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The rate of rotation of a solid disk with a radius of #1 m# and mass of #5 kg# constantly changes from #16 Hz# to #22 Hz#. If the change in rotational frequency occurs over #12 s#, what torque was applied to the disk?

Answer 1

Olivia Askew

The torque is #=15.7Nm#

The torque is the rate of change of angular momentum

#tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt#

where #I# is the moment of inertia

For a solid disc, #I=(mr^2)/2#

#I=5*1^2/2=2.5kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(22-16)/12*2pi#

#=(pi) rads^(-2)#

So the torque is #tau=2.5*(pi) Nm=5piNm=15.7Nm#

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Answer 2

Isabella Alvarez

Torque = moment of inertia × angular acceleration Moment of inertia (I) for a solid disk = (1/2) × mass × radius^2

Angular acceleration (α) = Change in angular frequency / Time Change in angular frequency = Final angular frequency - Initial angular frequency

Calculate the torque using the given values:

Initial angular frequency (ω₁) = 2π × Initial frequency
Final angular frequency (ω₂) = 2π × Final frequency

Substitute the values into the torque formula.

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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.