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Polar coordinates and mechanics question.

  1. Oct 21, 2007 #1
    Alright, the problem here is that I seem unable to grasp an example given in class. I am not sure if this is due to not copying it down correctly, or if there's something I am just missing. Either way, I know I am not the only one who has had a bit of trouble with this. I'm hoping that someone with a greater knowledge of physics can spot what is going on.

    1. The problem statement, all variables and given/known data
    Bead threaded on a smooth rod (i.e. no friction) which rotates with constant angular velocity [tex]\omega[/tex] about one end. Investigate the motion.

    Where r is a radial vector,

    [tex]\dot{r}[/tex]=0, [tex]\dot{\theta}[/tex]=[tex]\omega[/tex], and r=r[tex]_{0}[/tex] at t=0

    The dot above the letters and symbols, as far as I am aware, just denotes the first derivative (in case this isn't a standard notation of some sort). Also, the subscript doesn't seem to be working in the preview, so in case it does not turn out right, that is meant to be a subcript zero, not r^0. It's just denoting a particular distance up the rod that the bead is.

    2. Relevant equations
    Now this is the working out that I copied.

    Newtons second law:
    [tex]\ddot{r}[/tex] -r[tex]\dot{\theta}[/tex]^2 = 0.

    (I realise Newton's second law should include a multiplication by mass, but I'm just writing the information I have). r dot dot is the second derivative I believe.



    3. The attempt at a solution

    [tex]\theta[/tex] = [tex]\omega[/tex]t

    [tex]\dot{\theta}[/tex] = [tex]\omega[/tex] ; [tex]\ddot{r}[/tex] = [tex]\omega[/tex]*r[tex]^{2}[/tex]

    Solutions:
    e^wt and e^-wt

    Oh, the omega is not meant to be superscript throughout the question by the way, I don't know why it is; and the w in the 'solution' represents an omega. So it is e to the power of omega*time.

    All I need to know is what on Earth the solutions are for and how did they materialise?

    EDIT: I really made a mess of this whole latex thing.
     
    Last edited: Oct 21, 2007
  2. jcsd
  3. Oct 21, 2007 #2
    Let ω=omega θ=theta, r' = dr/dt, r'' = d^2r/dt^2 (second derivative) Let R be the vector , and r be the length.
    I'm not sure how you got to Newton's second law but here's how my professor proved it:
    R can be broken down into x and y components as such:
    R = rcosθ*i + rsinθ*j (i,j, are unit vectors along x and y)

    Let's take the derivative twice (Keep in mind that this is with respect to t)
    R' = -rsinθ*θ'*i + rcosθ*θ*'j
    R'' = -rcosθ*θ'^2*i - rsinθ*θ'^2*j

    Hey! That looks like our original R almost:
    R'' = -Rθ''
    R'' + Rθ'^2 = 0 <---matches your notes except for the sign..
    Keep in mind that θ' = ω so:
    R'' + R(ω^2) = 0

    I haven't taken differential equations and I'm guessing you haven't either. But in general if you want to solve for R but you know that its second derivative is itself times a constant, then there are only a few functions that do that: sine, cosine, and e^x. You "guess" that the solution is e^ωt. If you take the derivative twice, indeed, it is itself times a constant (ω^2).

    The solutions are for R. Sorry I don't know why I had issues with the signs, but they should be right otherwise.
     
  4. Oct 22, 2007 #3
    Ah thanks for the help, because my professor gave no explanation whatsoever on how he got from one step to the other.
     
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