Q 11-06-142JEE MainJEE Main 2022 (27 Jul, Shift 2)Medium
A solid cylinder is suspended symmetrically through two massless strings wound on it, as shown in the figure. The distance from the initial rest position, the cylinder should be unbinding the strings to achieve a speed of $4\ \text{m s}^{-1}$, is ______ cm.
(take $g = 10\ \text{m s}^{-2}$)
Numerical value type. Enter your answer.
Answer: 120
For the falling cylinder ($I = \frac12MR^2$), with total string tension $T_{\text{tot}}$:
$$Mg - T_{\text{tot}} = Ma, \qquad T_{\text{tot}}R = \frac12MR^2\cdot\frac aR$$
So $T_{\text{tot}} = \dfrac{Ma}{2}$ and $a = \dfrac{2g}{3} = \dfrac{20}{3}\ \text{m s}^{-2}$.
$$s = \frac{v^2}{2a} = \frac{16}{2(20/3)} = 1.2\ \text{m} = 120\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics