JEE Main 2020 — Rotational Motion Question with Solution
From: JEE Main 2020 (Online) 9th January Evening Slot
Question
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 :
This question includes a diagram. The text above accompanies the figure.
Choose an option
Show full solutionCorrect option: D
Correct answer
D
Step-by-step explanation
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 =
=
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