Question

A cylinder of radius $r$ and mass $m$ rests on a curved path of radius $R$ as shown in the figure. It was found that the cylinder can oscillate about the bottom position when displaced and left to itself. Find the period of oscillation .Assume that the cylinder rolls without slipping :

• A. $T=2\pi \sqrt{\frac{3(R-r)}{2g}}$
• B. $T=2\pi \sqrt{\frac{g}{R-r}}$
• C. $T=2\pi \sqrt{\frac{8(R-r)}{g}}$
• D. $T=2\pi \sqrt{\frac{R-r}{g}}$

Answer 1

The correct option is A $T=2\pi \sqrt{\frac{3(R-r)}{2g}}$

1. To find the time period of oscillation of cylinder. In this process cylinder rotates about it axis by a max value .say it is ${\theta }_0$ at point B.So, its angular accn $\frac{d^2{\theta }_0}{d{t}^2}$ (about point B)
   

The velocity of cylinder about its centre $=r{\omega }_0|{\omega }_0$= angular velocity of cylinder itself]

The velocity of cylinder about centre of large sphere is$=(R-r)\omega$ [$\omega$ = angualr velocity of cylinder about centre of large sphere ]

So, we can have (R-r) $\theta =r{\theta }_0$

and $(R-r)\frac{d^2\theta }{d{t}^2}=r\frac{d^2{\theta }_0}{d{t}^2}$

$\theta$ angular position of cylinder about vertical .

${\theta }_0$ angle of rotation of cylinder about point of contact

Hence the restoring torque about B

T = -mg r $\sin \theta$

T $\simeq$ -mg r $\theta$

scince $\left|\theta \right|$ is small

Moment of inertial of cylinder about point B

$I_B=\frac{1}{2}m{p}^2+m{p}^2=\frac{3}{2}m{p}^2$

Again T = $I_B\frac{d^2{\theta }_0}{d{t}^2}$

So,$I_B\frac{d^2{\theta }_0}{d{t}^2}=-mgr\Rightarrow \frac{S}{2}m{p}^2\frac{d^2{\theta }_0}{d{t}^2}+mgr\theta =0\Rightarrow \frac{d^2{\theta }_0}{d{t}^2}+\frac{2g}{3r}\theta =0$

Putting value of $\theta$

$\Rightarrow \frac{d^2{\theta }_0}{d{t}^2}+\frac{2g}{3r}.\frac{r}{(R-r)}{\theta }_0=0\Rightarrow \frac{d^2{\theta }_0}{d{t}^2}+\frac{2g}{3(R-r)}{\theta }_0=0$

So, the angular frequency or angular velocity of the motion

${\omega }_0=\sqrt{\frac{2g}{3(R-r)}}T=\frac{2\pi }{{\omega }_0}=2\pi \sqrt{\frac{3(R-r)}{2g}}$