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 #27 Hz#. If the change in rotational frequency occurs over #8 s#, what torque was applied to the disk?

1 Answer
Dec 26, 2016

Answer:

THe torque is #=206.2Nm#

Explanation:

The torque is the rate of change of angular momentum

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

The moment of inertia of a solid disc is #I=1/2mr^2#

#=1/2*1*5^2= 25/2=12.5 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(27-6)/8*2pi#

#=((21pi)/4)pi rads^(-2)#

So the torque is #tau=12.5*(21pi)/4 Nm=65.625piNm=206.2Nm#