- #1

rc75

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The problem is about a particle constrained to move on a smooth spherical surface with radius R. The particle starts at the equator of the sphere with an angular velocity of [tex] \omega [/tex], and the particle is fast in the sense that [tex]\omega ^2 R >> g[/tex].

The problem is to show that the depth z below the level of the equator is

[tex]z \approx \frac{2g}{\omega ^2} \sin ^2 \frac{\omega t}{2}[/tex].

I found that the energy of the particle can be used to write:

[tex] \frac{1}{2} [ \frac{(1+z^2)\dot{z}^2}{R^2-z^2} + \frac{\omega ^2 R^4}{R^2-z^2}]-gz= \frac{R^2 \omega ^2}{2}[/tex]

using cylindrical coordinates.

But the solutions claim instead that

[tex]\frac{1}{2} [ \frac{R^2 \dot{z}^2}{R^2-z^2} + \frac{\omega ^2 R^4}{R^2-z^2}]-gz= \frac{R^2 \omega ^2}{2}[/tex].

I don't see why I'm getting a different answer, or how these could be the same.

Also the solutions claim that the condition [tex]\omega ^2 R >> g[/tex] leads to [tex]z<<R[/tex]. I can see intuitively why that would be true, but I'm not sure how to show it formally.

Any help would be greatly appreciated.

Thanks.