Question

A person of mass $M$ is sitting on a swing of length $L$ and swinging with and 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 $l \, \left(l < < L\right),$ is close to:

• A. $Mgl \, \left(\right.1-\theta _{0}^{2}\left.\right)$
• B. $Mgl\left(\right.1+\frac{\theta _{0}^{2}}{2}\left.\right)$
• C. $Mgl$
• D. $Mgl\left(\right.1+\theta _{0}^{2}\left.\right)$

Answer 1

Answer: (D)

From angular momentum conservation
$MV_{0}L=MV\left(\right.L-l\left.\right)$
$V=V_{0}\left(\frac{L}{L - l}\right)$
Now
$W=Mgl+\frac{1}{2}M\left[\left(V_{0}\right)^{2} \left(\frac{L}{L - l}\right)^{2} - \left(V_{0}\right)^{2}\right]$
Since $l < < L$
$W=Mgl+\frac{1}{2}M\left[\left(V_{0}\right)^{2} \left(1 - \frac{l}{L}\right)^{- 2} - \left(V_{0}\right)^{2}\right] $
$ =Mgl+\frac{1}{2}M\left[\left(V_{0}\right)^{2} \left(1 + 2 \frac{l}{L}\right) - \left(V_{0}\right)^{2}\right]$
$=Mgl+M\left[\left(V_{0}\right)^{2} \left(\frac{l}{L}\right)\right]$
From SHM
$V_{0}=\omega A=\sqrt{\frac{g}{L}}\times L\theta _{0}=\sqrt{g L}\theta _{0}$
Using this
$W=Mgl+M\left[g L \times \left(\theta _{0}\right)^{2} \times \left(\frac{l}{L}\right)\right] $
$ W=Mgl\left[1 + \left(\theta _{0}\right)^{2}\right]$