepted Answer

The force on the mass is the tension in the string, which acts along the radius towards the centre (the hole). Since the line of action of the force passes through the centre, the torque about the centre is zero ($	au = r 	imes F = 0$). 
According to the law of conservation of angular momentum, if the net external torque is zero, the angular momentum remains constant.
Initial angular momentum, $L_i = m v_0 R_0$
Final angular momentum, $L_f = m v_f \left(\frac{R_0}{2}ight)$
Equating initial and final angular momentum:
$m v_0 R_0 = m v_f \left(\frac{R_0}{2}ight)$
$v_f = 2v_0$
The final kinetic energy is given by:
$K_f = \frac{1}{2} m v_f^2 = \frac{1}{2} m (2 v_0)^2 = \frac{1}{2} m (4 v_0^2) = 2 m v_0^2$

Step-by-Step Solution

Practice All PHYSICS Questions