A 800 gram grinding wheel 29.5cm in diameter is in the shape of a uniform solid

Question

A 800 gram grinding wheel 29.5cm 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 250rpm about an axle perpendicular to its face through its center. When the power switch is turned off, you observe that the wheel stops in 48.5s 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 of the motion.

* I used T=Ia

where: I= .5*.8*.295^2=. 035

and: a= 250(2pi/60)/48.5= .54

T should = .02NM, but mastering physics says its wrong, Please help.

Explanation / Answer

T=i*a

T=0.5*0.8*0.295^2 *(2pi*250/60)/48.5

= 0.0188N-m

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