The rate of rotation of a solid disk with a radius of #9 m# and mass of #5 kg# constantly changes from #21 Hz# to #12 Hz#. If the change in rotational frequency occurs over #5 s#, what torque was applied to the disk?

1 Answer
Nov 10, 2017

Answer:

The torque was #=2290Nm#

Explanation:

The torque is the rate of change of angular momentum

#tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt#

where #I# is the moment of inertia

The mass of the disk is #m=5kg#

The radius is #r=9m#

For the solid disk, #I=1/2mr^2#

So, #I=1/2*5*(9)^2=202.5kgm^2#

The angular velocity is

#omega=2xxpixxf# where the frequency is #=f#

And the rate of change of angular velocity is

#alpha=(d omega)/dt=(Deltaomega)/t=(42pi-24pi)/5=(18/5pi)rads^-2#

The torque is

#tau=202.5*18/5pi=2290Nm#