A 6 kg grinding wheel of radius 0.2 m rotates at a constant rotational frequency of 4 rad/s when an object makes contact with the outer edge of the wheel. Friction causes the wheel to stop in 2 seconds. What is the average torque on the wheel by the frictional force?

Answer 1

Torque = (Moment of Inertia)*(Angular deceleration)

The moment of inertia is
I = (1/2)*M*R^2 = 3*(0.2)^2
= 0.12 kg*m^2

Angular deceleration = (4 rad/s)/2s = 2 rad/s^2

Torque = 0.12 * 2 = 0.24 Newton-meter

Answer 2

Is actualy A 6 kg grinding wheel of radius 0.2 m rotates at a constant rotational frequency of 4(pi) rad/s when an object makes contact with the outer edge of the wheel. Friction causes the wheel to stop in 2 seconds. What is the average torque on the wheel by the frictional force?
ANS = 0.754

Answer 3

.754