A 800 gram grinding wheel 20.0cm in diameter is in the shape of a uniform solid

Q: A 800 gram grinding wheel 20.0cm in diameter is in the shape of a uniform solid

T = I * alpha where I = m * R^2 / 2 = 0.800 * 0.2^2 / 2 = 0.016 alpha = 240 * 2pi/60 /50 = 0.5026 rad/s^2 NOw, T = 0. 016 * 0.5026 = 8.0425 * 10^-3 N -m

ID: 2302104 • Letter: A

Question

A 800 gram grinding wheel 20.0cm 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 240rpm about an axle perpendicular to its face through its center. When the power switch is turned off, you observe that the wheel stops in 50.0s with constant angular acceleration due to friction at the axle.

Part A

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

?= _____ N?m, in the direction opposite to the motion

Explanation / Answer

T = I * alpha

where I = m * R^2 / 2

= 0.800 * 0.2^2 / 2

= 0.016

alpha = 240 * 2pi/60 /50

= 0.5026 rad/s^2

NOw, T = 0. 016 * 0.5026

= 8.0425 * 10^-3 N -m <============= answer

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A 800 gram grinding wheel 20.0cm in diameter is in the shape of a uniform solid

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