A solid sphere of mass 500 g and radius 10 | vedclass

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(A) Given: Mass $M = 500 \ g = 0.5 \ kg$,Radius $R = 10 \ cm = 0.1 \ m$,Velocity $v = 20 \ cm/s = 0.2 \ m/s$.For a solid sphere,the moment of inertia

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$0.014$

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(A) Given: Mass $M = 500 \ g = 0.5 \ kg$,Radius $R = 10 \ cm = 0.1 \ m$,Velocity $v = 20 \ cm/s = 0.2 \ m/s$.For a solid sphere,the moment of inertia about its center is $I = \frac{2}{5} MR^2$.For pure rolling,the total kinetic energy is the sum of translational and rotational kinetic energy: $K.E._{total} = K.E._{trans} + K.E._{rot} = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2$.Since $v = \omega R$,we have $\omega = \frac{v}{R}$.Substituting $I$ and $\omega$ into the equation: $K.E._{total} = \frac{1}{2} Mv^2 + \frac{1}{2} (\frac{2}{5} MR^2) (\frac{v}{R})^2 = \frac{1}{2} Mv^2 + \frac{1}{5} Mv^2 = \frac{7}{10} Mv^2$.Substituting the values: $K.E._{total} = \frac{7}{10} 	imes 0.5 	imes (0.2)^2 = 0.7 	imes 0.5 	imes 0.04 = 0.35 	imes 0.04 = 0.014 \ J$.

$A$ solid sphere of mass $500 \ g$ and radius $10 \ cm$ rolls without slipping with a velocity of $20 \ cm/s$. The total kinetic energy of the sphere will be ........ $J$.

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