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A solid disk with a radius of #8 m# and mass of #8 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 #6 Hz#?

Answer 1

Julia Aguilar

#=12.73Nm#

Given

#m-> "Mass of solid disk"=8kg#

#r-> "Radius of solid disk"=8m#

#P-> "Power used to solid disk"=480W#

#n-> "Frequency of rotaion of solid disk"=6Hz#

We are to find out

Formula

#"Power"(P)="Torque"(tau)xx"Angular velocity"(w)#

#=>"Power"(P)="Torque"(tau)xx2pixx"Frequency of rotation"(n)#

#=>480="Torque"(tau)xx2pixx6#

#"Torque"(tau)=480/(2pixx6)=12.73Nm#

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Answer 2

Xavier Adame

To find the torque applied when the disk is rotating at 6 Hz, use the formula:

Torque = (Power)/(Angular velocity)

First, calculate the angular velocity using the formula:

Angular velocity = 2π * frequency

Substitute the given values:

Angular velocity = 2π * 6 Hz = 12π rad/s

Now, use the given power and calculated angular velocity to find the torque:

Torque = 480 W / 12π rad/s ≈ 40.42 Nm

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.