A wheel free to rotate about its axis that is not frictionless is initially at rest...?

Question

A constant external torque of +52 Nm is applied to the wheel for 22 s, giving the wheel an angular velocity of +530 rev/min. The external torque is then removed, and the wheel comes to rest 120 s later. (Include the sign in your answers.)

(a) Find the moment of inertia of the wheel.
kg x m^2
(b) Find the frictional torque, which is assumed to be constant.
N x m

Answer 1

Ok, you need to remember this :

Torque = F.d = I.a

I = moment of inertia

a = angular acceleration

F = force

d = distance.

The first torque made the wheel rotate from the rest until the speed gained an angular velocity of 530 rev / min

The torque is 52Nm

52 Nm = I.a

I = moment of inertia of the wheel

a = angular aceleration

let's find a, with : Wf = 530 rev / min, time = 22s, initial velocity = 0 m/s

530*2pi / 60 = a*22

a = 2.52 rad/s^2

then : 52 = I*2.52 >>> I = 20.63 kg*m^2

b), Let's find the frictional torque with :

Torque = 20.63.a'

a' = final angular acceleration ( when the wheel stops)

initial speed before the braking : 530*2pi / 60 rad / s

final velocity = 0 rad / s, time = 120 s

0 = 530*2pi/60 - a'120

a' = 0.46 rad / s^2

Torque = 20.63*0.46 = 9.54 Nm

Hope that helped

Answer 2

Why don't you consider frictional torque in step a)?