Physics Friday 41

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 m1 and m2 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:


Tags: , , , , , ,

One Response to “Physics Friday 41”

  1. Physics Friday 42 « Twisted One 151’s Weblog Says:

    […] 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 […]

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: