JEE Main · 2011 · 10 Apr 2011 · Offline · Rotational Mechanics

A pulley of radius 2 m is rotated about its axis by a force

F = (20 t - 5 t^{2})

Newton (where

t

is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is

10 kg m^{2}

, the number of rotations made by the pulley before its direction of motion is reversed, is :

Answer: (A)

Solution

The tangential force gives a torque

τ = r F = 2 (20 t - 5 t^{2}) = 40 t - 10 t^{2}

. With

I = 10 kg m^{2}

, the angular acceleration is

α = τ / I = 4 t - t^{2}

. Integrating,

ω = 2 t^{2} - \frac{t ^{3}}{3}

, which returns to zero — the instant the rotation reverses — at

t = 6 s

. The angle turned up to that moment is

θ = \int_{0}^{6} ω d t = [\frac{2 t ^{3}}{3} - \frac{t ^{4}}{12}]_{0}^{6} = 144 - 108 = 36 rad

, so the number of revolutions is

\frac{36}{2 π} \approx 5.7

, i.e. more than

3

but less than

6

JEE Main · 2011 · 10 Apr 2011 · Offline · Single correct option