Q 11-06-113JEE MainJEE Main 2023 (13 Apr, Shift 2)Easy
A light rope is wound around a hollow cylinder of mass $5\ \text{kg}$ and radius $70\ \text{cm}$. The rope is pulled with a force of $52.5\ \text{N}$. The angular acceleration of the cylinder will be ______ $\text{rad s}^{-2}$.
Numerical value type. Enter your answer.
Answer: 15
$I=MR^2=5\times0.49=2.45\ \text{kg m}^2$, $\tau=FR=52.5\times0.7=36.75\ \text{N m}$.
$\alpha=\dfrac\tau I=\dfrac{36.75}{2.45}=15\ \text{rad s}^{-2}$.
Solution by Sreeraj P, M.Sc Physics