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Question

Using conservation of energy.

$m g h=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}$

$m g h=\frac{1}{2} m v^{2}+\frac{1}{2} m k^{2} \omega^{2}$&nb

Vedclass · Accepted Answer

$21 : 28 : 30$

Vedclass · Answer

Using conservation of energy.

$m g h=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}$

$m g h=\frac{1}{2} m v^{2}+\frac{1}{2} m k^{2} \omega^{2}$           where $k:$ radius of gyration

$m g h=\frac{\overline{1}}{2} m v^{2}+\frac{\overline{1}}{2} m v^{2} \frac{k^{2}}{r^{2}}$

$K E=\frac{P E}{1+k^{2} / r^{2}}$

$\Longrightarrow K E_{	ext {ring}}: K E_{	ext {coin}}: K E_{	ext {solid} b a l l}=\frac{1}{1+1}: \frac{1}{1+1 / 2}: \frac{1}{1+2 / 5}$

$K E_{	ext {ring}}: K E_{	ext {coin}}: K E_{	ext {solid} b a l l}=21: 28: 30$

Starting from the rest, at the same time, a ring, a coin and a solid ball of same mass roll down an incline without slipping .The ratio of their translational kinetic energies at the bottom will be

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