I know the meaning of (instantaneous) “center of rotation” and “center of zero acceleration” for a rigid body. I know that they do not coincide (in general) and that their positions change with the motion of the rigid body.

Then I know that for each point of the body we can define an instantaneous “center of curvature” related to the trajectory. My question is, how are these centers of curvature related to the “center of rotation” and “center of zero acceleration” of the whole rigid body? Is there any relation among them, in the most general case?

  • $\begingroup$ I know for a fact that the ICR and ICC are the same. I can prove it by, but since I don't know how does the two relate to ICA then i'll skip the answer. $\endgroup$ Sep 16 '19 at 20:27
  • $\begingroup$ @SamFarjamirad The ICR is defined by the motion of any two points on the rigid body. The ICC is defined by the motion of a point through space. look at the diagram here. The ICR of A is P. The ICC of A is P1. They are entirely independent in general. $\endgroup$
    – Phil Sweet
    Sep 17 '19 at 2:18
  • $\begingroup$ @PhilSweet Suppose you know the velocity of the points A and B, then by drawing a perpendicular to that velocities we find ICR, we know that the velocity vectors are tangent to the instantaneous radius of curvature, so their perpendicular again intersect on the same point. That's my reasoning. Glad to hear where do I make mistakes, and since this hard to decipher tone online, I'm not condescending here at all :) $\endgroup$ Sep 18 '19 at 21:16
  • $\begingroup$ @SamFarjamirad The ICR is defined by the velocity vector of two points at one instant. The ICC is defined by the velocity vector trace of one point through time. So when that one point is undergoing acceleration, it is the acceleration that determines the ICC. But the acceleration has no bearing on the ICR. The acceleration can be any vector at all and produce an ICC of any magnitude at all without changing the ICR. The ICC vector will always be co-linear with the ray to the ICR. $\endgroup$
    – Phil Sweet
    Sep 18 '19 at 22:08

Thought experiment -

Given a free body $B$ that is being acted on by forces, find its instantaneous center of rotation $P1$ and its instantaneous center of zero acceleration $P2$.

Does this let us solve for the center of curvature for any point $P$ on the body?

To do so, we need the acceleration vector at $P$ and the velocity vector at $P$. That lets us define the osculating circle to the path of $P$ at that instant.

Given the coordinates of $P$, we know the direction of the velocity vector is perpendicular to the segment $P$ to $P1$, but what is it's magnitude? We need the magnitude of the velocity at some other point (not $P$) to calculate it. The only other point is $P2$. But we know only it's velocity direction also, not it's magnitude. Similarly, we know the direction of the acceleration vector is perpendicular to the segment $P$ to $P2$, but not it's magnitude, and we don't know the acceleration magnitude of $P1$ either, so we can't derive it.

So knowing the ICR and center of zero acceleration does not provide sufficient constraints to determine the ICC. We know the vector towards the ICC will be co-linear with segment $\overline{PP1}$. But we don't know the distance.


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