What the heck my teacher did in the solution!

  • Thread starter Levi Tate
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Homework Statement



The problem is problem 2 of the attached picture


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The Attempt at a Solution



I got the problem wrong, and I am trying to understand what my teacher did. Particularly, on the right side of his solution, he took the θ acceleration to be zero, because there is no force in that direction, that is obvious. What is not obvious at all is the next step he took, d/dt(r^2ω)= 0, and from there he goes to equating the initial angular velocity and radius with the final angular velocity and radius. Everything else from the solution falls into place from there, but as far as I see, there is a PDE, which I have no idea how to solve, and I have no idea how he implied that time derivative from that information.

Any help would be greatly appreciated and thank you in advance, I plan to repay the 'karma' when I know how to solve some physics problems with confidence on this site.
 

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  • #2
Andrew Mason
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Homework Statement



The problem is problem 2 of the attached picture


Homework Equations





The Attempt at a Solution



I got the problem wrong, and I am trying to understand what my teacher did. Particularly, on the right side of his solution, he took the θ acceleration to be zero, because there is no force in that direction, that is obvious.
I don't think that is obvious. A figure skater spins with increasing angular speed as he/she brings their arms in even though there is no angular force.

What is not obvious at all is the next step he took, d/dt(r^2ω)= 0, and from there he goes to equating the initial angular velocity and radius with the final angular velocity and radius.
d/dt(r^2ω)= 0 because this is a purely central force so there is no external torque.

The initial angular momentum: L0 = mvr = mω0r02

What happens to angular momentum as the weight drops?


AM
 
  • #3
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We haven't gotten to angular momentum or central force, thanks by the way. So is what he is doing is pulling the mass out as a constant and asserting angular momentum is conserved, and if that is the case, the final angular momentum (although it's not really angular momentum because he's just using r^2w) is equal to the initial angular momentum. Man I still don't get that step, is there any way to explain it without torques and central forces and angular momentum, because we haven't even covered that yet?
 
  • #4
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It's just that one step, what is r^w and how did he imply the derivate of it is zero, which when differentiated, doesn't produce anything previously from the problem. This here is driving me nuts.
 
  • #5
Integral
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If you want more help you should but a little more effort into your images. Officially we frown on posting images, but in this case I don't see a good alternative. Please improve the images, I mean, rotate the dang thing at a minimum.
 
  • #6
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They're sideways? Sorry about that.
 
  • #7
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I cannot reattach the file..
 
  • #8
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This here might work, it is showing up the proper way when I load it, sorry about that.
 

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  • #9
George Jones
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[itex]d/dt \left( r^{2} \omega \right)= 0[/itex] follows from [itex]0 = r \ddot{\theta} + 2 \dot{r} \dot{\theta}[/itex].
 
  • #10
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But if I were given a different PDE, how would I recognize that, what is the mathematics? Also how does that information translate to the next line, where ro^2w=r^2w. Everything else in the problem just falls into place, but it seems like there is a large part of information missing in that step.

Thank you very much.
 
  • #11
George Jones
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Also how does that information translate to the next line, where ro^2w=r^2w.
[itex]d/dt \left(r^2 \omega \right) = 0[/itex] means that the value of [itex]r^2 \omega[/itex] stays constant. Consequently, its initial value, [itex]r_0^2 \omega_0[/itex], is equal to the value of [itex]r^2 \omega[/itex] at any other time.
 
  • #12
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Right, I do understand that part now, mainly now it is a question of the mathematical technique used to figure out how i extrapolated d/dt(r^2w) from that partial differential equation in the first place. If I see something else, I cannot simply guess and figure out what derivative of a function is equal to the equation, you know?
 
  • #13
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Also, when I take the time derivative of r^2w I get = 2(dr/dt)w+dw/dt(r^2)

Which isn't equal to the initial equation, unless I am doing something wrong?
 
  • #14
George Jones
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[itex]d/dt \left( r^2 \right)[/itex] is not equal to [itex]2 dr/dt[/itex].
 
  • #15
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Oh yeah sorry, the main problem is the second term there, the r^2, you know?
 
