A uniform sphere of mass 500,g rolls without | vedclass

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Question

(A) The total kinetic energy $(K)$ of a rolling sphere is the sum of its translational kinetic energy and rotational kinetic energy.$K = K_{	ext{tran

Vedclass · Accepted Answer

$1.4 	imes 10^{-4}\,J$

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(A) The total kinetic energy $(K)$ of a rolling sphere is the sum of its translational kinetic energy and rotational kinetic energy.$K = K_{	ext{trans}} + K_{	ext{rot}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$For a solid sphere,the moment of inertia $I = \frac{2}{5}mR^2$.Since the sphere rolls without slipping,$v = R\omega$,which implies $\omega = \frac{v}{R}$.Substituting these into the energy equation:$K = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2$$K = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$Given: $m = 500\,g = 0.5\,kg$ and $v = 0.02\,m/s$.$K = \frac{7}{10} 	imes 0.5 	imes (0.02)^2$$K = 0.7 	imes 0.5 	imes 0.0004 = 0.35 	imes 0.0004 = 0.00014\,J = 1.4 	imes 10^{-4}\,J$.

$A$ uniform sphere of mass $500\,g$ rolls without slipping on a plane surface such that its centre moves at a speed of $0.02\,m/s$. The total kinetic energy of the rolling sphere would be (in $J$):

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