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Particle along a spiral

  1. Sep 8, 2007 #1
    1. The problem statement, all variables and given/known data
    A particle moves outward along a spiral. Its trajectory is given by r=Aθ, where A is a constant. A=1/π m/rad. θ increases in time according to θ=αt^2/2, where α is a constant.
    Show that the radial acceleration is zero when θ=1/√2 rad. At what angles do the radial and tangential accelerations have equal magnitude?


    2. Relevant equations

    [tex]\frac{d\mathbf{r}}{dt} = \dot r\hat{\mathbf{r}} + r\dot\theta\hat{\boldsymbol\theta}[/tex]
    [tex]\frac{d^2\mathbf{r}}{dt^2} = (\ddot r - r\dot\theta^2)\hat{\mathbf{r}} + (r\ddot\theta + 2\dot r \dot\theta)\hat{\boldsymbol\theta}[/tex]

    3. The attempt at a solution

    I tried plugging things into the second equation. However, I don't know what t and α are, which is why I am stuck. I also tried substitution, but that didn't really get me anywhere either.
     
    Last edited: Sep 9, 2007
  2. jcsd
  3. Sep 8, 2007 #2

    learningphysics

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    What are the values of a and A?
     
  4. Sep 8, 2007 #3
    a is supposed to be alpha, an unknown constant. A is equal to 1/(pi) m/rad.
     
  5. Sep 9, 2007 #4

    learningphysics

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    The angular acceleration is: [tex]\frac{d^2R}{dt^2} - R(\frac{d\theta}{dt})^2[/tex]. substitute your values for R and [tex]\theta[/tex] into this equation and simplify as much as you can.

    Finally substitute in the value for theta 1/sqrt(2).
     
  6. Sep 9, 2007 #5
    Is angular acceleration the same as radial acceleration? Also, what is the equation for tangential acceleration? I just changed my original comment. The relevant equations are

    [tex]\frac{d\mathbf{r}}{dt} = \dot r\hat{\mathbf{r}} + r\dot\theta\hat{\boldsymbol\theta}[/tex]
    [tex]\frac{d^2\mathbf{r}}{dt^2} = (\ddot r - r\dot\theta^2)\hat{\mathbf{r}} + (r\ddot\theta + 2\dot r \dot\theta)\hat{\boldsymbol\theta}[/tex]
     
    Last edited: Sep 9, 2007
  7. Sep 9, 2007 #6

    learningphysics

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    Oops, I meant to write radial acceleration... yeah, that formula I posted is radial acceleration.

    tangential acceleration is: 2*dr/dt*w+r*d(w)/dt, where w = dtheta/dt... I looked this up here: http://aemes.mae.ufl.edu/~uhk/DYNAMICS.html because I couldn't remember it.
     
  8. Sep 9, 2007 #7
    What is this part of the equation: [tex]\((r\ddot\theta + 2\dot r \dot\theta)\hat{\boldsymbol\theta}[/tex]?
     
  9. Sep 9, 2007 #8

    learningphysics

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    That's tangential acceleration.

    This is radial acceleration:
    [tex](\ddot r - r\dot\theta^2)\hat{\mathbf{r}}[/tex]
     
  10. Sep 9, 2007 #9
    Thanks for the help. I think I figured it out.
     
  11. Sep 9, 2007 #10

    learningphysics

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    Cool. no prob.
     
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