Bouncing a Quarter Segment

Relevant wiki: Analyzing Elastic Collisions

The quarter sphere can be defined in polar coordinates by the inequalities $0 \le r \le R \hspace{1cm} 0 \le \theta \le \pi \hspace{1cm} -\tfrac14\pi \le \phi \le \tfrac14\pi$ Assuming that the quarter sphere has uniform density $\rho$ and total mass $M$ , then $M \; = \; \tfrac13\pi R^3 \rho$ and the centre of mass $G$ of the quarter sphere lies on the $x$ -axis with coordinates $(d,0,0)$ , where $\begin{aligned} Md & = \int_0^R\,dr \int_0^\pi\,d\theta \int_{-\frac14\pi}^{\frac14\pi}\,d\phi\,\rho r^2\sin\theta \times x \\ & = \rho \int_0^R\,dr \int_0^\pi\,d\theta \int_{-\frac14\pi}^{\frac14\pi}\,d\phi\,r^2 \sin\theta \times r \sin\theta \cos\phi \\ & = \tfrac14\rho R^4 \int_0^\pi\,d\theta \int_{-\frac14\pi}^{\frac14\pi} \sin^2\theta \cos\phi \,d\phi\; =\; \tfrac14\rho R^4 \times \tfrac12\pi \times \sqrt{2} \\ & = \frac{\pi}{4\sqrt{2}}\rho R^4 \end{aligned}$ so that $d \,=\, \tfrac{3}{4\sqrt{2}}R$ .

If $\hat{I}$ is the moment of inertia of the quarter sphere about the $y$ -axis, then $\begin{aligned} \hat{I} & = \int_0^R\,dr \int_0^\pi\,d\theta \int_{-\frac14\pi}^{\frac14\pi} \,d\phi\, \rho r^2\sin\theta (x^2 + z^2) \\ & = \rho \int_0^R\,dr \int_0^\pi\,d\theta \int_{-\frac14\pi}^{\frac14\pi} \,d\phi\, r^2\sin\theta (r^2\sin^2\theta\cos^2\phi + r^2\cos^2\theta) \\ & = \tfrac15\rho R^5 \int_0^\pi\,d\theta \int_{-\frac14\pi}^{\frac14\pi} \,d\phi\,(\sin^3\theta \cos^2\phi + \sin\theta\cos^2\theta) \\ & = \tfrac15\rho R^5 \int_{-\frac14\pi}^{\frac14\pi} \big(\tfrac43\cos^2\phi + \tfrac23\big)\,d\phi \\ & = \tfrac{2}{15}\rho R^5 \int_{-\frac14\pi}^{\frac14\pi}(\cos2\phi + 2)\,d\phi \; = \; \tfrac{2}{15}(\pi+1)\rho R^5 \\ & = \frac{2(\pi+1)}{5\pi}M R^2 \end{aligned}$

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 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: $MW \; = \; MV - J \hspace{2cm} Jd \; = \; I\omega$ 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 $\begin{aligned} \Delta T & = \tfrac12MV^2 - \tfrac12MW^2 - \tfrac12I\omega^2 \; = \; \tfrac12MV^2 - \tfrac12MW^2 - \tfrac12M(V-W)d\omega \\ & = \tfrac12M(V-W)(V+W - d\omega) \end{aligned}$ Since the collision is perfectly elastic, $\Delta T = 0$ , and hence $V + W = d\omega$ . Thus $\begin{aligned} I\omega & = Jd \; = \; Md(V-W) \; = \; Md(2V - d\omega) \\ (I + Md^2)\omega & = 2MVd \end{aligned}$ By the Parallel Axes Theorem, $\hat{I} = I + Md^2$ , so we deduce that $\begin{aligned} \omega & = \frac{2MVd}{\hat{I}} \; = \; 2MVd \times \frac{5\pi}{2(\pi+1)MR^2} \; = \; \frac{5\pi Vd}{(\pi+1)R^2} \; = \; \frac{15\pi V}{4(\pi+1)\sqrt{2} R} \end{aligned}$ Finally, $V = \sqrt{2gR}$ , and so $\omega \; = \; \frac{15\pi}{4(\pi+1)}\sqrt{\frac{g}{R}}$ Thus $\alpha \; = \; \frac{15\pi}{4(\pi+1)} \; = \; \boxed{2.84455}$