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A solid disk with a radius of #3 m# and mass of #12 kg# is rotating on a frictionless surface. If #480 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#?

Answer 1

Charles Aeschliman

The torque is #=19.1Nm#

mass of disc #=12kg#

radius of disc #=3m#

The power is related to the torque by the equation

#P=tau *omega#

#P=480W#

angular velocity, #omega=4*2pi rads^-1#

torque, #tau=P/omega#

#=480/(8pi)=60/pi=19.1Nm#

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Answer 2

Amelia Adams

To find the torque applied when the disk is rotating at 4 Hz, we can use the formula:

[ \text{Power} = \tau \cdot \omega ]

Where:

Power (P) = 480 W
Angular velocity (( \omega )) = 2πf (f is the frequency)

[ \omega = 2\pi \times 4 = 8\pi \text{ rad/s} ]

Rearranging the formula to solve for torque (( \tau )):

[ \tau = \frac{P}{\omega} = \frac{480}{8\pi} ]

[ \tau ≈ 60.48 \text{ Nm} ]

So, the torque applied when the disk is rotating at 4 Hz is approximately 60.48 Nm.

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Answer from HIX Tutor

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