Angular Motion Conceptual Question

[FeelTheFire]

Hey all,
in physics right now I am learning about rigid body rotations about a fixed axis. One of the questions we had for homework goes like this:

"The combination of a friction force and applied force produces a constant total torque of 36 N*m on a wheel rotating about a fixed axis. The applied force acts for 6 seconds. During this time, the angular speed of the wheel increases from 0 to 10 rad/s."

There are multiple questions that go along with that setup, one being to find the total number of revolutions the wheel undergoes during the acceleration period.

I am able to achieve the correct answer by finding the constant α and integrating it up to get θ as a function of time and converting from radians to rotations.

However, I started to think about if and how I could somehow integrate with respect to *ω* rather than time. My train of thought went something like this: (and curiously the answer I got came out to be bigger than the correct answer by a factor of 10)

ω = rad / sec → ω / 2∏ = rot / sec → ω * sec / 2∏ = rotations
So if ω were constant, we could plug it in above and find the number of rotations for a given time.

(Info from the above problem begins to be substituted in here)
Since ω isn't constant, I integrated ω from its initial value to its final value →

(6 sec / 2∏) ∫0→10 ω dω = (3 / ∏) (50) = 150/∏

The correct answer is 15/∏. I'd like to know what I'm doing wrong. I asked my physics teacher and he said that the integral is meaningless, omega is changing with time so it should be dt and the limits of integration shouldn't be with respect to omega. But that's the technique I used to get the right answer before by integrating the acceleration! I'm trying to do it differently here.

I'm just wondering if there's some correct answer in my line of thinking somewhere and if you can arrive at the correct answer by integrating with respect to ω using limits of 0 to 10 rather than integrating with respect to time and using limits from 0 to 6.

Hopefully this makes sense! Thanks and I really look forward to your responses.

 

[haruspex]

Integrating ω wrt ω does make physical sense, but not the sense you wanted.
Your constant ω case gives you θ = ωt. In a period dt, that's dθ=ωdt, leading, when ω varies, to the usual integral. Or if you consider a small change in ω, dω, then you can write dθ=tdω, and integrate t wrt ω.
If you consider energy, you can write L dθ = Iω dω, where I is the moment of inertia and L the torque. Now you get an integral of ω wrt ω, but this doesn't help if you don't know I.

 

[FeelTheFire]

Integrating ω wrt ω does make physical sense, but not the sense you wanted.
Your constant ω case gives you θ = ωt. In a period dt, that's dθ=ωdt, leading, when ω varies, to the usual integral. Or if you consider a small change in ω, dω, then you can write dθ=tdω, and integrate t wrt ω.
If you consider energy, you can write L dθ = Iω dω, where I is the moment of inertia and L the torque. Now you get an integral of ω wrt ω, but this doesn't help if you don't know I.

Thanks for the reply. I don't know anything about the wheel so I can't calculate I. As for integrating time wrt ω, I'm having trouble applying this concept to the problem.

Do I have to figure out time as a function of ω in order to complete that integral? And do the coefficients I have outside of the integral remain? Or should I just throw all of that thinking aside?

 

[haruspex]

Thanks for the reply. I don't know anything about the wheel so I can't calculate I.

You can get it from the other data you have, but that's a long way round to finding θ.

As for integrating time wrt ω, I'm having trouble applying this concept to the problem. Do I have to figure out time as a function of ω in order to complete that integral?

Yes. You have ω = at, so t = ω/a. ∫t.dω = ∫(ω/a)dω = ω²/(2a) = at²/2

 

[FeelTheFire]

You can get it from the other data you have, but that's a long way round to finding θ.

Yes. You have ω = at, so t = ω/a. ∫t.dω = ∫(ω/a)dω = ω²/(2a) = at²/2

It worked! Eureka
Thanks so much haruspex!