Q: A pulley of radius 2 m is rotated about its axis by a force F = (20t – 5t2) N (where, t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , then the number of rotations made by the pulley before its direction of motion is reversed, is
Sol: Given force, F = 20t – 5t2
$\large \alpha = \frac{F R}{I} $
$\large \alpha = \frac{(20 t – 5t^2)2}{10} $
= 4 t – t2
$\large \frac{d\omega}{dt} = 4 t – t^2 $
$\large \int_{0}^{\omega} d\omega = \int_{0}^{t} (4 t – t^2 ) dt $
$\large \omega = 2 t^2 – \frac{t^3}{3}$
When direction is reversed
ω = 0, i.e., t = 0 to 6s Now,
$\large d\theta = \omega dt $
$\large \int_{0}^{\theta} d\theta = \int_{0}^{6}(2 t^2 – \frac{t^3}{3}) dt $
$\large \theta = [\frac{2t^3}{3} – \frac{t^4}{12}]_{0}^{6} $
θ = 144 – 108 = 36 rad
Number of rotations, $\large n = \frac{\theta}{2\pi} = \frac{36}{2\pi} < 6 $