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

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(C) When a body rolls without slipping,its total kinetic energy $(K.E.)$ is the sum of its translational kinetic energy and rotational kinetic energy.

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

Vedclass · Answer

(C) When a body rolls without slipping,its total kinetic energy $(K.E.)$ is the sum of its translational kinetic energy and rotational kinetic energy.$K.E. = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$For a solid sphere,the moment of inertia $I = \frac{2}{5}mr^2$ and the angular velocity $\omega = \frac{v}{r}$.Substituting these values:$K.E. = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2$$K.E. = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$Given $m = 1 \ kg$ and $v = 1 \ m/s$:$K.E. = \frac{7}{10} 	imes 1 	imes (1)^2 = \frac{7}{10} \ J$

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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