A 795 gram grinding wheel 21.0cm in diameter is in the shape of a uniform solid

Question

A 795 gram grinding wheel 21.0cm 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 230rpm about an axle perpendicular to its face through its center. When the power switch is turned off, you observe that the wheel stops in 42.5s

with constant angular acceleration due to friction at the axle.

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

t= ______________ N*m in the direction opposite to the motion

Explanation / Answer

Torque= I*alpha

I = 0.5*m*r^2 = 0.5*0.795*0.105^2 = 0.00438 kgm^2

alpha = W2-W1/t = (0-(230*2*3.24/60))/42.5 = 0.566 rad/s^2

T = I*alpha = 0.002479 Nm

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