Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Rotating systems

  1. Nov 25, 2004 #1
    A motorcycle is driving along the walls of a round ring, which is rotating at [tex]\omega_0[/tex]. Its speed is constant V, in the same direction of the ring itself. Its mass is m, and the radius of the ring is R. I need to find the normal force that the walls exert on the motorcycle.

    If we look at the problem from the motorcycle's reference frame, then it is stationary in a frame that is rotating at [tex]\omega[/tex] = ([tex]\omega_0[/tex] + V/R), so the normal force is equal to the centrifugal force:

    [tex]N = m\omega^2R = m(\omega_0 + \frac{v}{R})^2R[/tex]

    Now, if we look at this from the ring's reference frame, then the motorcycle is moving at a constant speed V inside the ring which is rotating at [tex]\omega_0[/tex]. Then we also need to take into account coriolis effect (right?), and the forces in the radial axis are:

    [tex]N + 2m\omega_0v - m\omega_0^2R = 0[/tex]

    But the normal force is the same, no matter which frame we use, so:

    [tex]m\omega_0^2R - 2m\omega_0v = m(\omega_0 + \frac{v}{R})^2R[/tex]
    [tex]\omega_0^2 - 2\omega_0\frac{v}{R} = (\omega_0 + \frac{v}{R})^2[/tex]

    And that's obviously incorrect... so can someone please point out my mistakes? :smile:

  2. jcsd
  3. Nov 25, 2004 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Why? There is no radial component of velocity.
  4. Nov 25, 2004 #3
    Ahh... I see... so the normal is just:
    [tex]N = m(\omega_0 + \frac{v}{R})^2R[/tex]
  5. Nov 25, 2004 #4

    Doc Al

    User Avatar

    Staff: Mentor

    That's what I would say. :wink:
  6. Nov 25, 2004 #5
    Whoop-dee-do, it's right. Thanks Doc. :smile: Why on earth wouldI try to overcomplicate this? :confused:
  7. Nov 26, 2004 #6
    The solution was correct, but I'm not sure about one thing. We found that:

    [tex]N = m(\omega_0 + \frac{v}{R})^2R[/tex]

    Now, if we use the ring as the reference frame, then in the radial direction we have the normal force, the centrifugal force caused by the rotation of the ring, and the centrifugal force caused by the motion of the motorcycle itself. So we get:

    [tex]N = m\omega _0^2R + m\frac{v^2}{R} = m\omega _0^2R + m{\frac{v}{R}}^2R = m(\omega _0 + \frac{v}{R})^2R - 2m\omega _0v[/tex]

    And that's not the same as the N we found above (which is correct). And it just so happens, that the extra expression looks just like the coriolis force... so can you please enlighten me? :smile:

  8. Nov 26, 2004 #7

    Doc Al

    User Avatar

    Staff: Mentor

    Sorry Chen, but my earlier statement was nonsense. :blushing: I was thinking in terms of an inertial frame (the easy way to solve this problem) compared to the bike's frame (in which its speed is zero). From the ring frame, there is a coriolis force, of course.

    Let's do it from all three frames:

    Inertial frame:
    [tex]ma = N[/tex]
    [tex]m(\omega_0 + v/R)^2 R = N[/tex]

    Ring frame (involves both centrifugal and coriolis terms):
    [tex]ma = N -m\omega_0^2R -2m\omega_0 v[/tex]
    [tex]m(v/R)^2 R = N -m\omega_0^2R -2m\omega_0 v[/tex]
    [tex]N = mR[(v/R)^2 +2m\omega_0 (v/R) + \omega_0^2][/tex]
    [tex]N = m(\omega_0 + v/R)^2 R[/tex]

    Bike frame (no acceleration; no coriolis term; centrifugal term only)
    [tex]ma = N -m(\omega_0 + v/R)^2R[/tex]
    [tex]0 = N -m(\omega_0 + v/R)^2R[/tex]
    [tex]N = m(\omega_0 + v/R)^2 R[/tex]

    Your original error was not including the acceleration as measured in the ring frame.

    (Sorry if I added to the confusion!)
    Last edited: Nov 26, 2004
  9. Nov 26, 2004 #8
    Now it all makes sense. Thanks. :smile:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook