An object with a mass of # 2 kg# is traveling in a circular path of a radius of #2 m#. If the object's angular velocity changes from # 3 Hz# to # 8 Hz# in # 1 s#, what torque was applied to the object?

Answer 1

Emily Abbate

The torque is #=251.3Nm#

The torque is the rate of change of angular momentum

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

where #I# is the moment of inertia

For the object, #I=mr^2#

The mass is #m=2 kg#

The radius of the path is #r=2m#

So the moment of inertia is

,#I=2*(2)^2=8kgm^2#

The angular acceleration is

#alpha=(2pif_1-2pif_0)/t=(16-6)pi/1=10pirads^-2#

So,

The torque is

#tau=Ialpha=8*10pi=251.3m#

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Answer 2

Benjamin Asiedu

To calculate the torque applied to the object, you can use the formula:

Torque = Moment of inertia * Angular acceleration

The moment of inertia (I) for a point mass rotating about an axis at a distance (r) is given by:

I = m * r^2

Where: m = mass of the object r = radius of the circular path

Given: m = 2 kg r = 2 m

Calculate I:

I = 2 * (2^2) = 8 kg*m^2

Angular acceleration (α) can be found using the formula:

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

Given: Initial angular velocity (ω_i) = 3 Hz Final angular velocity (ω_f) = 8 Hz Time (t) = 1 s

Calculate α:

α = (8 Hz - 3 Hz) / 1 s = 5 Hz/s

Now, calculate the torque:

Torque = I * α = 8 kgm^2 * 5 Hz/s = 40 Nm

So, the torque applied to the object is 40 Newton meters.

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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.