Question

A disc is rolling uniformly on ground surface as shown in the figure. Velocity of centre is $V_0$ and angular velocity is $\omega$. Then acceleration of point P on the disc is:

• A. zero
• B. r\omega^2 vertically upward
• C. r\omega^2 vertically downward
• D. None

Answer 1

Answer: (B)

r\omega^2 vertically upward

Answer 2

The disc is rolling uniformly, which means the acceleration of the center of the disc ($\vec{a}_C$) is zero and the angular acceleration ($\vec{\alpha}$) is also zero. The acceleration of a point P on the disc is given by $\vec{a}_P = \vec{a}_C + \vec{a}_{P,rot}$. Since $\vec{a}_C = 0$, we have $\vec{a}_P = \vec{a}_{P,rot}$. The rotational acceleration is given by $\vec{a}_{P,rot} = \vec{\alpha} \times \vec{r}_{P/C} + \vec{\omega} \times (\vec{\omega} \times \vec{r}_{P/C})$. As $\vec{\alpha} = 0$, this simplifies to $\vec{a}_P = \vec{\omega} \times (\vec{\omega} \times \vec{r}_{P/C})$. This is the centripetal acceleration. For point P at the bottom of the disc, the position vector relative to the center is $\vec{r}_{P/C}$ pointing downwards. The term $\vec{\omega} \times (\vec{\omega} \times \vec{r}_{P/C})$ results in an acceleration directed towards the center of rotation. In this case, with $\vec{\omega}$ directed into the page (for clockwise rotation) and $\vec{r}_{P/C}$ pointing downwards, the resultant acceleration is $r\omega^2$ vertically upward.