A solid disk with a radius of #5 m# and mass of #4 kg# is rotating on a frictionless surface. If #15 W# of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at #5 Hz#?

Question

A solid disk with a radius of #5 m# and mass of #4 kg# is rotating on a frictionless surface.  If #15 W# of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at #5  Hz#?

Nora Arzate · Accepted Answer

The torque is #=0.48Nm#

The mass of the disc is #m=4 kg#

The radius of the disc is #r=5m#

The power and the torque are related by the formula

#P=tau omega#

The angular velocity is #omega=2pif=2pi*5=10pirads^-1#

The power is #P=15W#

Therefore,

The torque is

#tau=P/omega=15/(10pi)=0.48Nm#

Axel Acevedo · Answer

undefined To find the torque applied when the disk is rotating at 5 Hz, first calculate the angular velocity using the formula $ \omega = 2\pi f $, where $ \omega $ is the angular velocity in radians per second, and $ f $ is the frequency in hertz. Then use the formula for power in rotational motion: $ P = \tau \omega $, where $ P $ is power in watts, $ \tau $ is torque in newton-meters, and $ \omega $ is angular velocity in radians per second. Rearrange the formula to solve for torque: $ \tau = \frac{P}{\omega} $. Plug in the given values: $ \omega = 2\pi(5) $ and $ P = 15 $ W. Calculate $ \omega $, then plug it into the torque formula to find $ \tau $.