A mass $\mathrm{m}$ moves in a circle on a smooth horizontal plane with velocity $\mathrm{v}_{0}$ at a…

A mass $\mathrm{m}$ moves in a circle on a smooth horizontal plane with velocity $\mathrm{v}_{0}$ at a radius $\mathrm{R}_{0} .$ The mass is attached to string which passes through a smooth hole in the plane as shown. The tension in the string is increased gradually and finally $\mathrm{m}$ moves in a circle of radius $\frac{\mathrm{R}_{0}}{2} .$ The final value of the kinetic energy is

1. $\frac{1}{4} \mathrm{mv}_{0}^{2}$
2. $2 \mathrm{mv}_{0}^{2}$
3. $\frac{1}{2} \mathrm{mv}_{0}^{2}$
4. $\mathrm{mv}_{0}^{2}$

Solution

Applying angular momentum conservation


$\mathrm{mV}_{0} \mathrm{R}_{0}=(\mathrm{m})\left(\mathrm{V}^{1}\right)\left(\frac{\mathrm{R}_{0}}{2}\right)$
$\therefore \quad \mathrm{v}^{1}=2 \mathrm{~V}_{0}$
Therefore, new $\mathrm{KE}=\frac{1}{2} \mathrm{~m}\left(2 \mathrm{~V}_{0}\right)^{2}=2 \mathrm{mv}_{0}^{2}$ ~

Asked in: JEE Mains - Rotational Motion - Test 4

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