A wheel free to rotate about its axis that is not frictionless is initially at r

Question

A wheel free to rotate about its axis that is not frictionless is initially at rest. A constant external torque of+45 N-m is applied to the wheel for 24 s, giving the wheel an angular velocity of+580 rev/min. The external torque is then removed, and the wheel comes to rest 120 s later. (Include the sign in your answers.) Find the moment of inertia of the wheel. Find the frictional torque, which is assumed to be constant. You can apply Newton's second law in rotational form to both the speeding up and slowing down motions of the wheel to obtain two equations in the moment of inertia of the wheel and the frictional torque that you can solve simultaneously for these quantities. Assume that both the speeding-up and slowing-down of the wheel took place under constant-acceleration conditions.

Explanation / Answer

Canonical to newton's second law for linear forces, the resultant torque =I = 45 Nm - f (where is the angular acceleration and I is the moments of inertia and f is the frictional torque). Hence the resultant angular acceleration =(45- f)/I

Using the angular displacement form of the kinematic equation: t=f - i --> = ((580*2)/60)/24 =2.5307s-2

(45- f)/I = 2.5307 ----- Equation (1)

Now consider the decceleration part:

using kinematic equation   2t = f - i --> 2= -i/t =

resultant decceleration torque= - f = I 2 --> 2=-f / I --> I =ft/i

I= 120/((580*2)/60)f = 1.9757f ---------Equation (2)

solve the 2 simultaneous equaition:

f =7.500 Nm

I = 14.818 kg m2

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