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Speed of a point on a wheel as a function of θ

  1. Oct 29, 2014 #1
    1. The problem statement, all variables and given/known data
    A point P on a wheel of a bike radius R, P is a distance r from the axle.
    The speed of the bike is v, what is the speed of the point P as a function of θ?

    2. Relevant equations
    [tex]x=R\theta - r \sin\theta[/tex]
    [tex]y=R - \cos\theta[/tex]
    [tex]\rho=\frac{r}{R}[/tex]
    Distance travelled by the point in one revolution
    [tex]D =R{\displaystyle \int_{0}}^{2\pi}\sqrt{\left(1+\rho^{2}\right)-2\rho\cos\theta}\,\textrm{d}\theta[/tex]

    3. The attempt at a solution
    Speed is distance per unit time, but not sure where to begin as no function contains time.
    Can I make
    [tex]\theta=\omega t[/tex]? If so how can I solve the integral?
    James
     
  2. jcsd
  3. Oct 29, 2014 #2

    phinds

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    Seems like an oddly worded question, to me. The speed of the rim is going to be the same all the way around, not dependent on some particular point, as specified by an angle. Of course, you haven't actually said what θ IS so maybe I'm misinterpreting the problem.

    Oh ... wait ... I'm thinking of the speed relative to the AXLE. I guess the problem is to specify a point by an angle and then get the speed of that point relative to the ground, yes?
     
  4. Oct 29, 2014 #3
    θ is the angle of rotation of the wheel.
    I think you are right, as v is the speed of the bike itself.
    James
     
  5. Oct 29, 2014 #4

    phinds

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    Well, in any case, you're going to need a defined reference position for θ. Maybe you're supposed to take the axle as the XY origin and the X axis as zero degrees?
     
  6. Oct 29, 2014 #5

    vela

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    The y-equation can't be right. Looks like you forgot ##r##.

    Instead of ##R\theta##, try just writing ##vt##.
     
  7. Oct 29, 2014 #6

    LCKurtz

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    Where did you get these equations? Did you derive them or copy them from a book? Either way, what is your definition of ##\theta##? Also, as vela pointed out, the ##y## equation can't be correct.

    What does that equation have to do with it?

    The problem asks for the rate of change with respect to ##\theta##. Once you have the position vector ##\vec P(\theta)## you want ##v(\theta) = |\vec P~'(\theta)|##.
     
  8. Oct 29, 2014 #7
    Hi, I missed the r from the expression for y

    [tex]y=R-r\cos\theta[/tex]

    The expressions came from the question sheet.
    Thanks Vela,LCKurtz.
     
  9. Oct 29, 2014 #8

    LCKurtz

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    There are a couple of other things you might like to observe about this problem. Take the case when ##R=r##. Look at the speed of the point on the edge of the wheel when it touches the ground vs. when it is at the top of the wheel(whatever values of ##\theta## those are in your setup. You will see the speed isn't constant like it would be for just a spinning wheel. Yet the acceleration is of constant magnitude and directed towards the center in both cases.
     
  10. Oct 30, 2014 #9
    What I get is

    [tex]v(\theta) = |\vec P~'(\theta)|=\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}=\sqrt{(R-r\cos)^2+r^2\sin^2\theta}[/tex]


    [tex]v(\theta) = R\sqrt{(1+\rho^2)-2\rho\cos\theta}[/tex]
     
  11. Oct 30, 2014 #10

    vela

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    Your expression for ##v(\theta)## has units of length. The speed of the particle is actually given by
    $$v_P(\theta) = \left| \frac{d\vec{P}}{dt} \right| = \left| \frac{d\vec{P}}{d\theta}\frac{d\theta}{dt} \right|.$$ The factor of ##d\theta/dt## is where the information about the speed ##v## of the bicycle enters.
     
  12. Oct 30, 2014 #11
    Yes I see thanks, but how do I get [tex]d\theta/dt[/tex], or can't I with the given information?
     
  13. Oct 30, 2014 #12
    I think I have it
    [tex]d\theta/dt = \omega[/tex]

    Which has units s-1.
     
  14. Oct 30, 2014 #13

    vela

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    Right. Now you just need to relate ##\omega## to ##v##.
     
  15. Oct 30, 2014 #14
    Ok I think this is it

    [tex]
    v(\theta) = |\vec P~'(\theta)|=\sqrt{\omega\left(\frac{dx}{d\theta}\right)^2+\omega\left(\frac{dy}{d\theta}\right)^2}=\sqrt{\omega (R-r\cos)^2+\omega r^2\sin^2\theta}[/tex]

    [tex]v(\theta) = \omega R\sqrt{(1+\rho^2)-2\rho\cos\theta}[/tex]

    which has the units of speed.
     
    Last edited: Oct 30, 2014
  16. Oct 31, 2014 #15
    Cannot seem to be able to edit the above post

    [tex]
    v(\theta) = |\vec P~'(\theta)|=\sqrt{\left(\frac{dx}{d\theta}\omega\right)^2+\left(\frac{dy}{d\theta}\omega\right)^2}=\sqrt{\omega^2 (R-r\cos)^2+\omega^2 r^2\sin^2\theta}
    [/tex]

    James
     
  17. Oct 31, 2014 #16

    vela

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    So if the speed of the bike is ##v##…
     
  18. Oct 31, 2014 #17
    Then the speed of the point would be

    [tex]v+v(\theta)[/tex]
     
  19. Oct 31, 2014 #18

    vela

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    That's not right. In any case, you seemed to have missed my point. If I gave you numbers for ##R##, ##r##, and ##v##, and ##\theta##, could you give me a numerical answer for ##v(\theta)##? Until you get to that point, you're not done. In other words, your answer should only depend on those variables.
     
  20. Oct 31, 2014 #19
    Yes, for [itex]r=0[/itex] the speed is [itex]\omega R[/itex] and independant of [itex]\theta[/itex] and for [itex]r=R[/itex], ([itex]\theta=0[/itex] is the point in contact with the ground) the speed is 0 for [itex]\theta=0[/itex] and [itex]v=2\omega R[/itex] for [itex]\theta=\pi[/itex].
     
  21. Oct 31, 2014 #20

    vela

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    Ok, so what's the speed of P if θ=π/4, R=0.5 m, r=0.2 m, and v=3 m/s?
     
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