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Finding angle as a function of time

  1. Nov 14, 2011 #1
    1. The problem statement, all variables and given/known data

    I don't know exactly how to integrate in this case.
    This problem is about the track (spiral) of a spinning CD in a player.
    The radius of the spiral depends on time/revelations and is given by:

    [itex]r=r_i+\frac{h\theta}{2\pi}[/itex] , which means that the radius increases with h per revelation.

    v ist the constant speed with which the disc surface passes the laser. The rate of change of the angle is given by:
    [itex]\omega = \frac{d\theta}{dt} = \frac{v}{(r_i + \frac{h\theta}{2\pi} )}[/itex]

    I am looking for the angle as a function of time.

    [itex]\int \! \omega \, dt = \int \! \frac{v}{r_i+\frac{h\theta}{2\pi}} \, dt = ???[/itex]


    2. Relevant equations
    -


    3. The attempt at a solution
    Well, I tried to substitute [itex]\theta[/itex] with [itex]\omega t[/itex] but this is definately wrong because neither the angular velocity nor the angular acceleration is constant.
    Now I'm really stuck ...I'm somewhat slow on the uptake tomorrow. Any hints? How can I solve this?
     
  2. jcsd
  3. Nov 14, 2011 #2

    gneill

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    Staff: Mentor

    Re-arrange your equation to isolate dt on one side. Note that v is a constant so treat it as such. You should be able to integrate both sides of the resulting equation, one with respect to dt and the other with respect to dθ, with suitable integration limits.
     
  4. Nov 14, 2011 #3

    ehild

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    Differentiate equation (1) and substitute v/r for dθ/dt.You get a separable equation for r(t). Solve and use in (1) to get θ(t).

    ehild

    @gneill: The same thread at the same time again...
     
  5. Nov 14, 2011 #4
    Well, if I rearrange this equation and differentiate it ...I get zero? I still dont get it =/
     
  6. Nov 14, 2011 #5

    gneill

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    Can you show the rearranged equation?
     
  7. Nov 14, 2011 #6
    Okay, I think I'm now a little step further.

    [itex]k = \frac{h}{2\pi}[/itex]

    [itex]\frac{d\theta}{dt}=\frac{v}{r_i+k\theta}[/itex]

    [itex]dt=\frac{r_i+k\theta}{v}d\theta [/itex]

    [itex]t(\theta)=\int_0^t \, dt= \int_0^\theta \! (\frac{k\theta}{v}+\frac{r_i}{v} ) \, d\theta = \frac{k\theta^2}{2v}+\frac{r_i\theta}{v} [/itex]

    But what now?

    edit: solving the quadratic equation gives me

    [itex]-\frac{2\pi(r_i+\sqrt{r_i^2-hvt\pi^{-1}})}{h}[/itex]
     
    Last edited: Nov 14, 2011
  8. Nov 14, 2011 #7

    gneill

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    If you've solved for t as a function of θ, then you also have θ as a function of t by suitable rearrangement.
     
  9. Nov 14, 2011 #8
    [itex]\theta(t) = -\frac{2\pi(r_i+\sqrt{r_i^2-hvt\pi^{-1}})}{h}[/itex]

    Can this be right? Seems kinda odd to me, but this is the only way I see to solve for θ
     
  10. Nov 14, 2011 #9

    gneill

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    It looks reasonable to me.

    [EDIT]: There are two roots to the quadratic, so you should verify that you've chosen the one that makes physical sense. Perhaps plug some representative values into the variables.
     
    Last edited: Nov 15, 2011
  11. Nov 14, 2011 #10
    Great :) Then I think I finally got the clue... just need a little more experience :) Thanks!
     
  12. Nov 14, 2011 #11
    Oh, I got another question.

    To find the angular acceleration, I need to differentiate the angular velocity. But how do I do this here?
     
  13. Nov 15, 2011 #12

    gneill

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    You've just derived an expression for θ(t), the angular position with respect to time. What's the relationship between position, velocity, and acceleration?

    By the way, I added a note to post #9; you should verify that you've selected the right root of the quadratic when you solved for θ.
     
  14. Nov 15, 2011 #13
    Okay, I got it now. I differentiated θ(t) two times :) Thanks for your help!
     
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