A solid disk with a radius of #5 m# and mass of #2 kg# is rotating on a frictionless surface. If #72 W# of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at #4 Hz#?

1 Answer

#2.865\ \text{Nm}#

Explanation:

frequency of rotation of solid disk is #f=4\ Hz# hence its angular velocity #\omega# is given as follows

#\omega=2\pi f=2\pi\cdot 4=8\pi \ text{rad/s}#

Now, the power #P=72\ W# imparted to the solid disk by applying torque #T# to create final angular velocity #\omega=8\pi \text{rad/s}# is given as follows

#P=T\times \omega#

#T=P/\omega#

#=72/{8\pi}#

#=9/\pi#

#=2.865\ \text{Nm}#