The rate of rotation of a solid disk with a radius of #6 m# and mass of #5 kg# constantly changes from #2 Hz# to #3 Hz#. If the change in rotational frequency occurs over #1 s#, what torque was applied to the disk?

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Answer 1 · 2017-03-09T06:34:33

The torque was #=565.5Nm#

The torque is the rate of change of angular momentum

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

The moment of inertia of the solid disc is

#I=1/2*mr^2#

#=1/2*5*6^2= 90 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(3-2)/1*2pi#

#=(2pi) rads^(-2)#

So the torque is #tau=90*(2pi) Nm=180piNm=565.5Nm#