Suppose we have a thin ring of radius *a* and mass *M*. The ring is rotationally symmetric about the line through the center of the ring and perpendicular to the plane of the ring. Let us place a mass *m* on this axis at a distance *z* from the center of the ring. What then, is the gravitational potential energy between the two (that is, the potential energy not including the “internal” gavitational potential energy of the ring itself) as a function of *z*?

The gravitational potential energy for two point masses *m*_{1} and *m*_{2} separated by a distance *r* is . If we consider a small element *dM* of the ring, the distance between that element and our mass *m* is . Thus the potential energy due to this part of the ring is . As this is the same for all points on the ring, adding up all the elements simply gives:

.

As expected, the center of the ring gives the minimum energy; a cursory examination of the gravitational force will show that the mass *m* will be pulled toward the ring. As required by the rotational symmetry, the force **F** on the mass *m* on the axis is entirely along the axis.

Thus, we can compute this force from the above energy; with a positive sign indicating the direction of increasing *z*, the force is:

.

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Tags: Friday Physics, Gravity, physics, Potential Energy, Ring, symmetry, Symmetry Axis

This entry was posted on October 10, 2008 at 7:50 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
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October 17, 2008 at 12:01 am |

[…] Friday 42 Last week, we considered a thin ring of radius a and mass M, and the energy of a mass m on the perpendicular […]