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JEE Main 2017 (8 Apr, Shift 1) · Rotational Kinematics & Dynamics

Rotational Kinematics & Dynamics question from JEE Main 2017 (8 Apr, Shift 1)

Q64JEE Main 2017 · 8 APR · SHIFT 1Medium

Rotational Kinematics & Dynamics

A uniform disc of radius R and mass M is free to rotate only about its axis. A string is wrapped over its rim and a body of mass m is tied to the free end of the string as shown in the figure. The body is released from rest. Then the acceleration of the body is:

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Answer: (a)

Block: $mg - T = ma$ . Disc: $TR = \frac{1}{2}MR^2\cdot\frac{a}{R} \Rightarrow T = \frac{Ma}{2}$ .
$mg = a\left(m + \frac{M}{2}\right)$
So the answer is (a) $a = \frac{2mg}{2m + M}$ .

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