First slide

Law of conservation of angular momentum

Question

A particle of mass m moves in a circle on a smooth horizontal plane with velocity v₀ at a radius R₀. 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 it moves in a circle of radius R₀/2. The final value of the kinetic energy is

Moderate

A

$\large \frac 12mv_0^2$
B

$\large mv_0^2$
C

$\large \frac 14mv_0^2$
D

$\large 2mv_0^2$

Solution

According to law of conservation of angular momentum about the centre of the circle, L_i = L_fFinal K.E of the particle
$\large {K_f} = \frac{1}{2}m{v^2} = \frac{1}{2}m{(2{v_0})^2} = 2mv_0^2$

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