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 $\frac{\omega }{2}$. The ring and disc are separated by frictionless ball bearings. The system is in the x - z place. The point P on the inner disc is at distance R from the origin O, where OP makes an angled 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 $(3-\frac{\sqrt{3}}{4})R\omega \hat{i}+\frac{1}{4}R\omega \hat{k}$

Answer 1

Answer: (A), (B)

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

If the ring of radius '3R' does pure rolling, its centre moves with velocity,

${\overrightarrow{V}}_0=3\omega R\hat{i}$ i.e., towards right.

Now as the disc and bearings are contained in the ring, the whole system moves as one ring with the above velocity.

${\overrightarrow{V}}_{\frac{P}{G}}={\overrightarrow{V}}_{\frac{P}{O}}+{\overrightarrow{V}}_{\frac{O}{G}}$

where, G $\to$ ground

O $\to$ centre of the disc

Clearly, $|{\overrightarrow{V}}_{\frac{P}{O}}|=\frac{\omega R}{2}$ as shown,

${\overrightarrow{V}}_{\frac{P}{G}}=\frac{\omega R}{2}cos{30}^{\circ }\hat{k}+3\omega R\hat{i}-\frac{\omega R}{2}sin{30}^{\circ }\hat{i}$

${\overrightarrow{V}}_{\frac{P}{G}}=\frac{11}{4}R\omega \hat{i}+\frac{\sqrt{3}}{4}R\omega \hat{k}$

Now the velocity of 'P' with respect to ground can be formulated as.