Q 11-06-204JEE MainJEE Main 2019 (10 Jan, Shift 1)Medium
A homogeneous solid cylindrical roller of radius $R$ and mass $M$ is pulled on a cricket pitch by a horizontal force. Assuming rolling without slipping, angular acceleration of the cylinder is:
Answer: (C) $\dfrac{2F}{3MR}$
The force $F$ acts at the axle. With friction $f$ backwards at the ground:
$$F - f = Ma,\qquad fR = \frac{MR^2}{2}\alpha,\qquad a = R\alpha$$
So $f = \dfrac{MR\alpha}{2}$ and $F = MR\alpha + \dfrac{MR\alpha}{2} = \dfrac32MR\alpha$:
$$\alpha = \frac{2F}{3MR}$$
Solution by Sreeraj P, M.Sc Physics