# A disc is rolling (without slipping) on a horizontal surface: Kinematics

**Problem:**
A disc is rolling (without slipping) on a horizontal surface. P and Q are two points equidistant from the centre C. Let v_{P}, v_{Q} and v_{C} be the magnitude of velocities of points P, Q and C respectively, then,
(IIT JEE 2004)

- $v_Q>v_C>v_P$
- $v_Q< v_C < v_P$
- $v_Q=v_P$, $v_C=v_P /2$
- $v_Q< v_C > v_P$

**Solution:**
The point of contact A is instantaneously at rest in rolling without slipping i.e., $\vec{v}_A=\vec{0}$.

The magnitude of velocities of point C, Q, and P are
\begin{align}
&v_C=|\vec\omega\times\vec{r}_\text{AC}|=\omega r_\text{AC}, \nonumber \\
&v_Q=|\vec\omega\times\vec{r}_\text{AQ}|=\omega r_\text{AQ}, \nonumber \\
& v_P=|\vec\omega\times\vec{r}_\text{AP}|=\omega r_\text{AP}. \nonumber
\end{align}
Since $r_\text{AP} < r_\text{AC} < r_\text{AQ}$, we get $v_Q>v_C>v_P$. You are encouraged to draw velocity vectors to see the results.

## More Problems on Kinematics of Rigid Bodies

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