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:

  • Option 1)

    Mgl(1-\theta_{0}^{2})

  • Option 2)

    Mgl(1+\frac{\theta_{0}^2}{2})

  • Option 3)

    Mgl

  • Option 4)

    Mgl(1+\theta_{0}^{2})

 

Answers (1)

 

Law of conservation of angular moment -

\vec{\tau }= \frac{\vec{dL}}

- wherein

If net torque is zero

i.e. \frac{\vec{dL}}= 0

\vec{L}= constant

angular momentum is conserved only when external torque is zero .

 

 

Apply Conservation of angular momentum

MV_{0}L=MV_{1}(L-l)\Rightarrow V_{1}=V_{0}\left ( \frac{L}{L-l} \right )

        w_{g}+w_{p}=\Delta K.E

\Rightarrow w_{p}-mgl=\frac{1}{2}m\left ( v_{1}^{2}-v_{0}^{2} \right )

\Rightarrow w_{p}=mgl+\frac{1}{2}mv_{0}^{2}\left[\left ( \frac{L}{L-l} \right )^{2} -1\right ]

\Rightarrow w_{p}=mgl+\frac{1}{2}mv_{0}^{2}\left[\left ( 1-\frac{l}{L} \right )^{-2} -1\right ]

It is given that L>>>>l

so by binamial theorem,

w_{p}=mgl+\frac{1}{2}mv_{0}^{2}\left[\left ( 1+\frac{2l}{L} \right ) -1\right ]

       =mgl+\frac{1}{2}mv_{0}^{2}\left( \frac{2l}{L} \right )=mgl+mv_{0}^{2}\frac{l}{L}...........(1)

v_{0}=w\times A \rightarrow maximum velocity

v_{0}=\sqrt{\frac{g}{L}}\cdot (\theta_{0}L)=\theta_{0}\sqrt{gl}.................putting in eq (1)

\Rightarrow w_{p}=mgl+m \theta_{0}^{2}gl\times\frac{l}{L}

w_{p}=mgl\left ( 1+\theta_{0}^{2} \right )

 

 


Option 1)

Mgl(1-\theta_{0}^{2})

Option 2)

Mgl(1+\frac{\theta_{0}^2}{2})

Option 3)

Mgl

Option 4)

Mgl(1+\theta_{0}^{2})

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