Question

The figure shows a system consisting of (i) a ring of outer radius 3R rolling clockwise without slipping on a horizontal surface with angular speed $\omega$ and (ii) an inner disc of radius 2R 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 OP makes an angle of $30^{\circ }$ with the horizontal. Then with respect to the horizontal surface,

• A. the point $O$ has a linear velocity $3R\omega \hat{i}$
• B. the point $P$ has a linear velocity $\frac{11}{4}R\omega \hat{i}+\frac{\sqrt{3}}{4}R\omega \hat{k}$
• C. the point P has a linear velocity $\frac{13}{4}R\omega \hat{i}-\frac{\sqrt{3}}{4}R\omega \hat{k}$
• D. the point P has a linear velocity $\left(3-\frac{\sqrt{3}}{4}\right)R\omega \hat{i}+\frac{1}{4}R\omega \hat{k}$

Answer 1

Answer: (B)

Velocity of point $O$ is
$v_0=(3r\omega )\hat{i}$
$v_{PO}$ is $\frac{R.\omega }{2}$ in the direction shown in figure.
In vector form
$V_{PO}=-\frac{R\omega }{4}\sin 30^{\circ }\hat{i}+\frac{R\omega }{2}\cos 30^{\circ }\hat{k}$
$V_{PO}=-\frac{R\omega }{4}\sin 30^{\circ }\hat{i}$
$=-\frac{R\omega }{4}\hat{i}+\frac{\sqrt{3}R\omega }{4}\hat{k}$
But $V_{PO}=v_P-v_O$
$\therefore V_P=v_{PO}-v_O$
$=\left(-\frac{R\omega }{4}\hat{i}+\frac{\sqrt{3}R\omega }{4}\hat{k}\right)+3R\omega \hat{i}$
$=\frac{11}{4}R\omega \hat{i}+\frac{\sqrt{3}}{4}R\omega \hat{k}$