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Particle on a sphere

  1. Aug 5, 2005 #1
    Imagine we have a particle sitting on top of a sphere of radius R. The sphere is inertially fixed. A small disturbance force sets the particle in motion from its unstable equilibrium point atop the sphere. Theta is measured from the vertical to the position of the particle. Assume this angle is formed from the center of the circle. At what angle will the particle separate from the sphere? I really can't think of a geometrical condition that must be true for this problem. I've determined the velocity as a function of theta, assuming the only forces are the normal and weight (no friction - the sphere is smooth).

    Perhaps this occurs when the horizontal component of the velocity is greater than the rate of change of the sphere's curvature in the x direction?

    Any hints, ideas? Thanks
     
  2. jcsd
  3. Aug 5, 2005 #2

    Doc Al

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    Staff: Mentor

    Think in terms of forces, not geometry. Apply Newton's 2nd law to the particle as it slides down the sphere. Realize that for the particle to maintain contact with the sphere it must centripetally accelerate, and the only force available to provide such acceleration is the radial component of the particle's weight. At some point the particle will be going too fast to maintain contact. (Hint: Find the point where the normal force just equals zero.)
     
  4. Aug 6, 2005 #3
    Hmmm, I seem to be having some trouble finding a soltuion. If i define u_r to be radially outward, and u_theta, to be tangential to the circle I obtain the follwoing EOMS.

    N - mgcos(theta) = -m*omega^2*r (u_r equation)
    mgsin(theta) = m*omega_dot*r (u_theta equation)

    The condition that when the two separate is when N = 0, so I have the set of equations

    g/r*cos(theta)=(d_theta/dt)^2
    g/r*sin(theta) = d2_theta/dt^2

    I tried solving the bottom one in simulink. Its actually quite a simple model. I just let r = 9.8 m so that the ratio g/r = 1 s^-2. Then I plotted

    g/r*cos(theta)-(d_theta/dt)^2 as a function theta. (This is the equation for the normal force), however it never is zero.

    Are my EOM's incorrect? I believe I did the dynamics of this problem correctly...
     
    Last edited: Aug 6, 2005
  5. Aug 6, 2005 #4
    nevermind, thank you. I'm an idiot, had my parathesis around the omega term so it was like (-yout(:,2)).^2..... So as you can see the negative sign in that term will be real useful..... I get a reasonable answer of 48 degrees with the model and seeing with the assumption of having a radius of 9.8 m, I guess this seems reasonable. Thanks for the hint!

    BTW, is there an analytical solution to the set of equations, or is the only way to solve them using numerical methods such as all the built in junk in MATLAB?
     
  6. Aug 6, 2005 #5

    Doc Al

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    Staff: Mentor

    You don't need numerical methods--or special assumptions--to solve this problem!

    That's one equation that you need, but I'll rewrite it in terms of tangential speed instead of omega (taking N = 0 as the condition for breaking contact):
    [tex]mg \cos \theta = m v^2 / r [/tex]

    Combine this with an expression for speed (or kinetic energy) as a function of theta (use conservation of energy) and you'll get a simple analytical solution.
     
  7. Aug 6, 2005 #6
    wow, that was easy, I gotta stop starting this stuff at 1 AM when my mind decides it doesn't want to think anymore. Thanks!
     
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