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

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

A disc is rolling without

  1. $v_Q>v_C>v_P$
  2. $v_Q< v_C < v_P$
  3. $v_Q=v_P$, $v_C=v_P /2$
  4. $v_Q< v_C > v_P$

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

A disc is rolling without slipping
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.

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