The rate of rotation of a solid disk with a radius of #9 m# and mass of #5 kg# constantly changes from #32 Hz# to #15 Hz#. If the change in rotational frequency occurs over #6 s#, what torque was applied to the disk?

1 Answer
Jan 7, 2017

Answer:

The torque is #=3604.98Nm#

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=1/2*mr^2#

#=1/2*5*9^2= 405/2 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(32-15)/6*2pi#

#=((17pi)/3) rads^(-2)#

So the torque is #tau=405/2*(17pi)/3 Nm=3604.98Nm#