JEE MAIN - Physics (2020 - 9th January Evening Slot - No. 11)
A uniformly thick wheel with moment of inertia
I and radius R is free to rotate about its centre
of mass (see fig). A massless string is wrapped
over its rim and two blocks of masses m1 and
m2 (m1 $$ > $$ m2) are attached to the ends of the
string. The system is released from rest. The
angular speed of the wheel when m1 descents
by a distance h is :
_9th_January_Evening_Slot_en_11_1.png)
$${\left[ {{{2\left( {{m_1} + {m_2}} \right)gh} \over {\left( {{m_1} + {m_2}} \right){R^2} + I}}} \right]^{{1 \over 2}}}$$
$${\left[ {{{{m_1} + {m_2}} \over {\left( {{m_1} + {m_2}} \right){R^2} + I}}} \right]^{{1 \over 2}}}gh$$
$${\left[ {{{\left( {{m_1} - {m_2}} \right)} \over {\left( {{m_1} + {m_2}} \right){R^2} + I}}} \right]^{{1 \over 2}}}gh$$
$${\left[ {{{2\left( {{m_1} - {m_2}} \right)gh} \over {\left( {{m_1} + {m_2}} \right){R^2} + I}}} \right]^{{1 \over 2}}}$$
Explanation
By conservation of energy :
Loss in energy = Gain in enrgy
m1gh = m2gh + $${1 \over 2}$$ m1V2 + $${1 \over 2}$$ m2V2 + $${1 \over 2}I{\omega ^2}$$
$$ \Rightarrow $$ (m1 – m2)gh = $${1 \over 2}$$ (m1 + m2)V2 + $${1 \over 2}I{\omega ^2}$$
$$ \Rightarrow $$ (m1 – m2)gh = $${1 \over 2}$$ (m1 + m2)$${\left( {\omega R} \right)^2}$$ + $${1 \over 2}I{\omega ^2}$$
$$ \Rightarrow $$ (m1 – m2)gh = $${1 \over 2}$$ (m1 + m2)$${\left( {\omega R} \right)^2}$$ + $${1 \over 2}I{\omega ^2}$$
$$ \Rightarrow $$ (m1 – m2)gh = $${{{\omega ^2}} \over 2}\left[ {\left( {{m_1} + {m_2}} \right){R^2} + I} \right]$$
$$ \Rightarrow $$ $$\omega $$ = $${\left[ {{{2\left( {{m_1} - {m_2}} \right)gh} \over {\left( {{m_1} + {m_2}} \right){R^2} + I}}} \right]^{{1 \over 2}}}$$
Loss in energy = Gain in enrgy
m1gh = m2gh + $${1 \over 2}$$ m1V2 + $${1 \over 2}$$ m2V2 + $${1 \over 2}I{\omega ^2}$$
$$ \Rightarrow $$ (m1 – m2)gh = $${1 \over 2}$$ (m1 + m2)V2 + $${1 \over 2}I{\omega ^2}$$
$$ \Rightarrow $$ (m1 – m2)gh = $${1 \over 2}$$ (m1 + m2)$${\left( {\omega R} \right)^2}$$ + $${1 \over 2}I{\omega ^2}$$
$$ \Rightarrow $$ (m1 – m2)gh = $${1 \over 2}$$ (m1 + m2)$${\left( {\omega R} \right)^2}$$ + $${1 \over 2}I{\omega ^2}$$
$$ \Rightarrow $$ (m1 – m2)gh = $${{{\omega ^2}} \over 2}\left[ {\left( {{m_1} + {m_2}} \right){R^2} + I} \right]$$
$$ \Rightarrow $$ $$\omega $$ = $${\left[ {{{2\left( {{m_1} - {m_2}} \right)gh} \over {\left( {{m_1} + {m_2}} \right){R^2} + I}}} \right]^{{1 \over 2}}}$$
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