Question 252: 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 + $$	heta $$02)

Answer

mg$$\ell $$

Answer

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

Answer

mg$$\ell $$(1 - $$	heta $$02)

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: