Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

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)

best_answer

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)

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

Posted by

solutionqc

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024 session 2 result date announced; Steps to check NTA session 2 scores

anam-khan

JEE Main 2024 session 2 final answer key soon at jeemain.nta.ac.in; result on April 25

anam-khan

JEE Main 2024 Session 2: Last day to challenge answer key today

Latest Question