A solid sphere of mass 5 ,kg rolls on a plane | vedclass

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Question

(A) The total kinetic energy $(KE)$ of a rolling body is the sum of its translational and rotational kinetic energies, given by the formula:$KE = \fra

Vedclass · Accepted Answer

$56$

Vedclass · Answer

(A) The total kinetic energy $(KE)$ of a rolling body is the sum of its translational and rotational kinetic energies, given by the formula:$KE = \frac{1}{2} mv^2 + \frac{1}{2} I\omega^2$For a solid sphere, the moment of inertia about its centre is $I = \frac{2}{5} mR^2$ and the angular velocity is $\omega = \frac{v}{R}$.Substituting these values, we get:$KE = \frac{1}{2} mv^2 + \frac{1}{2} (\frac{2}{5} mR^2)(\frac{v}{R})^2 = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \frac{7}{10} mv^2$Given $m = 5 \,kg$ and $v = 4 \,m/s$:$KE = \frac{7}{10} 	imes 5 	imes (4)^2 = \frac{7}{10} 	imes 5 	imes 16 = 7 	imes 8 = 56 \,J$Thus, the correct option is $A$.

$A$ solid sphere of mass $5 \,kg$ rolls on a plane surface. Find its kinetic energy at an instant when its centre moves with speed $4 \,m/s$. (in $\,J$)

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