Question

A pulley, with a rotational inertia of 1.7 ✕ 10−3 kg · m2 about its axle and a radius of 11 cm, is acted on by a force applied tangentially at its rim. The force magnitude varies in time as F = 0.50t + 0.30t2, where F is in newtons and t in seconds. The pulley is initially at rest.

At t = 4.0 s what is its angular velocity (in rad/s)?

Sample submission: 456

Answer 1

Inertia of pulley = 1.7 x 10^-3 kg.m²

Radius of the pulley = r = 11 cm = 0.11 m

Force acting on the pulley = F = 0.5t + 0.3t²

Initial angular velocity = ₁ = 0 rad/s

Angular velocity after 4 sec = ₂

Time period = T = 4 sec

Torque =

Angular acceleration =

= Fr

= I

Fr = I

(0.5t + 0.3t²)(0.11) = (1.7x10^-3)

= 32.353t + 19.411t²

d = (32.353t + 19.411t²)dt

Integrating we get,

= 64.706t² + 58.233t³ + C

At t=0sec, =₁=0 rad/s

Therefore C=0

= 64.706t² + 58.233t³

At t=4sec =₂

₂ = 64.706(4)² + 58.233(4)³

₂ = 4762.21 rad/s

Angular velocity at t=4sec = 4762.21 rad/s