Question

A small sphere of radius $r$ and mass $m$ oscillates back and forth on a smooth concave surface of radius $R$ as shown in figure. Find the time period of small oscillations.

• A. $2\pi \sqrt{\frac{7R}{5g}}$
• B. $2\pi \sqrt{\frac{3g}{7R}}$
• C. $2\pi \sqrt{\frac{7(R-r)}{5g}}$
• D. $2\pi \sqrt{\frac{5(R-r)}{7g}}$

Answer 1

The correct option is C $2\pi \sqrt{\frac{7(R-r)}{5g}}$

When you displace the sphere through a small angle $\theta$.


Total energy of system
$\text{TE}=\text{KE}$ of sphere + $\text{PE}$ of sphere
$=(\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2)+mgh$
[Taking mean point as a reference point]
$=[\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}m{r}^2\right){\left(\frac{v}{r}\right)}^2]+mg(R-r)(1-\text{cos}\theta )$
$\Rightarrow TE=\frac{7}{10}m{v}^2+mg(R-r)(1-\text{cos}\theta )$
For SHM, $\frac{d}{dt}\left(TE\right)=0$
$\Rightarrow \frac{7}{10}m.2v\times \frac{dv}{dt}+mg(R-r)[0+\text{sin}\theta ]\frac{d\theta }{dt}=0$
$\Rightarrow \frac{7}{5}mva+mg(R-r)\text{sin}\theta \times {\omega }_0=0‘$
$[\because \frac{d\theta }{dt}={\omega }_0=\frac{v}{R-r}\&a=\alpha (R-r\left)\right]$
$\Rightarrow \frac{7}{5}m{\omega }_0(R-r)(R-r)\alpha +mg(R-r)\theta {\omega }_0=0$
$[\because \sin \theta \approx \theta \text{for small angle}]$
$\Rightarrow \alpha =\frac{-5g\theta }{7(R-r)}[\because \alpha =\frac{a}{R-r}]$
So ${\omega }^2=\frac{5g}{7(R-r)}[\because \alpha =-{\omega }^2\theta \text{for SHM}]$
So, $T=2\pi \sqrt{\frac{1}{{\omega }^2}}=2\pi \sqrt{\frac{7(R-r)}{5g}}$