An object with a mass of # 7 kg# is traveling in a circular path of a radius of #9 m#. If the object's angular velocity changes from # 2 Hz# to # 23 Hz# in # 8 s#, what torque was applied to the object?

Answer 1

Charlotte Aaron

The torque was #=9352Nm#

The torque is the rate of change of angular momentum

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

where #I# is the moment of inertia

For the object, #I=mr^2#

So, #I=7*(9)^2=567kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(23-2)/8*2pi#

#=(21/4pi) rads^(-2)#

So the torque is #tau=567*(21/4pi) Nm=9352Nm#

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Answer 2

William Audy

To find the torque applied to the object, you can use the equation:

Torque = Moment of inertia × Angular acceleration

First, calculate the moment of inertia (I) of the object using the formula:

I = m × r^2

where: m = mass of the object (7 kg) r = radius of the circular path (9 m)

Then, calculate the angular acceleration (α) using the formula:

α = (final angular velocity - initial angular velocity) / time

where: initial angular velocity (ω1) = 2 Hz final angular velocity (ω2) = 23 Hz time (t) = 8 s

Finally, substitute the calculated values into the torque formula to find the torque applied to the object.

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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.