Question

A smooth ring of mass $m$ and radius $R=1m$ is pulled at $P$ with a constant acceleration $a=4m$ $s^{-2}$ on a horizontal surface such that the plane of the ring lies on the surface. Find the angular acceleration of the ring at the given position. (in $rad/s^2$)

Answer 1

As the ring purely rolls on the ground, thus the acceleration of point P is twice the acceleration of its centre of mass i.e. $a=2a_c$

$\therefore$ $2a_c=4$

$⟹$ $a_c=2m/s^2$

Radius of ring $R=1m$

Also, the acceleration of point A is zero i.e. $a_A=0$ (pure rolling)

$\therefore$ $a_c-R\alpha =0$

We get, $\alpha =\frac{a_c}{R}$

$⟹$ $\alpha =\frac{2}{1}=2$ $rad/s^2$