Question

A ball of mass m is connected by a light inextensible cord and is rotated in a circle of radius R on a smooth fixed horizontal table. Initially the angular velocity of the ball was ${\omega }_0$ and pulling force applied was $T_1$. Now the pulling force is increased to $T_2$, until the radius of rotation of the ball becomes $\frac{R}{4}$. Then ratio $\frac{T_2}{T_1}$ is?

• A. $64$
• B. $16$
• C. $48$
• D. $256$

Answer 1

The correct option is A $64$

$I_0$ of ball $=m{R}^2$

$I^{\prime }$ of ball $=\frac{m{R}^2}{16}$

By conservation of angular momentum,

$m{R}^2{\omega }_0=\frac{m{R}^2}{16}{\omega }^{\prime }{\omega }^{\prime }=16{\omega }_0$

$T$ is force resposible for centripetal force

$T_1={\omega }_0^{2}k$ and $T_2=m{\omega }^{\prime 2}{R}^{\prime }$

$\therefore$ $T_2=m16{\omega }_0^2\frac{R}{4}=64m{\omega }_0^{2}R=64T_1$

$\therefore$ $\frac{T_1}{T_2}=64$