A solid disk with a radius of #8# #m# and mass of #2# #kg# is rotating on a frictionless surface. If #640# #W# of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at #16# #Hz#?

1 Answer
Mar 28, 2017

Answer:

The torque is #=6.4Nm#

Explanation:

mass of disc #=2kg#

radius of disc #=8m#

The power is related to the torque by the equation

#P=tau *omega#

#P=640W#

Frequency of rotation is #f=16Hz#

#omega=f*2pi#

angular velocity, #omega=16*2pi rads^-1#

torque, #tau=P/omega#

#=640/(32pi)=6.4Nm#