Question

A centrifuge rotor rotating at 10,000 rpm is shut off and is eventually brought to rest by a frictional force of 1.20m n. if the mass of the rotor is 4.80 kg and it can be approximated as a solid cylinder of radius 0.0710m, through how many revolutions will the rotor turn before coming to rest, and how long will it take?

Answer 1

Answer: The moments of inertia are listed on p. 223, and a uniform cylinder through its center is: I = 1/2mr2 so I = 1/2(4.80 kg)(.0710 m)2 = 0.0120984 kgm2 Since there is a frictional torque of 1.20 Nm, we can use the angular equivalent of F = ma to find the angular deceleration: t = Ia -1.20 Nm = (0.0120984 kgm2)a a = -99.19 rad/s/s Now we have a kinematics question to solve: wo = (10,000 Revolutions/Minute)(2p radians/revolution)(1 minute/60 sec) = 1047.2 rad/s w = 0 a = -99.19 rad/s/s Let's find the time first: w = wo + at : wo = 1047.2 rad/s; w = 0 rad/s; a = -99.19 rad/s/s t = 10.558 s = 10.6 s And the displacement (Angular) Now the formula I want to use is only in the formula packet in its linear form, but it works just as well in angular form s = (u+v)t/2 Which is q = (wo+w)t/2 : wo = 1047.2 rad/s; w = 0 rad/s; t = 10.558 s q = (125.7 rad/s+418.9 rad/s)(3.5 s)/2 = 952.9 radians But the problem wanted revolutions, so let's change the units: q = (5528.075087 radians)(revolution/2p radians) = 880. revolutions