# A pulley of radius 2 m is rotated about its axis by a force F = (20t – 5t^2) N (where, t is measured in seconds) applied tangentially…

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, $\large n = \frac{\theta}{2\pi} = \frac{36}{2\pi} < 6$