Question

A spherical ball of radius $R$ and mass $M$ submerged inside the water. Ball has non-uniform mass density so that it's centre of mass is at $R/8$ distance from centre shown in figure.
$I_C=$ Moment of inertia about $\text{COM}$


If we tilt a ball slightly about its centre, what is the time period of angular oscillation of the ball?
(density of water = density of ball)

• A. $T=2\pi \sqrt{\frac{4I_C}{MgR}}$
• B. $T=2\pi \sqrt{\frac{8I_C}{MgR}}$
• C. $T=2\pi \sqrt{\frac{I_C}{MgR}}$
• D. $T=2\pi \sqrt{\frac{2I_C}{MgR}}$

Answer 1

The correct option is B $T=2\pi \sqrt{\frac{8I_C}{MgR}}$

After giving slight tilting,


Torque on ball about point $\text{C}$,

$\tau =Bd$

$(B=\text{Buoyancy}=Mg)$

$\therefore$ Restating torque will be

$\tau =-B\left(R/8\right)\sin \theta$

For small angle, $\sin \theta \approx \theta$

$\tau =-B\left(R/8\right)\sin \theta \approx -\frac{BR}{8}\theta$

$I_C\alpha =I_C\frac{d^2\theta }{d{t}^2}=-\frac{MgR}{8}\theta$

$\therefore \frac{d^2\theta }{d{t}^2}+\frac{MgR}{8I_C}\theta =0$

Compare with general equation of SHM

${\omega }^2=\frac{MgR}{8I_C}$

$\therefore T=2\pi \sqrt{\frac{8I_C}{MgR}}$

Hence, option $\left(\text{B}\right)$ is the right answer.

Why this question ? Tip :- In simple harmonic motion of floating body, first try to find out force couple responsible for restoring torque and by equating torque with $I\alpha$, and then find out time period of oscillation.