The rate of rotation of a solid disk with a radius of #4 m# and mass of #5 kg# constantly changes from #21 Hz# to #12 Hz#. If the change in rotational frequency occurs over #8 s#, what torque was applied to the disk?

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Answer 1 · 2017-03-23T06:48:03

THe torque was #=282.7Nm#

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*4^2= 40 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(21-12)/8*2pi#

#=(9/4pi) rads^(-2)#

So the torque is #tau=40*(9/4pi) Nm=(90pi)Nm=282.7Nm#