If a solid sphere of mass 1 kg and radius 0. | vedclass

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Question

(C) The total kinetic energy of a body rolling without slipping is the sum of its translational kinetic energy and rotational kinetic energy.$K_{total

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$\frac{7}{10} \ J$

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(C) The total kinetic energy of a body rolling without slipping is the sum of its translational kinetic energy and rotational kinetic energy.$K_{total} = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$For a solid sphere,the moment of inertia $I = \frac{2}{5}mr^2$.Since it rolls without slipping,$\omega = \frac{v}{r}$.Substituting these into the equation:$K_{total} = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2$$K_{total} = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$Given $m = 1 \ kg$ and $v = 1 \ m/s$:$K_{total} = \frac{7}{10} 	imes 1 	imes (1)^2 = 0.7 \ J = \frac{7}{10} \ J$.

If a solid sphere of mass $1 \ kg$ and radius $0.1 \ m$ rolls without slipping at a uniform velocity of $1 \ m/s$ along a straight line on a horizontal floor,the kinetic energy is

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