The rate of rotation of a solid disk with a radius of #1 m# and mass of #5 kg# constantly changes from #16 Hz# to #27 Hz#. If the change in rotational frequency occurs over #12 s#, what torque was applied to the disk?

1 Answer
Mar 27, 2017

Answer:

The torque was #=14.4Nm#

Explanation:

The torque is the rate of change of angular momentum

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

mass, #m=5kg#

radius, #r=1m#

The moment of inertia of a solid disc is

#I=1/2*mr^2#

#=1/2*5*1^2= 2.5 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=((27-16))/12*2pi#

#=(11/6pi) rads^(-2)#

So the torque is #tau=2.5*(11/6pi) Nm=14.4Nm#