Question

Inside a fixed sphere of radius $R$ and uniform density p, there is spherical cavity of radius $\frac{R}{2}$ such that surface of the cavity passes through the centre of the sphere as shown in figure. A particle of mass $m_0$ is released from rest at centre B of the cavity. Calculale velocity with which particle strikes the centre A of the sphere. Neglect earth's gravity. Initially sphere and particle are at rest.

Answer 1

Applying conservation of mechanical energy, Increase in kinetic energy=decrease in gravitational potential energy
or $\frac{1}{2}{m}_0{v}^2=U_B-U_A=m_0(V_B-V_A)$
$\therefore$ $v=\sqrt{2(V_B-V_A)}$ (i)
Potential at A
$V_A=$potential due to complete sphere-potential due to cavity
$=-\frac{1.5GM}{R}-[-\frac{Gm}{R/2}]$
$=\frac{2Gm}{R}-\frac{1.5GM}{R}$
Here, $m=\frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3\rho =\frac{\pi \rho R^3}{6}$
and $M=\frac{4}{3}\pi R^3\rho$
Substituting the valuest we get
$V_A=\frac{G}{R}[\frac{\pi \rho R^3}{3}-2\pi \rho R^3]=-\frac{5}{3}\pi G\rho R^2$
Potential at B
$V_B=-\frac{GM}{R^3}[1.5R^2-0.5{\left(\frac{R}{2}\right)}^2]+\frac{1.5Gm}{R/2}$
$=-\frac{11}{8}\frac{GM}{R}+\frac{3Gm}{R}$
$=\frac{G}{R}[\frac{\pi \rho R^3}{2}-\frac{11}{6}\pi \rho R^3]$
$=-\frac{4}{3}\pi G\rho R^2$
$\therefore$ $V_B-V_A=\frac{1}{3}\pi G\rho R^2$
So, from Eq. (i)
$v=\sqrt{\frac{2}{3}\pi G\rho R^2}$