Q 11-05-193JEE MainJEE Main 2019 (12 Apr, Shift 1)Hard
A person of mass $M$ is sitting on a swing of length $L$ and swinging with an angular amplitude $\theta_0$. If the person stands up when the swing passes through its lowest point, the work done by him, assuming that his centre of mass moves by a distance $l$ ($l \ll L$), is close to:
Answer: (D) $Mgl\left(1 + \theta_0^2\right)$
At the lowest point the speed is $v^2 = 2gL(1 - \cos\theta_0) \approx gL\theta_0^2$.
To raise his centre of mass by $l$ (towards the pivot), he must push against gravity and supply the centripetal force:
$$W \approx \left(Mg + \frac{Mv^2}{L}\right)l = Mgl\left(1 + \theta_0^2\right)$$
(Since $l \ll L$, the speed hardly changes over the short distance $l$.)
Solution by Sreeraj P, M.Sc Physics