JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Elasticity & SHM

The moment the person stands up, the only quick change is in the distance of the centre of mass from the pivot, and the push is along the string, so angular momentum about the pivot is conserved. Writing $L_i=M{V}_{0}L$ before standing and $L_f=M{V}_1(L-\ell )$ after, we get $V_1=V_0\frac{L}{L-\ell }$. Now apply the work-energy theorem, $W_g+W_p=\Delta KE$, where gravity does $W_g=-Mg\ell$ as the centre of mass rises. Hence the person's work is $W_p=Mg\ell +\frac{1}{2}M(V_1^2-V_0^2)$. Using the binomial approximation $(1-\ell /L{)}^{-2}\approx 1+2\ell /L$ for $\ell \ll L$, the kinetic part becomes $\frac{1}{2}M{V}_0^2\cdot \frac{2\ell }{L}=M{V}_0^2\frac{\ell }{L}$. At the lowest point of a swing of angular amplitude ${\theta }_0$ the speed is $V_0={\theta }_0\sqrt{gL}$, so $M{V}_0^2\frac{\ell }{L}=Mg\ell {\theta }_0^2$. Collecting terms, $W_p=Mg\ell (1+{\theta }_0^2)$.