Kinematics of a Body: The Instantaneous Axis
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A rigid body carries exactly one restriction: the distances between its points never change. We differentiate that restriction to obtain the velocity relation v_B = v_A + omega cross r, then watch the body move under its translation and rotation terms. A rolling wheel makes the instantaneous centre visible and shows why a different material point occupies it from one instant to the next. Finally, one continuous three-dimensional chapter turns the point at rest into a tilted instantaneous axis, sweeps the family of points along it, and follows a material point through Chasles' screw motion. Live limiting cases reduce the screw to pure rotation and pure translation.
Kinematics of a rigid body begins with one promise: distances inside the body never change. That promise will give us its complete velocity law. Here is the body itself, a collection of material points locked together. Pick any two of them and call them A and B. The vector r runs from A to B, and rigidity says its length never changes. Write that fixed length as r dotted with itself, equal to the constant L squared. Differentiate. The constant disappears, leaving r dotted with its own rate of change equal to zero. So that rate points square to r. Here is the whole family. Change the length and even reverse the arrow; it stays perpendicular. Every member can be written as omega crossed with r. For a rigid body, one angular velocity omega works for every pair of points at once. It belongs to the body, not to A or B. Now compose the positions. Start at the position of A, add r, and land at the position of B. Differentiate that addition. The first velocity is v A. The second term is the rate of r, which becomes omega cross r. Their sum is v B. A translates with v A while the body turns about A with angular velocity omega. B carries both effects, so its green arrow follows the sum we just built. Now let the body move through the translation and rotation we just composed. Translation plus rotation is the entire instantaneous freedom of a rigid body. Every construction that follows is this one relation read in a different way.
In the plane, the rigid-body relation has a consequence you can watch. Here is a wheel rolling along the ground without slipping. The material point touching the ground is not sliding. At this instant C has velocity zero, so it is momentarily at rest. That is the instantaneous centre. Now take A in the relation to be C. Its velocity term is zero. What remains says that every point P moves as omega crossed with the vector from C to P. The hub is one radius from C, so it moves at omega R. The top material point is two radii away, so its speed is twice the hub speed. At Q, the same rule gives a velocity perpendicular to C Q. Its length grows in direct proportion to Q's distance from C. At one instant, then, the whole velocity field looks exactly like a wheel pinned at C and rotating about it. But C is not one fixed material point. Watch C zero leave the ground as the wheel rolls. A moment later, a different material point is touching down, and that new point is the one at rest. The same idea gives a ruler construction. Here is a moving bar with the velocity direction already attached at A and B. Each point circles C, so draw a perpendicular to each velocity. The centre must lie on both lines. The two lines meet here, at C. From that one point, every speed is omega times the distance out to the material point. There is one limiting case. If the two velocities are equal and parallel, their perpendiculars are parallel too. They never meet, so C is at infinity and the body is purely translating.
Lift the body into three dimensions. This solid extends through all three coordinate directions, with its material points locked into one shape. Which of those points, if any, are momentarily at rest? Ask the rigid-body relation. Choose A on the body, draw its velocity, and keep the body's one angular velocity omega in view. We are hunting for a point P whose velocity is zero. The cross-product term must cancel v A exactly. It can do that only with the part square to omega, because every omega cross r is perpendicular to omega. Here the magenta arrow is that exact cancellation. For the moment v A is entirely square to omega. Solving the cancellation gives this particular displacement r zero. But r zero is only one answer. Add any multiple lambda omega and the cross product does not change. Watch lambda sweep through its values: every point it reaches is another solution, so the solutions fill a line. That tilted line is the instantaneous axis of rotation. Every green location on it has zero velocity at this instant. A point off the axis swings around it in a circle lying square to omega. Its speed is omega times its perpendicular distance from the axis. The same construction at B uses the same omega. Turn the view and the geometry separates cleanly: the red axis is one tilted line in space, and both green velocities stand square to it. Omega belongs to the whole body. A and B do not get different angular velocities. What changes from point to point is r, and therefore the cross-product contribution. Now remove the assumption we just used. If v A is not square to omega, split it into a perpendicular piece and a parallel piece. Let those two pieces land. Complete their parallelogram, and its green diagonal is the original velocity v A. The cross product can cancel the perpendicular piece, exactly as before. It can never touch the parallel piece, whose projection formula is this. So the axis still exists, but its points now slide along omega instead of resting. This yellow arrow is the velocity shared by every point on it. Take a point away from the axis. Its velocity is the sum of a turn around the axis and that same slide along it. Follow one material point. The combined motion traces this yellow helix, winding around the tilted axis while advancing along it. This is Chasles' theorem. At every instant, rigid-body motion is a screw: a rotation about an axis together with a translation along that axis. Watch the numbered point perform both parts. The slide per unit turn is the pitch. First limiting case: let the slide shrink to nothing. The advancing helix closes into a circle, the axis velocity vanishes, and the motion becomes pure rotation. Second limiting case: restore the slide and let omega shrink to nothing. The winding opens into a straight path, the red omega arrow disappears, and equal green velocities show pure translation. Put turn and slide back together. The tilted axis, the helix and the travelling point return as the one picture that contains the general case. Three ideas carry the lecture. Keep the moving screw beside them while we read the list. First, any two points share one omega, and their velocities differ by omega cross the vector between them. Second, plane motion has an instantaneous centre, but the material point occupying it changes as the body moves. Third, spatial motion has an instantaneous axis. Rotation about it plus translation along it is the general screw motion, every instant.
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