As your intuition might suggest, a rigid body is a system of particles in which the relative positions of the component particles remains constant even under the application of torques and forces (of not too great an entity). Does my intuition really tell me this?, you may wonder. Well, maybe not, but the word rigid surely suggests the idea of a body which is not deformed. For example a rubber band is not a rigid body: it can be easily deformedby pulling it. A steel block, on the other hand, is a rigid body...AND the gyroscope (rmember it's a flywheel) is also a rigid body!
3.1 Spinning rigid bodies and symmetry exploitation
When a rigid body is spun about an axis passing through it, becuase the particles are "locked" in their relative positions, all the particles along the axis will rotate about themselves, and any of the others at an arbitrart distance R from it will describe circular paths about the axis. The velocities vi of the particles at different Rs will, of course, differ from each other: but the angular velocity w will be the same. So each particle will possess an angular momentum (wrt a point O on the axis of rotation):
Li= ri x mi vi
To obtain the total angular momenturm L we need to sum from i=1 to i=N, where N is the total number of particles making up the rigid body. The sum is complicated (almost impossible)...would resolving into the cartesian components of L help? Well yes, but things are even simpler if we suppose, furthermore, that the rigid body is spinning about an axis of symmetry. We then note, in fact, that for every particle on one side of the axis with L1, there's one on the other side with L2, so that the horizontal components of L1 and L2 are equal and opposite. Thus when we sum the total AM is then along the axis of rotation because the horizontal contributions cancel in pairs (Fig.4). So is this axial component really that simple?
Fig.4: The symmetry of the flywheel entails L points along the rotation axis.
Let's see...(we will work with magnitudes: it's easier, and we already know L points along the axis): for a particle the axial component is
Li(axis)= mi vi ri cos(h) (ri and vi are perpendicular), where h is the angle between Li and the axis. But vi =wRi and, also, ri cos(h) = ri cos(90-h)=ri sin(p) = Ri where p is the angle between ri and the axis. So that in the end: Li(axis) = miwRi Ri = w miRi2
and summing over N particles :
L(axis) = Si(w miRi2)a = Iw a = Iw
where I=Si(miRi2) is what we call moment of inertia of a rigid body, a is a unit vector pointing along the axis, and w = wa is the usefully defined angular velocity vector (obviously pointing along the axis as well!).
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