Question

A hollow sphere of radius r rolls without slipping in a hemisphere of radius 4r. Calculate the frequency of small oscillations, about point P as shown in figure.

• A. $\frac{1}{2\pi }\sqrt{\frac{g5}{r}}$
• B. $\frac{1}{2\pi }\sqrt{\frac{g}{5r}}$
• C. $2\pi \sqrt{\frac{r}{g5}}$
• D. None of these

Answer 1

The correct option is B

$\frac{1}{2\pi }\sqrt{\frac{g}{5r}}$

The hollow sphere is purely rolling on the curvature. Let's assume the sphere is at some $\theta$ from the vertical. At this instant its C.O.M has a velocity V. It will be rolling with angular velocity $\omega$ such that $\omega =\frac{v}{r}$

The centre of the sphere is doing circular motion about the centre of the curvature 0. Centre of sphere is moving with velocity v at distance 3r so its angular velocity about 0 is ${\omega }^{\prime }=\frac{v}{3r}$ - - - - - - (1)

Let's write the energy of the sphere at this instant. It will have rotational + translational energy + potential energy.

$\underset{}{\frac{1}{2}m{v}^2}$ + $\underset{}{\frac{1}{2}l{\omega }^2}$ + $\underset{}{mg3r(1-cos\theta }=E$

Translational Rotational Potential

Energy Energy Energy

Now $\therefore$ energy = constant

$\frac{dE}{dt}=0$

$\frac{1}{2}m2v\frac{dv}{dt}+\frac{d\left(\frac{12}{23}m{r}^2{\omega }^2\right)}{dt}-mg3Rsin\theta \frac{d\theta }{dt}=0$ (negative sign as $\frac{d\theta }{dt}$ is decreasing)

$(\therefore r^2{\omega }^2=v^2)\left(\begin{array}{c}\therefore \frac{d\theta }{dt}={\omega }^{\prime }\\ {\omega }^{\prime }=\frac{v}{3r}\end{array}\right)$

$\frac{1}{2}m2v\frac{dv}{dt}+\frac{1}{2}\frac{2}{3}m.2v\frac{dv}{dt}-mg3rsin\theta \frac{d\theta }{dt}=0$

$mva+\frac{2}{3}mva-mg3rsin\theta \frac{v}{3r}$

$\frac{5}{3}mva-mgsin\theta v=0$

$\frac{5}{3}ma=mgsin\theta$

$a=\frac{3gsin\theta }{5}$

If $\theta$ was small then angular acceleration of center of mass with respect to center of curvature o will be

$\alpha =\frac{3gsin\theta }{5}\left(\frac{1}{3r}\right)$

$\alpha =\frac{-g\theta }{5r}$

$\omega =\sqrt{\frac{g}{5r}}$

$f=\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{g}{5r}}$