Question

An aluminum flywheel (density=2.7g/cm^3) has a radius of 1.5m and moment of inertia 25kg. The flywheel is initially spinning at 1200 rpm, and is to be stopped by a friction brake that runs on the curved outside surface. If the coefficient of friction is 0.600, how hard must the brake be pushed onto the flywheel to bring it to rest in 4.0s?

Answer 1

Moment of inertia of flywheel, I = 0.5*M*R^2

= 0.5*25*1.5^2

= 28.1 kg.m^2

wi = 1200 rpm

= 1200*2*pi/60 rad/s

= 126 rad/s

wf = 0

t = 4.0 s

angular acceleration, alfa = (wf - wi)/t

= (0 - 126)/4

= -31.5 rad/s^2

|alfa| = 31.5 rad/s^2

let F is the force exerted by the brake on the flywheel.

use, Net torque = I*|alfa|

fk*R = 28.1*31.5

mue_k*F*R = 28.1*31.5

F = 28.1*31.5/(mue_k*R)

= 28.1*31.5/(0.6*1.5)

= 984 N <<<<<<<<------------Answer