1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What the heck my teacher did in the solution!

  1. Sep 24, 2013 #1
    1. The problem statement, all variables and given/known data

    The problem is problem 2 of the attached picture


    2. Relevant equations



    3. 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.
     

    Attached Files:

  2. jcsd
  3. Sep 24, 2013 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    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.

    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
     
  4. Sep 24, 2013 #3
    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?
     
  5. Sep 24, 2013 #4
    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.
     
  6. Sep 24, 2013 #5

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
  7. Sep 24, 2013 #6
    They're sideways? Sorry about that.
     
  8. Sep 24, 2013 #7
    I cannot reattach the file..
     
  9. Sep 24, 2013 #8
    This here might work, it is showing up the proper way when I load it, sorry about that.
     

    Attached Files:

  10. Sep 24, 2013 #9

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    [itex]d/dt \left( r^{2} \omega \right)= 0[/itex] follows from [itex]0 = r \ddot{\theta} + 2 \dot{r} \dot{\theta}[/itex].
     
  11. Sep 24, 2013 #10
    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.
     
  12. Sep 24, 2013 #11

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    [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.
     
  13. Sep 24, 2013 #12
    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?
     
  14. Sep 24, 2013 #13
    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?
     
  15. Sep 24, 2013 #14

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    [itex]d/dt \left( r^2 \right)[/itex] is not equal to [itex]2 dr/dt[/itex].
     
  16. Sep 24, 2013 #15
    Oh yeah sorry, the main problem is the second term there, the r^2, you know?
     
  17. Sep 24, 2013 #16

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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]
     
  18. Sep 24, 2013 #17
    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?
     
  19. Sep 24, 2013 #18
    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.
     
  20. Sep 24, 2013 #19
    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.
     
  21. Sep 25, 2013 #20

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    With practice, you should be able to see stuff like


    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted