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  4. A circular ring of mass m and radius r is rolling on a smooth horizontal surface with speed v. Its kinetic energy is

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Q. A circular ring of mass m and radius r is rolling on a smooth horizontal surface with speed v. Its kinetic energy is

Haryana PMTHaryana PMT 2003System of Particles and Rotational Motion

Solution:

KE of rotation = $\frac{1}{2}\frac{mr^2}{2}\times \omega^2$
$=\frac{1}{2}mr^2\times \frac{v^2}{r^2}$
$=\frac{1}{2}mv^2$