Acentrifuge rotor rotating at 10,300 rpm is shut off and is eventually brought uniformly to rest by a frictional torque of 1.2 m*n. if the mass of the rotor is 4.8 kg and it can be approximated as a solid cylinder of radius .0710 m, through how many revolutions will the rotor turn before coming to rest, and how long will it take? show all work and formulas for best rating.

The exercise is solved based on the consideration of Torque through the moment of inertia and angular acceleration, that is,

Where ,

I= Moment of inertia

Angular acceleration.

Our torque is 1.2mN, so we can find easily the Angular Acceleration.

But first we will find the Inertia for a solid cylinder given by,

Where m is the mass of 4.8Kg and R the radius 0.071m

Replacing in the first equation we have,

A) With the acceleration found now we can know how many revolutions the body had, through the equation of rotational dynamics when time is not counted,

B) Using a similar formula of rotational dynamics we can find the time, that is,

layer a is the exosphere and layer b is the thermosphere.   is the answer i think but i hope this .

yup its correct its: c) b b