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I Friction in a hemispherical bowl

  1. Dec 26, 2016 #1
    I have been trying to find the speed of a point mass sliding with friction (f=μN) in a hemispherical bowl. Start at the top at rest. So far I have -μN+mgcosθ=ma and N=mgsinθ+mv2/R where θ is the angle below horizontal from the center of the sphere and R is the radius of the sphere. I know it likely does not have a simple solution but would be happy with an approximate solution for small μ. Capture.JPG
     
  2. jcsd
  3. Dec 26, 2016 #2

    mfb

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    For an exact solution you'll have to solve a differential equation.

    For an approximation for small μ, you can calculate the speed without friction, then determine friction and energy loss based on that speed, then calculate a new speed based on the initial value and the energy loss. That gives an easier integral instead of a differential equation.
     
  4. Dec 26, 2016 #3
    Right you are, and as I see it there are two coupled second-order, nonlinear equations. Good idea to use v(θ,μ)≈v((θ,0) to calculate energy lost. I'll try it.
     
  5. Dec 26, 2016 #4
    I think the suggestion by mfb was a good one. Using the frictionless v(θ)=√(2gRsinθ) I find v≈√(2gR(1-3μ)) at the bottom. For comparison, a path on a 450 incline with the same drop has v=√(2gR(1-μ)).
     
  6. Dec 27, 2016 #5

    mfb

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    Out of curiosity, I simulated the setup. With μ=0.001 and m=R=g=1 I got E=0.99701, in agreement with your result. μ=0.01 leads to E=0.9704. Even with μ=0.1, the approximation is not bad: E=0.732.

    The mass stops at the center for μ>0.60. Tested with 100, 200 and 500 steps: The critical value is somewhere between 0.603 and 0.605. There is no obvious mathematical constant in that range.
     
  7. Dec 27, 2016 #6
    This is beautiful. I will post your results on my web site. (I have had my wrist slapped here before by mentioning it explicitly, so will not give you the exact link!) I will link back here so that you will be properly acknowledged. Thanks.
     
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