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.
$\not M V_0 L = \not 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}mV_{0}^{2} \left(\left(\frac{L}{L-\ell}\right)^{2} -1\right) $
$ =mg\ell + \frac{1}{2} mV_{0}^{2}\left(\left( 1- \frac{\ell}{L}\right)^{-2} - 1 \right)$
Now, $ \ell << L $
By, Binomial approximation
$ = mg \ell +\frac{1}{2}mV_{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 + mv_{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) $