The rate of rotation of a solid disk with a radius of #8 m# and mass of #5 kg# constantly changes from #5 Hz# to #17 Hz#. If the change in rotational frequency occurs over #6 s#, what torque was applied to the disk?

Question

1072 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-02T07:46:36

The torque was #=2010.6Nm#

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, rotating about the center is

#I=1/2*mr^2#

#=1/2*5*8^2= 160 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(17-5)/6*2pi#

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

So the torque is #tau=160*(4pi) Nm=2010.6Nm#