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Homework Help: Dynamics - particle moves on smooth inside surface of hemisphere?

  1. Jan 10, 2010 #1
    1. The problem statement, all variables and given/known data
    I'm currently working on this question:

    A particle moves on the smooth inside surface of the hemisphere
    z = -(a^2 - r^2)^(1/2), r <= a,
    where (r, theta, z) denote cylindrical polar coordinates, with the z-axis vertically upward.

    Initially the particle is at z = 0, and it is projected with speed V in the theta-direction.

    1. Show that the particle moves between two heights in the subsequent motion, and find them.

    2. Show, too, that if the parameter b = (V^2)/4ga is very large then the difference between the two heights is approximately a/(2b).

    2. Relevant equations

    Conservation of energy 1/2 m(r'^2 + r^2 theta'^2 + z'^2) + mgz = constant
    rtheta' = V
    r^2 theta = aV
    theta ' = aV/r^2
    ( ' = dot, differentiation)


    3. The attempt at a solution

    I'm really stuck on question 1, I got to an answer using conservation of energy that ended with solving a quadratic, this is the particle moves between z=0 and z= (V^2 + sqrt(V^4 + 16g^2a^2))/4g but when I sub in the parameter in (2) I get to the wrong answer, ie 2ab. So I guess I've done 1 wrong. Any ideas guys?
     
  2. jcsd
  3. Jan 10, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi Kate2010! Welcome to PF! :smile:

    (have a theta: θ and a square-root: √ and try using the X2 tag just above the Reply box :wink:)
    How did you get 2ab from that equation (I get a/2b)?

    (Did you use √(1 + X) ~ 1 + X/2 ?)
     
  4. Jan 10, 2010 #3
    I can get to a(b + √(b² +1))

    Then I said √(b² +1) is roughly equal to b, I think this is where I'm going wrong?
     
  5. Jan 10, 2010 #4

    tiny-tim

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    oops! :redface: i've just noticed that when i got a/2b, i misread your equation as z= (V^2 - sqrt(V^4 + 16g^2a^2))/4g (with a minus instead of a plus) …

    are you sure it isn't that (because the z should be negative, shouldn't it)?
     
  6. Jan 10, 2010 #5
    Yes you're right, z should be negative.

    So I get to a(b - √(b² +1))

    I still don't understand how this is approximately equal to a/2b.
     
  7. Jan 10, 2010 #6

    tiny-tim

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    he he :biggrin: … i'm accidentally right again! :redface:
    b is large, so you need to divide by b to get (1 + something-small) inside the √ …

    a(b - √(b² +1)) = ab(1 - √(1 +1/b²)).

    Now just use √(1 + X) ~ 1 + X/2. :wink:
     
  8. Jan 10, 2010 #7
    Got it! Thanks a lot :biggrin:
     
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