Question

A uniform disc mass $m$ and radius $r$ rotates along an axis passing through its centre of mass and normal to its plane. An unstretchable rope is found on the disc. Tangential acceleration of a point $P$ on the periphery of the disc when a uniform force $F$ is applied on the rope is

Answer 1

Torque,

$\tau =Fr=I\alpha$

$\Rightarrow \alpha =\frac{Fr}{I}=\frac{Fr}{\frac{1}{2}m{r}^2}=\frac{2F}{mr}$

Tangential acceleration at point P. $a=\alpha r=\frac{2F}{mr}r=\frac{2F}{m}$

Hence, tangential acceleration is $\frac{2F}{m}$ .