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Math Help - conservation of energy/cylindrical polar coordinates

  1. #1
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    conservation of energy/cylindrical polar coordinates

    A particle moves on the smooth inside surface of the hemisphere
    z = −sqrt(a^2r^2)
    where (
    r, θ, 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 θ-direction.
    Show that the particle moves between two heights in the subsequent motion, and find
    them.
    Show, too, that if the parameter
    β = V^2/4ga is very large then the difference between

    the two heights is approximately
    a/2β.

    Ok

    I've spent days trying to do this and I cant seem to get it!

    I end up with z^2 = z^2.v^2 - 2.g.z.a^2 - 2gz^3

    where bold denotes 'z dot' (derivative of z wrt time)

    factorising this I get z((z - ((v^2 + sqrt(v^2 - 4ga))/4g)(z - ((v^2 - sqrt(v^2 - 4ga))/2z^2 + a^2

    getting the particle lying between((v^2 + sqrt(v^2 - 4ga)) and ((v^2 - sqrt(v^2 - 4ga))

    is this correct?

    how should I go about the next bit?

    many thanks


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  2. #2
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    Here's a hint in those questions you need to write down the Lagrangian, and then solve Euler-Lagrange equations.

    I leave you to do the calculations.

    Cheers,
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  3. #3
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    we havent been lectured that yet.. I think the way I have given is how we are expected to solve it, but I think I must've made a mistake somewhere? thanks though
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  4. #4
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    One force acting on the particle is gravity: -mg in the z direction, but there is also a "passive force" acting perpendicular to the hemisphere to keep it on the hemisphere. Find the components of -mg perpendicular to and parallel to the hemisphere and then discard the componet perpendicular.
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  5. #5
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    Yeh..

    I get 1/2(r^2 +r^2.theta^2 + z^2) - gz = 1/2.V^2

    eliminating r in favour of z and theta, I get the equation written in my first post..

    where bold again denotes differentiation
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  6. #6
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    Quote Originally Posted by James0502 View Post
    Yeh..

    I get 1/2(r^2 +r^2.theta^2 + z^2) - gz = 1/2.V^2

    eliminating r in favour of z and theta, I get the equation written in my first post..

    where bold again denotes differentiation
    I'm not sure how you got the -gz - i get a positive gz. I'm stuck on this question too actually.
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  7. #7
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    Quote Originally Posted by DeFacto View Post
    I'm not sure how you got the -gz - i get a positive gz. I'm stuck on this question too actually.
    yeh.. sorry - I got +ve gz also.. The question is really starting to get to me now! like over a week since I first attempted it!
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