The figure shows a system consisting of (i) a ring of outer radius $3 R$ rolling clockwise without slipping…

The figure shows a system consisting of (i) a ring of outer radius $3 R$ rolling clockwise without slipping on a horizontal surface with angular speed $\omega$ and (ii) an inner disc of radius $2 R$ rotating anti-clockwise with angular speed $\omega / 2$. The ring and disc are separated by frictionless ball bearings. The system is in the $x-z$ plane. The point $P$ on the inner disc is at a distance $R$ from the origin, where $O P$ makes an angle of $30^{\circ}$ with the horizontal. Then with respect to the horizontal surface,
  1. the point $O$ has linear velocity $3 R \omega \hat{i}$
  2. the point $P$ has linear velocity $\frac{11}{4} R \omega \hat{i}+\frac{\sqrt{3}}{4} R \omega \hat{k}$.
  3. the point $P$ has linear velocity $\frac{13}{4} R \omega \hat{i}-\frac{\sqrt{3}}{4} R \omega \hat{k}$
  4. the point $P$ has linear velocity $\left(3-\frac{\sqrt{3}}{4}\right) R \omega \hat{i}+\frac{1}{4} R \omega \hat{k}$

Solution

Velocity at centre 'O' $\quad \therefore \quad \vec{v}_{o}=3 R \omega \hat{i}$
$\vec{V}_{P}=3 R \omega \hat{i}-\frac{R \omega}{2} \sin 30^{\circ} \hat{i}+\frac{R \omega}{2} \cos 30^{\circ} \hat{k}$ $\therefore \quad \vec{V}_{P}=\left[3 R_{\omega} \hat{i}-\frac{R_{\omega}}{4} \hat{i}\right]+\frac{\sqrt{3} R_{\omega}}{4} \hat{k}$ or, $\quad \vec{V}_{P}=\frac{11}{4} R_{\omega} \hat{i}+\frac{\sqrt{3}}{4} R_{\omega} \hat{k}$ !

Asked in: JEE Advanced 2012 (Paper 2)

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