Getting back to the old things #6

The quarter sphere has its axis horizontal, and can be defined in polar coordinates by the inequalities $0 \le r \le R \hspace{1cm} \tfrac12\pi \le \theta \le \pi \hspace{1cm} -\tfrac12\pi \le \phi \le \tfrac12\pi$ Assume that the sphere has uniform density and total mass $M$ , then the centre of mass $G$ of the quarter sphere now has coordinates $(c,0,-c)$ , where $c = \tfrac38R$ . If $\hat{I}$ is the moment of inertia of the quarter sphere about the $y$ -axis, then $\hat{I} \; = \; \tfrac25MR^2$ since this is just the moment of inertia of a slice of a sphere about its diameter (we have one quarter of a whole sphere, but also one quarter of the mass of a whole sphere, so the formula for the moment of inertia is the same as that of a whole sphere).

If the centre of mass of the quarter sphere is moving downwards with speed $V$ just before the collision, and is moving downwards with speed $W$ just after the collision, and if the quarter sphere also has angular velocity $\omega$ after the collision, having received an impulse of $J$ from the floor, then we have the impulsive equations of motion: $MW \; = \; MV - J \hspace{2cm} Jc \; = \; I\omega$ where $I$ is the moment of inertia of the quarter sphere about the axis passing through the centre of mass $G$ which is parallel to the $y$ -axis. Then the loss of kinetic energy during the collision is $\begin{aligned} \Delta T & = \tfrac12MV^2 - \tfrac12MW^2 - \frac12I\omega^2 \; = \; \tfrac12MV^2 - \tfrac12MW^2 - \tfrac12M(V-W)c\omega \\ & = \tfrac12M(V-W)(V+W - c\omega) \end{aligned}$ Since the collision is perfectly elastic, $\Delta T = 0$ , and hence $V + W = c\omega$ . Thus $\begin{aligned} I\omega & = Jc \; = \; Mc(V-W) \; = \; Mc(2V - c\omega) \\ (I + Mc^2)\omega & = 2MVc \end{aligned}$ By the Parallel Axes Theorem, $\hat{I} = I + 2Mc^2$ , so we deduce that $I + Mc^2 \; = \; \hat{I} - Mc^2 \; = \; \tfrac25MR^2 - \tfrac{9}{64}MR^2 \; =\; \tfrac{83}{320}MR^2$ and so $\begin{aligned} \omega & = \frac{2MVc}{I + mc^2} \; = \; 2MVc \times \frac{320}{83MR^2} \; = \; \frac{640Vc}{83R^2} \; = \; \frac{240 V}{83 R} \end{aligned}$ Finally, $V = \sqrt{2gR}$ , and so $\omega \; = \; \frac{240}{83}\sqrt{\frac{2g}{R}}$ making the answer $240 + 83 + 2 = \boxed{325}$ .

@Joe Wsp Therapper please post your solution.