Question 248: Centre of Mass and Rotational Motion

Question

A person of mass M is, sitting on a swing of length L and swinging with an angular amplitude $	heta $
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
MV
0
L = MV
1
(L Ã¢ÂÂ $\ell $ )
${V_1} = {V_0}\left( {{L \over {L - \ell }}} ight)$
${w_g} + {w_p} = \Delta KE$
$ - mg\ell  + {w_p} = {1 \over 2}m\left( {V_1^2 - V_0^2} ight)$
${w_p} = mg\ell  + {1 \over 2}mV_0^2\left( {{{\left( {{L \over {L - \ell }}} ight)}^2} - 1} ight)$
$ = mg\ell  + {1 \over 2}mV_0^2\left( {{{\left( {1 - {L \over {L - \ell }}} ight)}^{ - 2}} - 1} ight)$
Now $\ell  \ll L$
By, Binomial approximation
$ = mg\ell  + {1 \over 2}mV_0^2\left( {{{\left( {1 + {L \over {L - \ell }}} ight)}^{ - 2}} - 1} ight)$
$ = mg\ell  + {1 \over 2}mV_0^2\left( {{{2\ell } \over L}} ight)$
${w_p} = mg\ell  + mV_0^2{\ell  \over L}$
Here, V
0
= maximum velocity = $\omega  	imes A = \left( {\sqrt {{g \over L}} } ight)\left( {{	heta _0}L} ight)$
So, ${w_p} = mg\ell  + m{\left( {{	heta _0}\sqrt {gL} } ight)^2}{\ell  \over L}$
= $mg\ell \left( {1 + 	heta _0^2} ight)$

Answer

mg$\ell $(1 + $	heta $
0
2
)

Answer

mg$\ell $

Answer

mg$\ell $(1 + ${{	heta _0^2} \over 2}$)

Answer

mg$\ell $(1 - $	heta $
0
2
)

Centre of Mass and Rotational Motion

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;

Choose the correct answer: