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

By

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)

A sphere is rolling without

  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.

More Problems on Kinematics of Rigid Bodies

See Our Book

  1. 300 Solved Problems on Rotational Mechanics by Jitender Singh and Shraddhesh Chaturvedi
  2. IIT JEE Physics by Jitender Singh and Shraddhesh Chaturvedi