JEE MAIN - Physics (2019 - 12th April Morning Slot - No. 13)
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;
mg$$\ell $$(1 + $$\theta $$02)
mg$$\ell $$
mg$$\ell $$(1 + $${{\theta _0^2} \over 2}$$)
mg$$\ell $$(1 - $$\theta $$02)
Explanation
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, $${w_p} = mg\ell + m{\left( {{\theta _0}\sqrt {gL} } \right)^2}{\ell \over L}$$
= $$mg\ell \left( {1 + \theta _0^2} \right)$$
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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