- JEE Mains 2019
- Hard
- MCQ
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 \(l\, ( l < < L)\), is close to
Answer & Solution
(A) \(Mgl\,\left( {1 + \theta _0^2} \right)\)
Step-by-step Solution
Detailed explanation
\(\mathrm{MV}_{0} \mathrm{L}=\mathrm{MV}_{1}(\mathrm{L}-\ell)\)
\(V_{1}=V_{0}\left(\frac{L}{L-\ell}\right)\)
\(\mathrm{w}_{\mathrm{g}}+\mathrm{w}_{\mathrm{p}}=\Delta \mathrm{KE}\)
\(-m g \ell+w_{p}=\frac{1}{2} m\left(V_{1}^{2}-V_{0}^{2}\right)\)
\(w_{p}=m g \ell+\frac{1}{2} m V_{0}^{2}\left(\left(\frac{L}{L-\ell}\right)^{2}-1\right)\)
\(=m g \ell+\frac{1}{2} m V_{0}^{2}\left(\left(1-\frac{L}{L-\ell}\right)^{-2}-1\right)\)
Now, \(\ell<<\mathrm{L}\)
By, Binomial approximation
\(=m g \ell+\frac{1}{2} m V_{0}^{2}\left(\left(1+\frac{L}{L-\ell}\right)^{-2}-1\right)\)
\(=m g \ell+\frac{1}{2} m V_{0}^{2}\left(\frac{2 \ell}{L}\right)\)
\(\mathrm{W}_{\mathrm{P}}=\mathrm{mg} \ell+\mathrm{mV}_{0}^{2} \frac{\ell}{\mathrm{L}}\)
Here, \(\mathrm{V}_{0}=\) maximum velocity \(=\omega \times \mathrm{A}=(\sqrt{\frac{\mathrm{g}}{\mathrm{L}}})\left(\theta_{0} \mathrm{L}\right)\)
\(\mathrm{So}, \mathrm{w}_{\mathrm{p}}=\mathrm{mg} \ell+\mathrm{m}\left(\theta_{0} \sqrt{\mathrm{gL}}\right)^{2} \frac{\ell}{\mathrm{L}}\)
\(=m g \ell\left(1+\theta_{0}^{2}\right)\)
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