  • #16
George Jones
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I have been trying not to give the game away, but ...

multiplying each side of [itex]0 = r \ddot{\theta} + 2 \dot{r} \dot{\theta}[/itex] by [itex]r[/itex] gives [itex]0 = r^2 \ddot{\theta} + 2r \dot{r} \dot{\theta} = d/dt \left( r^2 \omega \right)[/itex]
 
  • #17
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Right, but if I were to come across a similar equation, is there a technique to use to figure out what the derivative of the differential equation is without guessing?

Thank you, I don't know if this will help me in subsequent work, it's not something i'd recognize, it's like a guess and check type thing, it seems there is a method of a solution, is there one, or should I just guess if i come across another problem like this on the exam?
 
  • #18
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Basically right now, I understand this problem due to you multiplying an r in there, but all I have is a memory of one problem, I don't like that, I still do not understand the method which leads to the derivative of the function being equal to the function itself. In ODEs I know you can use an integrating factor, but this is a PDE, which I have no knowledge of.
 
  • #19
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Well thanks a lot, that did increase my understanding of that problem. It is still shaky to be why I needed the derivative of the function to equal the function (and also zero because the function did), but it illuminated that the initial conditions don't change with time, so I can solve that problem, I don't know about the next one. Thank you very much though everybody.
 
  • #20
George Jones
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With practice, you should be able to see stuff like

multiplying each side of [itex]0 = r \ddot{\theta} + 2 \dot{r} \dot{\theta}[/itex] by [itex]r[/itex] gives [itex]0 = r^2 \ddot{\theta} + 2r \dot{r} \dot{\theta} = d/dt \left( r^2 \omega \right)[/itex]

There isn't a "turn the crank" solution to every problem.

This is, however, a "turn the crank" solution to this differential equation. "Multipltying" by [itex]dt[/itex] gives [itex]0 = r d \omega + 2 \omega d r[/itex]. Have you taken a course in ordinary differential equations? If you have, then make this equation exact, i.e., find an integrating factor.
 
  • #21
Andrew Mason
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Well thanks a lot, that did increase my understanding of that problem. It is still shaky to be why I needed the derivative of the function to equal the function (and also zero because the function did), but it illuminated that the initial conditions don't change with time, so I can solve that problem, I don't know about the next one. Thank you very much though everybody.
You are not supposed to understand why [itex]\vec{a} = (\ddot{r} - r\dot{\theta}^2)\hat{r} + (r\ddot{\theta} + 2\dot{r}\dot{\theta})\hat{\theta}[/itex].

You are just supposed to figure out that the total force on m is provided by the tension in the string and if the acceleration is as given above, that the second term [itex](r\ddot{\theta} + 2\dot{r}\dot{\theta})\hat{\theta}[/itex] is equal to 0.

That second term is the acceleration perpendicular to the force (T), so we can conclude that it must be 0. If you are then clever enough to realize that the second term is (1/r) multiplied by the time rate of change of r2ω you can then conclude that r2ω is constant (ie. it is always equal to r02ω0). After replacing r by r0-Vt the rest is just algebra.

I expect that your teacher may be leading up to angular momentum. In any central force, angular momentum (designated as L = mvr = mωr2), is conserved. The quantity r2ω = L/m is, therefore, a constant if there is only a central force. This is a very important and useful principle.

AM
 
  • #22
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Yes, Beautiful! You and George. My teacher did in fact relate that problem to central forces and angular momentum today in lecture, but that isn't going to be on the exam. Great stuff though. Love the physics.

George, you told me exactly what I needed to know, the only thing is though is that I don't know how to find an integrating factor for the PDE, as opposed to exp^integral(something), which I forgot but could easily look up, but everything is good, because he said we aren't going to be solving any partial differential equations that are not seperable. I am taking a course called methods of theoretical physics where we do some PDEs later on.

Yeah this was a tough problem, I had it all ****ed up, I expressed it in the wrong variables.

Anyways thanks a lot for sorting me out here fellows. Much appreciated.
 
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