Q 11-06-226JEE MainJEE Main 2019 (11 Jan, Shift 2)Medium
A string is wound around a hollow cylinder of mass $5$ kg and radius $0.5$ m. If the string is now pulled with a horizontal force of $40$ N, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string)
Answer: (B) $16\ \text{rad/s}^2$
Let friction $f$ act forwards at the contact point. With $I = mR^2$ and $a = R\alpha$:
$$F + f = ma,\qquad (F - f)R = mR^2\alpha = mRa \Rightarrow F - f = ma$$
Adding: $2F = 2ma$, so $a = \dfrac Fm = \dfrac{40}{5} = 8\ \text{m/s}^2$ (and $f = 0$).
$$\alpha = \frac aR = \frac{8}{0.5} = 16\ \text{rad/s}^2$$
Solution by Sreeraj P, M.Sc Physics