A solid disk with a radius of #12 m# and mass of #8 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 #12 Hz#?

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Answer 1 · 2017-07-18T11:41:51

The torque is #=8.49Nm#

The mass of the disc is #m=8 kg#

The radius of the disc is #r=12m#

The power and the torque are related by the formula

#P=tau omega#

The angular velocity is #omega=2pif=2pi*12=24pirads^-1#

The power is #P=640W#

Therefore,

The torque is

#tau=P/omega=640/(24pi)=8.49Nm#