A particle of mass m is constrained to move on the surface of a torus (solid thi

Question

A particle of mass m is constrained to move on the surface of a torus (solid thick ring), shown below, but otherwise moves freely. (There is no gravity in this problem. The torus is floating somewhere in intergalactic space.) Points on the surface can be represented by a pair of angles (psi, theta) such that x = (a + b cos theta) cos psi y = (a + b cos theta) sin psi z = b sin theta where a > b. The particle is initially traveling around the outermost equatorial circle (theta = 0) with velocity v. But then it is given a very, very tiny impulse in the theta-direction, so that it starts to oscillate while it continues to transverse the equatorial circle (as shown by the wiggling line on the torus below). Find the frequency of these oscillations in theta, treating theta as small.

Explanation / Answer

Circular orbits are possible in both horizontal and vertical planes. We use a cylindrical coordinate system (r, , z), with the origin at the center of the torus and the z axis vertically upwards, as shown below. A point on the surface of the torus can also be described by two angular coordinates, one of which is the azimuth in the cylindrical coordinate system. The other angle we define as measured with respect to the plane z = 0 in a vertical plane that contains the point as well as the axis.

We seek motion at constant angular velocity about the z axis, which suggests that we consider a frame that rotates with this angular velocity. In this frame, the particle (whose mass we take to be unity) is at rest at angle 0, and is subject to the downward force of gravity g and the outward centrifugal force 2 (a + b cos 0).

The resultant force, which we call geff, must be perpendicular to the surface of the torus. Hence, the angle 0 of the steady circular orbits must obey

tan 0 = g /2 (a + b cos 0) (1)

Since the right hand side of eq. (1) is positive, we see that there are two solutions, one at angle 1 in the first quadrant, and another at angle 2 in the third quadrant

It seems “obvious” that only the motion at angle 1 in the first quadrant is stable, in the sense of supporting small oscillations about the steady motion when perturbed slightly. It is tempting to analyze these oscillations in the rotating frame as simple pendulum motion subject to an effective gravity geff. This would imply that the frequency of the small oscillations is

= sqroot (geff /b) = squareroot(squareroot of g 2 + 4 (a + b cos 0) 2/ b) .

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