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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 :
Multiple Choice Options
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Step-by-Step Verified Solution
By conservation of energy :
Loss in energy = Gain in enrgy
m1gh = m2gh + m1V2 + m2V2 +
(m1 – m2)gh = (m1 + m2)V2 +
(m1 – m2)gh = (m1 + m2) +
(m1 – m2)gh = (m1 + m2) +
(m1 – m2)gh =
=
Loss in energy = Gain in enrgy
m1gh = m2gh + m1V2 + m2V2 +
(m1 – m2)gh = (m1 + m2)V2 +
(m1 – m2)gh = (m1 + m2) +
(m1 – m2)gh = (m1 + m2) +
(m1 – m2)gh =
=