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gorica Group Title

A smooth surface of revolution is hyperbolic with equation z=a^2/r, the axis Oz pointing vertically downwards and r, θ and z being cylindrical polar coordinates. A small particle mass m slides on the interior of the surface. Calculate potential energy.

  • one year ago
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  1. theEric Group Title
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    I'm not sure if I can help, but here are my thoughts that you and others can dispute. Is this potential energy just the potential energy due to gravity? It looks like mass is the only data on this particle. Then \(E_{potential}=m\ g\ z\) if you consider the particle to have \(0\ [J]\) at \(z=0\). And \(a^2\) and \(r\) are unknown? Then the potential energy could not be known, but could be put in terms of \(a\) and \(r\).

    • one year ago
  2. theEric Group Title
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    As for the surface, I guess \(a^2\) is constant and so we'd have a surface like|dw:1374850120317:dw|It's inside that, on the interior, I guess. That is because it is a hyperbola whose \(z\)-value does not depend on \(\theta\), but only \(r\). So \(\large z=\frac{a^2}{r}\) at every value for \(\theta\) in those cylindrical coordinates.

    • one year ago
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