Question

A 750 gram grinding wheel 28.0 cm in diameter is in the shape of a uniform solid disk. (We can ignore the small hole at the center.) When it is in use, it turns at a constant 205 rpm about an axle perpendicular to its face through its center. When the power switch is turned off, you observe that the wheel stops in 50.0 swith constant angular acceleration due to friction at the axle.

What torque does friction exert while this wheel is slowing down?

Answer 1

Initial angular velocity, wo = 205 x 2pi/60 = 21.47 rad/sec

Angular acceleration, a = (0 - 21.47)/50 = -0.429 rad/sec^2

Moment of inertia, I = 0.5 x 0.75 x 0.14^2 = 0.00735 kg m^2

Torque, T = I a = 0.00735 x 0.429 = 3.15 x 10^-3 Nm

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