Question

A uniform-density wheel of mass 8 kg and radius 0.42 m rotates on a low-friction axle. Starting from rest, a string wrapped around the edge exerts a constant force of 14 N for 0.84 s. (a) What is the final angular speed?

(a) ωf = radians/s

(b) What was the average angular speed?

ωaverage = radians/s

(c) Through how big an angle did the wheel turn?

θ = radians

(d) How much string came off the wheel?

d = m

Answer 1

a)The moment of inertia of the wheen will be:

I = 1/2 mr^2 = 0.5 x 8 x 0.42^2 = 0.7056 kg-m^2

Torque due to force is:

Tau = F x R = 14 x 0.42 = 5.88 N-m

Also, Tau = I alpha

I alpha = 5.88 => alpha = 5.88/0.7056 = 8.33 rad/s^2

omega = omega0 + alpha t = alpha t

omega = 8.33 x 0.84 = 7 rad/s

Hence, omega = 7 rad/s

b)distance covered is

theta = omega0 t + 1/2 alpha t^2

theta = 0.5 x 8.33 x 0.84 x 0.84 = 2.94 rad

omega(av) = 2.94/0.84 = 3.59 rad/s

Hence, omega(av) = 3.59 rad/s

c)calculated above

theta = 2.94 rad

d) d = 2.94 x 0.42 = 1.23 m

Hence, d = 1.23 m