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QJEEM20Ph5a22: JEE Main 2020 Physics - Class 11 JEEM20Ph5a22 Solution: A uniformly thick wheel with moment of inertia I and radius R…

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 : JEE Main 2020 (Online) 9th January Evening Slot Physics - Rotational Motion Question 135 English

Multiple Choice Options

  • A [2(m1+m2)gh(m1+m2)R2+I]12{\left[ {{{2\left( {{m_1} + {m_2}} \right)gh} \over {\left( {{m_1} + {m_2}} \right){R^2} + I}}} \right]^{{1 \over 2}}}
  • B [m1+m2(m1+m2)R2+I]12gh{\left[ {{{{m_1} + {m_2}} \over {\left( {{m_1} + {m_2}} \right){R^2} + I}}} \right]^{{1 \over 2}}}gh
  • C [(m1−m2)(m1+m2)R2+I]12gh{\left[ {{{\left( {{m_1} - {m_2}} \right)} \over {\left( {{m_1} + {m_2}} \right){R^2} + I}}} \right]^{{1 \over 2}}}gh
  • D [2(m1−m2)gh(m1+m2)R2+I]12{\left[ {{{2\left( {{m_1} - {m_2}} \right)gh} \over {\left( {{m_1} + {m_2}} \right){R^2} + I}}} \right]^{{1 \over 2}}}

Step-by-Step Verified Solution

By conservation of energy :

Loss in energy = Gain in enrgy

m1gh = m2gh + 12{1 \over 2} m1V2 + 12{1 \over 2} m2V2 + 12Iω2{1 \over 2}I{\omega ^2}

⇒\Rightarrow (m1 – m2)gh = 12{1 \over 2} (m1 + m2)V2 + 12Iω2{1 \over 2}I{\omega ^2}

⇒\Rightarrow (m1 – m2)gh = 12{1 \over 2} (m1 + m2) (ωR)2{\left( {\omega R} \right)^2} + 12Iω2{1 \over 2}I{\omega ^2}

⇒\Rightarrow (m1 – m2)gh = 12{1 \over 2} (m1 + m2) (ωR)2{\left( {\omega R} \right)^2} + 12Iω2{1 \over 2}I{\omega ^2}

⇒\Rightarrow (m1 – m2)gh = ω22[(m1+m2)R2+I]{{{\omega ^2}} \over 2}\left[ {\left( {{m_1} + {m_2}} \right){R^2} + I} \right]

⇒\Rightarrow ω\omega = [2(m1−m2)gh(m1+m2)R2+I]12{\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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