The rate of rotation of a solid disk with a radius of #5 m# and mass of #5 kg# constantly changes from #12 Hz# to #10 Hz#. If the change in rotational frequency occurs over #9 s#, what torque was applied to the disk?

1 Answer
Apr 25, 2018

Answer:

The torque was #=87.27Nm#

Explanation:

The torque is the rate of change of angular momentum

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

where #I# is the moment of inertia

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

The radius of the disc is #r=5m#

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

So, #I=5*(5)^2/2=62.5kgm^2#

The initial angular velocity is

#omega_0=2pif_0=2*12*pi=24pirads^-1#

The final angular velocity is

#omega_1=2pif_1=2*10*pi=20pirads^-1#

The time is #t=9s#

The angular acceleration is

#alpha=(omega_1-omega_0)/t=(24pi-20pi)/9=4/9pirads^-2#

So,

The torque is #tau= 62.5*4/9pi=87.27Nm#