A solid sphere of mass m and radius r is allo | vedclass

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Question

(A) Using the Work-Energy Theorem $(WET)$: $W_g = K_f - K_i$ Since the sphere starts from rest,$K_i = 0$. The work done by gravity is $W_g = mgL \sin

Vedclass · Accepted Answer

$1: \sqrt{2}$

Vedclass · Answer

(A) Using the Work-Energy Theorem $(WET)$: $W_g = K_f - K_i$ Since the sphere starts from rest,$K_i = 0$. The work done by gravity is $W_g = mgL \sin 	heta$. The total kinetic energy $(K.E.)$ in pure rolling is given by: $K_f = \frac{1}{2} mv^2 + \frac{1}{2} I_{cm} \omega^2$ For a solid sphere,$I_{cm} = \frac{2}{5} mr^2$ and for pure rolling,$\omega = \frac{v}{r}$. $K_f = \frac{1}{2} mv^2 + \frac{1}{2} (\frac{2}{5} mr^2) (\frac{v^2}{r^2}) = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \frac{7}{10} mv^2$. Equating work done to change in kinetic energy: $mgL \sin 	heta = \frac{7}{10} mv^2$ $v^2 = \frac{10}{7} gL \sin 	heta$ This implies $v^2 \propto \sin 	heta$. Therefore,the ratio is: $\frac{v_1^2}{v_2^2} = \frac{\sin 30^{\circ}}{\sin 45^{\circ}} = \frac{1/2}{1/\sqrt{2}} = \frac{1}{\sqrt{2}}$. Thus,the ratio $v_1^2 : v_2^2$ is $1 : \sqrt{2}$.

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