Question

A spherical cavity of radius $\frac{R}{2}$ is removed from a solid sphere of radius $R$ and the sphere is placed on a rough horizontal surface as shown in the figure. The sphere is given a gentle push and the friction is large enough to prevent slipping. Find the time period of oscillation of the sphere.

• A. $2\pi \sqrt{\frac{20R}{17g}}$
• B. $2\pi \sqrt{\frac{4R}{7g}}$
• C. $2\pi \sqrt{\frac{147R}{17g}}$
• D. $2\pi \sqrt{\frac{177R}{10g}}$

Answer 1

The correct option is D $2\pi \sqrt{\frac{177R}{10g}}$

From the figure, we can deduce that position of centre of mass of solid sphere $(x_1,y_1)=(0,0)\text{units}$
Position of centre of mass of cavity $(x_2,y_2)=(0,\frac{R}{2})\text{units}$
Then, position of centre of mass of the sphere with cavity is given by
$y_{cm}=\frac{\rho V_1{y}_1-\rho V_2{y}_2}{\rho (V_1-V_2)}$
$\Rightarrow \frac{4}{3}\pi (R^3-\frac{R^3}{8})\rho y_{cm}=\frac{4}{3}\pi \frac{R^3}{8}\times \rho \times \frac{R}{2}$
$\Rightarrow \frac{7}{8}{y}_{cm}=-\frac{R}{16}$
$\Rightarrow y_{cm}=x\left(say\right)=-\frac{R}{14}\text{units}$


Moment of inertia of the cavitied sphere about an axis $\perp$ to plane of the figure, through point of contact $\left(P\right)$ is calculated as follows-
Let $M=$ mass of sphere with cavity
From definition ,
$M=\frac{4}{3}\pi R^3(1-\frac{1}{8})\rho \Rightarrow \rho =\frac{3M}{4\pi R^3}\times \frac{8}{7}\text{units}$
Mass of the removed sphere of radius $\frac{R}{2}$ is $m$
$\Rightarrow m=\rho \frac{4}{3}\pi \times \frac{R^3}{8}=\frac{M}{7}\text{units}$
Mass of sphere without cavity $M_0=m+M=\frac{8M}{7}\text{units}$
$\therefore$ Required moment of inertia
$I=\text{(M.I of complete sphere without cavity about an axis through P) - (M.I of the cavity about an axis through P)}$
$I=\frac{7}{5}{M}_0{R}^2-[\frac{2}{5}m{\left(\frac{R}{2}\right)}^2+m{\left(\frac{3R}{2}\right)}^2]$
$=\frac{7}{5}\times \frac{8M}{7}\times R^2-\left[\frac{47}{20}\frac{M}{7}{R}^2\right]=\frac{177}{140}M{R}^2$


Now, the sphere is displaced gently by $\theta$.
Restoring torque about point $P$ is given by
$\tau =Mgx\sin \theta$
When $\theta$ is very small $\sin \theta \approx \theta$
$\Rightarrow \tau =-\frac{R}{14}Mg\theta$
But we know that, $\tau =I\alpha$
$\therefore I\alpha =-\frac{R}{14}Mg\theta$
$\Rightarrow \frac{177}{140}M{R}^2\alpha =-\frac{R}{14}Mg\theta$
$\Rightarrow \alpha =-\frac{140}{177\times 14}\frac{g}{R}\theta$
Comparing this with $\alpha =-{\omega }^2\theta$, we get
$\omega =\sqrt{\frac{10}{177}\frac{g}{R}}$
$\Rightarrow$ Time period of oscillation
$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{177R}{10g}}$
Thus, option (d) is the correct answer.