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

Answer 1

Answer: (D)

From angular momentum conservation
$M{V}_{0}L=MV(L-l)$
$V=V_0\left(\frac{L}{L-l}\right)$
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
$V_0=\omega A=\sqrt{\frac{g}{L}}\times L{\theta }_0=\sqrt{gL}{\theta }_0$
Using this
$W=Mgl+M\left[gL\times {\left({\theta }_0\right)}^2\times \left(\frac{l}{L}\right)\right]$
$W=Mgl\left[1+{\left({\theta }_0\right)}^2\right]$