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 anticlockwise with angular speed $\frac{\omega }{2}.$ The ring and the disc are separated by frictionless ball bearings. The system is in the xz-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 this horizontal, where O is the centre of the disc. Then with respect to this horizontal surface

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

Answer 1

Answer: (A), (B)

The point P has linear velocity $(\frac{11}{4}R\omega \hat{i}+\frac{\sqrt{3}}{4}R\omega \hat{k})$


${\overrightarrow{V}}_0-\left(3R\right)\omega \hat{i}=0.$
$\therefore {\overrightarrow{V}}_0=3R\omega \hat{i}.$
Now, ${\overrightarrow{V}}_{P,O}=-\frac{R\omega }{4}\hat{i}+\frac{R\omega \sqrt{3}}{4}\hat{j}$
$\Rightarrow {\overrightarrow{V}}_p={\overrightarrow{V}}_{P,O}+\overrightarrow{V_0}$
$=-\frac{R\omega }{4}\hat{i}+\frac{R\omega \sqrt{3}}{4}\hat{j}+3R\omega \hat{i}$
$=\frac{11}{4}R\omega \hat{i}+R\omega \frac{\sqrt{3}}{4}\hat{j}.$
Hence, the correct options are (a) and (b).