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})

Most Viewed Questions

Preparation Products

Knockout JEE Main April 2021 (One Month)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 14000/- ₹ 4999/-
Buy Now
Knockout JEE Main May 2021

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 22999/- ₹ 9999/-
Buy Now
Test Series JEE Main May 2021

Unlimited Chapter Wise Tests, Unlimited Subject Wise Tests, Unlimited Full Mock Tests, Get Personalized Performance Analysis Report,.

₹ 6999/- ₹ 2999/-
Buy Now
Knockout JEE Main May 2022

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Live Classes, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 34999/- ₹ 24999/-
Buy Now
JEE Main Rank Booster 2021

Booster and Kadha Video Lectures, Unlimited Full Mock Test, Adaptive Time Table, 24x7 Doubt Chat Support,.

₹ 13999/- ₹ 6999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions