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Spiraling particle

  1. Mar 13, 2015 #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 = \frac{1}{\pi} m/rad##. ##θ## increases in time according to ##θ = \frac{\alpha t^2}{2}## where ##\alpha## is a constant.
    (a) Sketch the motion, and indicate the approximate velocity and acceleration at a few points.
    (b) Show that the radial acceleration is zero when ##\theta = \frac{1}{\sqrt{2}}##
    (c) At what angles do the radial and tangential accelerations have equal magnitude?


    2. Relevant equations

    $$\vec{r} = r \hat{r}$$
    $$\vec{v} = \dot{r} \hat{r} + r \dot{\theta} \hat{\theta}$$
    $$\vec{a} = (\ddot{r} - r \dot{\theta}^2) \hat{r} + (r \ddot{\theta} + 2 \dot{r} \dot{\theta}) \hat{\theta}$$

    3. The attempt at a solution

    Part (a): ##r = \frac{\theta}{\pi}## and ##\theta = \frac{\alpha t^2}{2}##, so ##\dot{\theta} = \alpha t## and ##\dot{r} = \frac{\alpha t}{\pi}##.
    ##\ddot{r} = \frac{\alpha}{\pi}## and ##\ddot{\theta} = \alpha##. Now, I can plug everything into the three equations above (position, velocity, and acceleration) and substitute ##\sqrt{\frac{2\theta}{\alpha}}## for ##t## so everything is in terms of ##\theta##.
    I'm confused, though. If I wish to keep ##r## positive (the convention used in my book), then I should only consider values of ##\theta \geq 0##, is this okay? Also, when exactly do I stop? ##2\pi##? ##4\pi##? It seems to me that it's not possible to restrict ##\theta## according to conventions like ##0 \leq \theta < 2\pi##. The spiral doesn't repeat, and it doesn't stop. What do I do in this case?
    Also, at what points do I indicate the approximate velocity and accelerations of the particle? For angles that are multiples of ##\pi##?
    Part (b): That's trivial. Plug in the value of ##\theta## and do the math to show that ##a_r = 0##.
    Part (c): Equate ##\ddot{r} - r \dot{\theta}^2## to ##r \ddot{\theta} + 2 \dot{r} \dot{\theta}## and solve for ##\theta##.
    EDIT: I just solved the equation in part (c) and ended up with a negative solution! This goes against the assumption I made in part (a). I'm really confused now. I thought that a convention must be followed at all times when analyzing motion in polar coordinates. Such as ##r \geq 0## and ##-\pi < θ \leq \pi## or ##r \geq 0## and ##0 \leq θ < 2\pi##.
     
    Last edited: Mar 13, 2015
  2. jcsd
  3. Mar 13, 2015 #2

    haruspex

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    It says in (c) that the two accelerations are to have equal magnitudes. That does not require them to be equal.
     
  4. Mar 13, 2015 #3
    This gives a modular equation. I just solved the equation and the solutions I got are (to three significant figures) are ##\theta = 0.186##, ##\theta = -2.67##, ##\theta = 2.67##, and ##\theta = -0.186##.
     
  5. Mar 13, 2015 #4

    haruspex

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    I get the 0.186, but the other root seems to be a little more than 2.67.
     
  6. Mar 14, 2015 #5
    I probably made an algebraic error somewhere. But how do I deal with the roots? Do I ignore the negative values of ##θ## according to the convention I followed in part (a)?
     
  7. Mar 14, 2015 #6

    haruspex

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    At what times would those negative values occur?
     
  8. Mar 14, 2015 #7
    At ##t = \sqrt{\frac{2θ}{\alpha}}##. I don't know whether ##\alpha## is positive or negative, and because of that, I don't know whether or not I should disregard negative values of ##θ##.
     
  9. Mar 14, 2015 #8

    haruspex

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    It's probably intended as positive. If it's negative then theta is always negative, so you would ignore the positive values. If you want to express the answer in closed form you could put in a factor ##\frac{\alpha}{|\alpha|}##.
     
  10. Mar 14, 2015 #9
    What about the range of values of ##θ##? Where does it end?
     
  11. Mar 14, 2015 #10

    haruspex

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    Why should there be a limit? There's no limit to time.
     
  12. Mar 14, 2015 #11
    I thought ##r## and ##\theta## always have to be restricted in some interval.
     
  13. Mar 14, 2015 #12

    haruspex

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    Not really. For any given (r, θ), you can normalise such that r>=0 and θ is in a range of length 2 pi. But these are not restrictions on the values of r and theta.

    Edit: I'lll reword that last bit. They are not fundamental restrictions on the ranges. They are restrictions you can choose to apply.
     
  14. Mar 14, 2015 #13
    Could you please elaborate?
     
  15. Mar 14, 2015 #14

    haruspex

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    As variables, r and theta can take any real values. But a given point in the plane can be represented by them in many different ways. Adding a multiple of 2pi to theta leaves you at the same point; adding an odd multiple of pi while also switching the sign of r also leaves you at the same point. So in choosing how to represent points in the plane in polar coordinates you can choose restricted ranges for the variables.
     
  16. Mar 17, 2015 #15
    But won't this produce discontinuities?
     
  17. Mar 17, 2015 #16

    haruspex

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    No, I was saying you can choose restricted ranges for representation of the points, not for calculating where the points are.
    Consider an arithmetic spiral, ##r=a\theta##. If you restrict the range of theta then that will only give you one turn of the spiral. But any given point on the spiral can be represented by ##(r', \theta')##, where those variables have restricted ranges.
     
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