Question

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 $\ell (\ell <<L)$, is close to :

• A. $Mg\ell$
• B. $Mg\ell (1+{\theta }_0^2)$
• C. $Mg\ell (1-{\theta }_0^2)$
• D. $Mg\ell (1+\frac{{\theta }_0^2}{2})$

Answer 1

Answer: (B)

Angular momentum conservation.
$/M{V}_{0}L=/M{V}_1(L-\ell )$
$V_1=V_0\left(\frac{L}{L-\ell }\right)$
$w_g+w_p=\Delta KE$
$-mg\ell +w_p=\frac{1}{2}m\left(V_1^2-V_0^2\right)$
$w_p=mg\ell +\frac{1}{2}m{V}_0^2\left({\left(\frac{L}{L-\ell }\right)}^2-1\right)$
$=mg\ell +\frac{1}{2}m{V}_0^2\left({\left(1-\frac{\ell }{L}\right)}^{-2}-1\right)$
Now, $\ell <<L$
By, Binomial approximation
$=mg\ell +\frac{1}{2}m{V}_0^2\left(\left(1+\frac{2\ell }{L}\right)-1\right)$
$=mg\ell +\frac{1}{2}m{V}_0^2\left(\frac{2\ell }{L}\right)$
$W_p=mg\ell +m{v}_0^2\frac{\ell }{L}$
here, $V_0$ = maximum velocity
$=\omega \times A\left(\sqrt{\frac{g}{L}}\right)\left({\theta }_{0}L\right)$
$V_0={\theta }_0\sqrt{gL}$
$W_p=mg\ell +m{\left({\theta }_0\sqrt{gL}\right)}^2\frac{\ell }{L}$
$=mg\ell \left(1+{\theta }_0^2\right)$