Question

A solid disk of radius r rolls without slipping on a horizontal floor with angular velocity $\omega$ and angular acceleration $\alpha$. The magnitude of the acceleration of the point of contact on the disc is

• A. Zero
• B. $r\alpha$
• C. $\sqrt{(r\alpha {)}^2+(r{\omega }^2{)}^2}$
• D. $r{\omega }^2$

Answer 1

The correct option is D $r{\omega }^2$


Motion of the body can be expressed as the sum of pure rotational and pure translation.
For no slip condition.
Translational acceleration of centre of mass $\left(a\right)=\alpha r$
Velocity of centre of mass $\left(v\right)=\omega r$

Now superimposing the two motions.

Linear acceleration of bottom point
$=\alpha r-\alpha r+r{\omega }^2$
Velocity of bottom point $=\omega r-\omega r=0$
$\therefore$ Net acceleration of bottom point $=r{\omega }^2$