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Gravitational Potential Due to a Thin Rod of Varying Density. |
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| Feb27-12, 07:04 PM | #1 |
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Gravitational Potential Due to a Thin Rod of Varying Density.
1. The problem statement, all variables and given/known data
GRAVITATIONAL POTENTIAL AND FIELD DUE TO A “THIN” ROD A thin rod of length L lies along the +y-axis, with one end at the origin (see diagram). Assume: • The rod has length only- no thickness in other directions. • The density of the rod increases proportionally to the y-coordinate: λ = ky, where k is a known constant and λ is in kg/m • Gravitational potential is zero at infinity: φ (∞) = 0 a) Find the gravitational potential φ ( x) at a point (x,0) by direct integration. b) Find the gravitational field g at a point (x,0) by direct integration. 2. Relevant equations dφ = -(G dm)/r 3. The attempt at a solution Still stuck on part a, so that's really the brunt of my question for now (though assistance with part b is more than welcome!). Using the given density function to solve for dm and substituting √(x^2+y^2) for r, I have an expression for dφ: dφ = -(Gk y dy)/√(x^2+y^2) ...but I have no idea how to manipulate this to get a soluble integral :( I've been messing around with partial derivatives and polar coordinates for hours, but nothing seems to work. PLEASE HELP! |
| Feb27-12, 07:55 PM | #2 |
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Ah, I think I may have figured it out... x isn't actually changing here, so my attempts to relate it to y were completely unnecessary? The answer then (I think) would be:
φ(x) = --Gk[√(x^2+L^2) - x] . Part b has me a bit stumped though.... help would be lovely. |
| Feb27-12, 08:44 PM | #3 |
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Alrighty, figured that bit out on my own as well... polar coordinates work out nicely.
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