# A sphere is rolling without slipping on a fixed horizontal plane: Kinematics

Problem: A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, A is the point of contact, B is the centre of the sphere and C is the topmost point. Then,  (IIT JEE 2009) 1. $\vec{V}_C-\vec{V}_A=2\big(\vec{V}_{B}-\vec{V}_C\big)$
2. $\vec{V}_C-\vec{V}_B=\vec{V}_B-\vec{V}_A$
3. $|\vec{V}_C-\vec{V}_A|=2\,|\vec{V}_B-\vec{V}_C|$
4. $|\vec{V}_C-\vec{V}_A|=4\,|\vec{V}_B|$

Solution: Let $\vec{\omega}$ be the angular velocity of the sphere. Let ${\text{AB}}=\vec{r}$ and hence ${\text{AC}}=2\vec{r}$. The velocities $\vec{V}_B$ and $\vec{V}_C$ are related to $\vec{V}_A$ by \begin{align} \vec{V}_B&=\vec{V}_A+\vec{\omega}\times\vec{r}, \\ \vec{V}_C&=\vec{V}_A+\vec{\omega}\times(2\vec{r}) \\ &=\vec{V}_A+2\vec{\omega}\times\vec{r}. \end{align} Using above equations, we get \begin{align} \vec{V}_C-\vec{V}_A & =2\vec{\omega}\times\vec{r} \\ &=-2(\vec{V}_B-\vec{V}_C), \\ \vec{V}_C-\vec{V}_B &=\vec{\omega}\times\vec{r} \\ &=\vec{V}_B-\vec{V}_A. \nonumber \end{align} Also, since sphere is rolling without slipping, $\vec{V}_A=\vec{0}$.

Thus options (B) and (C) are correct.