- #1
kaze
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A infinitely long rod, A, with linear mass density, say p, is placed along the z-axis. Another thin rod, B, of length L and mass density, say p_{2}, is placed orthogonally to A on the x-axis from x=a to x=a+L. What is the gravitational force experienced by rod B from rod A?
I have solved a similar problem which asks for the gravitational potential given a rod of finite length and a point a perpendicular distance away from the rod's midpoint (thornton and marion problem 5-7). I am trying to extend this but not having much luck. So if we were trying to find the gravitational potential of the end point on rod B closest to A we would get:
\phi =-G \int_{A} \frac{p}{r}ds =- p G \int_{A} \frac{1}{\sqrt{x^2+a^2}} dx = - p G \lim_{z\rightarrow \infty} \int_{-z}^{z} \frac{1}{\sqrt{x^2+a^2}} dx = - p G \lim_{z \rightarrow \infty} \ln(x+\sqrt{x^2+a^2})|_{-z}^{z}
This already goes to infinity. I was thinking that I could make \phi a function of a and then once I find a function for the gravitational potential, I'll integrate over rod B to get the entire potential and hence the force by F=-\nabla U = -\nabla (m U ) = -m \nabla U
However, I guess I am stuck at step one since I cannot get the integral to work out. If you have any suggestions, I'd really appreciate the help. Thanks.
I have solved a similar problem which asks for the gravitational potential given a rod of finite length and a point a perpendicular distance away from the rod's midpoint (thornton and marion problem 5-7). I am trying to extend this but not having much luck. So if we were trying to find the gravitational potential of the end point on rod B closest to A we would get:
\phi =-G \int_{A} \frac{p}{r}ds =- p G \int_{A} \frac{1}{\sqrt{x^2+a^2}} dx = - p G \lim_{z\rightarrow \infty} \int_{-z}^{z} \frac{1}{\sqrt{x^2+a^2}} dx = - p G \lim_{z \rightarrow \infty} \ln(x+\sqrt{x^2+a^2})|_{-z}^{z}
This already goes to infinity. I was thinking that I could make \phi a function of a and then once I find a function for the gravitational potential, I'll integrate over rod B to get the entire potential and hence the force by F=-\nabla U = -\nabla (m U ) = -m \nabla U
However, I guess I am stuck at step one since I cannot get the integral to work out. If you have any suggestions, I'd really appreciate the help. Thanks.
