Calculating Rotational Kinetic Energy and Instantaneous Rate of Change

[tachu101]

Homework Statement

A hoop of mass 1kg and radius 2m is rotating about its center with an angular speed of 3rad/sec. A force of 10N is applied tangentially at the rim.

a. The rotational kinetic energy of the hoop is?
b. The instantaneous rate at which the kinetic energy is changing is?

Homework Equations

KE= 1/2Iw^2

The Attempt at a Solution

For the first part I can find I= MR^2 for a hoop, so I= (1)(2^2) so I=4. So KE= 1/2Iw^2 = (1/2)(4)(3^2)= 18. What do I do with the extra 10N applied (do I just add it in or does it not factor into the rotational kinetic energy?) I don't know what equation to use for the second part.

 

[tiny-tim]

A hoop of mass 1kg and radius 2m is rotating about its center with an angular speed of 3rad/sec. A force of 10N is applied tangentially at the rim.

a. The rotational kinetic energy of the hoop is?
b. The instantaneous rate at which the kinetic energy is changing is?
…
KE= 1/2Iw^2
…
What do I do with the extra 10N applied (do I just add it in or does it not factor into the rotational kinetic energy?) I don't know what equation to use for the second part.

Hi tachu101!

KE is geometric … it only depends on the motion, not on what's causing it.

So you only need the 10N for the second part …

and for that, what equation do you know that connects force and rotation?

 

[tachu101]

For the second part would I use torque=Ia which would then be torque/I=a? Would that be (10)(2)/((1)(2^2))= 5 rad/sec^2 does that help me?

 

[tiny-tim]

(have an omega: ω and an alpha: α and a delta: ∆ )

For the second part would I use torque=Ia which would then be torque/I=a? Would that be (10)(2)/((1)(2^2))= 5 rad/sec^2 …

Yup!

That gives you dω/dt, from which you could find dKE/dt.

But there's a more direct way …

remember, ∆KE = ∆(work done) …

so what is the equation for work done by a torque ?

 

[tachu101]

W= torque* $$\Delta$$$$\Theta$$ which would then be (10)(2)*(3)= 60J ? And how would I get dKe/dt from dW/dt

 

[tiny-tim]

W= torque* $$\Delta$$$$\Theta$$ which would then be (10)(2)*(3)= 60J ? And how would I get dKe/dt from dW/dt

uh … W= torque*∆θ, where ∆θ is the total change in angle, is correct …

but 3 is not the total change in angle, it's only the rate-of-change of angle, dθ/dt …

you need dW/dt = d(torque*θ)/dt ​

 

[tachu101]

hmmm... am I missing an equation because I am not sure how to find the d(torque*θ)/dt? Also, given that I found 5 rad/sec^2 how could I use that to find a change in KE?

 

[tiny-tim]

hmmm... am I missing an equation because I am not sure how to find the d(torque*θ)/dt? Also, given that I found 5 rad/sec^2 how could I use that to find a change in KE?

The torque here is constant, so d(torque*θ)/dt = torque * dθ/dt = torque * ω.

(you could use the dω/dt = 5 if you differentiate 1/2 Iω² …

but it's more direct if you use the work done method)

 

[tachu101]

So is (10)(2)*(3)= 60J Correct?

 

[tiny-tim]

So is (10)(2)*(3)= 60J Correct?

This would be easier to understand if you wrote it out in full …

but it's not joules anyway, is it?

 

[tachu101]

Does the instantaneous change in KE equal torque * ω? Is torque (10)*(2)=20?

 

[tiny-tim]

Instantaneous rate of change in KE equal torque * ω.

And yes, torque (10)*(2)=20.

 

[tachu101]

In summary...

I think that part a is 9J because I think it is (1/2)(I)(w^2)= 9
Part b is then torque*W which is then (10)(2)*(3)= 60

 

[tiny-tim]

In summary...

I think that part a is 9J because I think it is (1/2)(I)(w^2)= 9
Part b is then torque*W which is then (10)(2)*(3)= 60

I thought you got 18J for a?

Yup, b is 60.

 

[tachu101]

I looked back and I think I missed a 1/2 in part a. It should be (1/2)((1/2)(1)(2^2))(3^2)= 9 I think I missed the 1/2 in the I value.

 

[tiny-tim]

I looked back and I think I missed a 1/2 in part a. It should be (1/2)((1/2)(1)(2^2))(3^2)= 9 I think I missed the 1/2 in the I value.

What 1/2? … it's a hoop, not a disc.

Going to bed now … :zzz: