Physics Friday 62

[Part 1 of ?]

Let us consider a rotating oblate spheroid with equatorial radius a and polar semi-axis c. We choose an inertial frame (x,y,z) and body coordinate system (x’,y’,z’), both with origin at the ellipsoid’s center of mass. Suppose we have a rotation about a principal axis with angular momentum , where Iz’ is the moment of inertia along the z’ principal axis and is the z’ unit vector).

Now, let us put a point mass M at a displacement d from the center of our ellipsoid. Then a mass element of our ellipsoid at position r ( is the density at that point in the object, not necessarily uniform). experiences a gravitational force due to the mass M of

Integrating the torque element due to this force over the volume V of the ellipsoid gives a total torque of

Now, suppose our point mass is far from our body, compared to its size, so that , where and . Now using
, we see that approximating to first order in r/d,
, so we can in this situation approximate our torque by:

Now, to find the first integral on the right, we note that
For a an object in volumeV of total mass m and density , the center of mass has position (vector)

(see here); as our origin is the center of mass, we thus see that the integral , and the first integral in our approximation is the zero vector, and

Consider the moment of inertia tensior I of our object (see here):
, where E3 is the 3×3 identity matrix and is the outer product of r with itself. Applying the tensor to d, we use and to get
Next, consider the cross product of d and this vector:
which is our remaining integral in the torque approximation:


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One Response to “Physics Friday 62”

  1. Physics Friday 63 « Twisted One 151’s Weblog Says:

    […] Friday 63 By twistedone151 [Part 2 of ?] In the previous part, we introduced a spinning oblate spheroid, and showed that a distant point mass M at displacement d […]

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