Rotational Motion

JEE Main level, single correct, medium

220 of 280

Question 220 of 280

A person of mass MM is sitting on a swing of length LL and swinging with and an angular amplitude θ0.{\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),l\left(l\lt \lt L\right), is close to:

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Answer: (D) Mgl(1+θ02)Mgl(1+{\theta }_{0}^{2})

From angular momentum conservation
MV0L=MV(L−l)V=V0(LL−l)\begin{matrix}M{V}_{0}L=MV(L-l) \\ V={V}_{0}\left(\dfrac{L}{L-l}\right)\end{matrix}
Now
W=Mgl+12M[V02(LL−l)2−V02]W=Mgl+\dfrac{1}{2}M\left[{{V}_{0}}^{2}{\left(\dfrac{L}{L-l}\right)}^{2}-{{V}_{0}}^{2}\right]
Since l≪Ll≪L
W=Mgl+12M[V02(1−lL)−2−V02]=Mgl+12M[V02(1+2lL)−V02]=Mgl+M[V02(lL)]\begin{matrix}W=Mgl+\dfrac{1}{2}M\left[{{V}_{0}}^{2}{\left(1-\dfrac{l}{L}\right)}^{-2}-{{V}_{0}}^{2}\right] \\ =Mgl+\dfrac{1}{2}M\left[{{V}_{0}}^{2}\left(1+2\dfrac{l}{L}\right)-{{V}_{0}}^{2}\right] \\ =Mgl+M\left[{{V}_{0}}^{2}\left(\dfrac{l}{L}\right)\right]\end{matrix}
From SHM
V0=ωA=gL×Lθ0=gLθ0{V}_{0}=\omega A=\sqrt{\dfrac{g}{L}}\times L{\theta }_{0}=\sqrt{gL}{\theta }_{0}
Using this
W=Mgl+M[gL×θ02×(lL)]W=Mgl[1+θ02]\begin{matrix}W=Mgl+M\left[gL\times {{\theta }_{0}}^{2}\times \left(\dfrac{l}{L}\right)\right] \\ W=Mgl\left[1+{{\theta }_{0}}^{2}\right]\end{matrix}

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