QJEEM19Phec8f: JEE Main 2019 Physics - Class 11 JEEM19Phec8f Solution: A person of mass M is, sitting on a swing of length L and…

Question

Verified Solution • 2025–26 Session moderate

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 center of mass moves by a distance

\ell

(

\ell

<< L), is close to;

Accepted Answer

Angular momentum conservation
MV₀L = MV₁(L –

\ell

)

{V_1} = {V_0}\left( {{L \over {L - \ell }}} \right)

{w_g} + {w_p} = \Delta KE

- mg\ell + {w_p} = {1 \over 2}m\left( {V_1^2 - V_0^2} \right)

{w_p} = mg\ell + {1 \over 2}mV_0^2\left( {{{\left( {{L \over {L - \ell }}} \right)}^2} - 1} \right)

= mg\ell + {1 \over 2}mV_0^2\left( {{{\left( {1 - {L \over {L - \ell }}} \right)}^{ - 2}} - 1} \right)

Now

\ell \ll L

By, Binomial approximation

= mg\ell + {1 \over 2}mV_0^2\left( {{{\left( {1 + {L \over {L - \ell }}} \right)}^{ - 2}} - 1} \right)

= mg\ell + {1 \over 2}mV_0^2\left( {{{2\ell } \over L}} \right)

{w_p} = mg\ell + mV_0^2{\ell \over L}

Here, V₀ = maximum velocity =

\omega \times A = \left( {\sqrt {{g \over L}} } \right)\left( {{\theta _0}L} \right)

So,

{w_p} = mg\ell + m{\left( {{\theta _0}\sqrt {gL} } \right)^2}{\ell \over L}

=

mg\ell \left( {1 + \theta _0^2} \right)

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