Question

A solid sphere (radius = R) rolls without slipping in a cylindrical through (radius = 5R). Find the time period of small oscillations.

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• A. $2\pi \sqrt{\frac{28R}{5g}}$
• B. $2\pi \sqrt{\frac{2R}{5g}}$
• C. $2\pi \sqrt{\frac{8R}{5g}}$
• D. $2\pi \sqrt{\frac{28R}{g}}$

Answer 1

The correct option is A

$2\pi \sqrt{\frac{28R}{5g}}$

For pure rolling to take place,

$v=R\omega {\omega }^{\prime }=\text{angular velocity of COM of sphere C about O}\frac{v}{4R}=\frac{R\omega }{4R}=\frac{\omega }{4}\therefore \frac{d{\omega }^{\prime }}{dt}=\frac{1}{4}\frac{d\omega }{dt}or{\alpha }^{\prime }=\frac{\alpha }{4}\alpha =\frac{a}{R}\text{for pure rolling}where,a=\frac{gsin\theta }{1+\frac{I}{m{R}^2}}=\frac{5gsin\theta }{7}as,I=\frac{2}{5}m{R}^2\therefore {\alpha }^{\prime }=\frac{5gsin\theta }{28R}Forsmall\theta ,sin\theta \approx \theta ,\text{being restoring in nature,}{\alpha }^{\prime }=-\frac{5g}{28R}\theta \therefore T=2\pi \sqrt{\left|\frac{\theta }{{\alpha }^{\prime }}\right|}=2\pi \sqrt{\frac{28R}{5g}}$