A wheel free to rotate about its axis that is not frictionless is initially at rest. A constant external torque of + 46 N·m is applied to the wheel for 17 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.) (a) Find the moment of inertia of the wheel. (b) Find the frictional torque, which is assumed to be constant.

Question

Marcos Townsend

Physics

28 October, 15:52

A wheel free to rotate about its axis that is not frictionless is initially at rest. A constant external torque of + 46 N·m is applied to the wheel for 17 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.) (a) Find the moment of inertia of the wheel. (b) Find the frictional torque, which is assumed to be constant.

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Agustin Lutz · Answer 1

Let Torque due to friction be

F

Net torque

= 46 - F

Angular impulse = change in angular momentum

= (46 - F) x 17 = I X 580

When external torque is removed, only friction creates torque reducing its speed to zero in 120 s so

Angular impulse = change in angular momentum

F x 120 = I X 580

(46 - F) x 17 = F x 120

137 F = 46 x 17

F = 5.7 Nm

b)

Putting this value in first equation

5.7 x 120 = I x 580

I = 1.18 kg m²