The rate of rotation of a solid disk with a radius of #1 m# and mass of #5 kg# constantly changes from #6 Hz# to #24 Hz#. If the change in rotational frequency occurs over #6 s#, what torque was applied to the disk?

1 Answer
Oct 4, 2017

Answer:

The torque was #=47.1Nm#

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

For a solid disc, #I=(mr^2)/2#

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

The radius of the disc is #r=1m#

So, the moment of inertia is #I=5*(1)^2/2=2.5kgm^2#

The rate of change of the angular velocity is

#(domega)/dt=(24-6)/6*2pi=6pirads^-2#

So,

The torque is

#tau=2.5*6pi=47.1Nm#