An object with a mass of # 2 kg# is traveling in a circular path of a radius of #4 m#. If the object's angular velocity changes from # 4 Hz# to # 8 Hz# in # 3 s#, what torque was applied to the object?

Answer 1

The torque was #=268.1Nm#

The torque is the rate of change of angular momentum

#tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt#
The moment of inertia of the object is #I=mr^2#
#=2*4^2= 32 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(8-4)/3*2pi#
#=((8pi)/3) rads^(-2)#
So the torque is #tau=32*(8pi)/3 Nm=256/3piNm=268.1Nm#
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Answer 2

Torque = Moment of Inertia * Angular Acceleration Angular Acceleration = (Final Angular Velocity - Initial Angular Velocity) / Time

Angular Acceleration = (8 Hz - 4 Hz) / 3 s Angular Acceleration = 1.33 Hz/s

Moment of Inertia (for a point mass rotating about an axis at a distance r) = mass * radius^2 Moment of Inertia = 2 kg * (4 m)^2 = 32 kg*m^2

Torque = 32 kgm^2 * 1.33 Hz/s = 42.56 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.

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.

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.

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.

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