A hollow sphere of mass 2 , kg and diameter 0 | vedclass

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(D) The total kinetic energy of a rolling body is given by $K = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.For a hollow sphere,the

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$0.42$

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(D) The total kinetic energy of a rolling body is given by $K = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.For a hollow sphere,the moment of inertia $I = \frac{2}{3}mR^2$ and $\omega = \frac{v}{R}$.Substituting these,$K = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2(1 + \frac{2}{3}) = \frac{1}{2}mv^2(\frac{5}{3})$.Given: $m = 2 \, kg$,$v = 0.5 \, m/s$.$K = \frac{1}{2} 	imes 2 	imes (0.5)^2 	imes \frac{5}{3} = 0.25 	imes \frac{5}{3} = 0.4166... \approx 0.42 \, J$.

$A$ hollow sphere of mass $2 \, kg$ and diameter $0.2 \, m$ is rolling on an inclined plane with a velocity of $0.5 \, m/s$. The kinetic energy of the sphere will be .......... $J$.

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