Question

A wheel free to rotate about its axis that is not frictionless is initially at rest. A constant external torque of +53 N·m is applied to the wheel for 21 s, giving the wheel an angular velocity of +624 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.

(b) Find the frictional torque, which is assumed to be constant.

Answer 1

Assuming :
I = Moment of inertia
alpha = angular acceleration
wf = Angular Velocity
Te = External torque = +53 N.m
Tf = Frictional Torque = constant for first case
Tn = Net torque
1 and 2 : they are subscript for before and after.

ATQ:
t1 = 21 s
t2 = 120 s
wf = 624 rev/min = 624* (2 pi radian / 1 revolution ) * (1 min/60s) = 65.345 rad/s

From Newton 2nd law for rotational motion :
Tn1 = Te - Tf
Tn1 = I * alpha1 ----------------------------------(1)

then From 1 eqn of rotational motion we have :
wf - 0 = alpha1 * t1 -------------------------------(2)

Also
Tn2 = -Tf = I alpha2 ----------------------------------(3)

and,From 1 eqn of rotational motion we have :
wf = alpha2* t2 ----------------------------(4)

From Above eqns we have :
alpha1 = wf / t1
alpha2 = wf / t2

Te = I (alpha1 - alpha2)
Te = I wf (1/t1 + 1/t2)
Te = (I*wf(t1 + t2)) / t1*t2
so
I = (Te*t1*t2) / (wf(t1+t2) ) ----------------------------(5)

Also,
Tf = - I alpha2 = I wf / t2
Tf = Te (t1) / (t1+t2) -------------------(6)

PART A: From 5,
I = Te *t1 * t2) / (wf(t1+ t2) )
I = 53*21*120/(64.345(21+120))
I = 14.72 kg.m^2

PART B: From eqn 6
Tf = Te (t1) / (t1+t2)
Tf = 53*21/(21+120)
Tf = 7.89 N.m or 7.9 N.m

('+' sign shows Counterclockwise direction)