# The figure shows a system consisting of a ring of outer radius 3R

Problem: 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 degree with the horizontal. Then with respect to the horizontal surface,  (IIT JEE 2012)

1. the point O has a linear velocity $3R\omega\hat{\imath}$.
2. the point P has a linear velocity $\frac{11}{4}R\omega\hat{\imath}+\frac{\sqrt{3}}{4}R\omega\hat{k}$.
3. the point P has a linear velocity $\frac{13}{4}R\omega\hat{\imath}-\frac{\sqrt{3}}{4}R\omega\hat{k}.$
4. the point P has a linear velocity $\left(3-\frac{\sqrt{3}}{4}\right)R\omega\hat{\imath}+\frac{1}{4}R\omega\hat{k}$.

Solution: Since outer ring is rolling without slipping, the point on the ring in contact with the ground (say point C) is at rest i.e., $\vec{v}_\text{C}=\vec{0}$. Thus, velocity of the point O is given by \begin{align} \vec{v}_\text{O}&=\vec{v}_\text{C}+\vec{\omega}_\text{o}\times\vec{r}_\text{CO}\\ &=\vec{0}+\omega\,\hat{\jmath}\times 3R\,\hat{k}\\ &=3R\omega\,\hat{\imath}.\nonumber \end{align} The point P lies on the inner disc having an angular velocity \begin{align} \vec{\omega}_\text{i}=-{\omega}/{2}\;\hat\jmath. \end{align} The position vector from O to P is \begin{align} \vec{r}_\text{OP}&=R\cos30,\hat\imath+R\sin30,\hat{k}\\ &={\sqrt{3}R}/{2}\;\hat\imath+{R}/{2}\;\hat{k}.\nonumber \end{align} Thus, velocity of the point P is given by \begin{align} \vec{v}_\text{P}&=\vec{v}_\text{O}+\vec{\omega}_\text{i}\times \vec{r}_\text{OP} \nonumber\\ &=3R\omega\,\hat\imath+\left(-\frac{\omega}{2}\,\hat\jmath\right)\times\left(\frac{\sqrt{3}R}{2}\,\hat\imath+\frac{R}{2}\,\hat{k}\right)\nonumber\\ &=\frac{11\omega R}{4}\,\hat\imath+\frac{\sqrt{3}\omega R}{4}\,\hat{k}.\nonumber \end{align}

Thus, options (A) and (B) are correct.