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

The correct option is B $Mg\ell (1+{\theta }_0^2)$

Angular momentum conservation.
$M{V}_{0}L=M{V}_1(L-\ell )$
$V_1=V_0\left(\frac{L}{L-\ell }\right)$

$\frac{1}{2}M{V}_1^2-\frac{1}{2}M{V}_o^2=W_p+Wg$

$\frac{1}{2}M(\frac{V_o^2{L}^2}{{(L-l)}^2}-V_o^2)=W_p-Mgl$

$\frac{1}{2}M{V}_o^2(\frac{1}{{(1-\frac{l}{L})}^2}-1)=W_p-Mgl$

$W_p=\frac{1}{2}M{V}_o^2[{(1-\frac{l}{L})}^{{}_{-2}}-1]+Mgl$

$=\frac{1}{2}M{V}_o^2[(1+\frac{2l}{L})-1]+Mgl$

$=\frac{\frac{1}{2}M{V}_o^{2}2l}{L}+Mgl$

$=\frac{1}{2}M{V}_o^2\left(gL\right)\left(\frac{2l}{L}\right)+Mgl=Mgl(1+{\theta }_o^2)$

Put $V_o$ in above eq. to get the final result

$\because V_o=Aw$

$={\theta }_o\sqrt{\frac{g}{L}}L$

$={\theta }_o\sqrt{gL}$