I was just reading Cengage's Rigid Body Dynamics, and the whole concept of Instantaneous Axis/Centre of Rotation whizzed past me. Can anyone help me understand ICOR's following properties?

  • Why is IAOR perpendicular to the plane of motion?

  • Why does ICOR has zero velocity?

  • How to find ICOR in various cases? Especially I observe, in a lot of cases, the ICOR is point of contact itself (like in a rolling ball) which is the most confusing to me.

Please help me with these basic understanding flaws. I will be thankful for the same


1 Answer 1


I think the orientation of the IAOR is mainly due to mathematical reasons. With the rotation equation: $\vec{v_A}=\vec{\omega}\times\vec{r_{CA}}$ I.e. for a pure rotation about the z-axis, there must not be a velocity along the z-axis, and the z-component of $\vec{v_A}$ is only zero, if $\vec{\omega}$ is along the z-axis.

The ICOR has by definition zero velocity at this moment, with respect to a certain frame of reference. Take for axample a rolling cylinder (wheel). If you take the non-moving floor as reference frame, the point of the wheel touching the floor must be the ICOR, because if that point was moving, slipping would occur. But if you take the moving wheel itself as the reference frame, the center of the wheel becomes the reference frame, as from it's perspective the wheel is only rotating and the ground is moving with the wheel, so slipping doesn't occur.

Consider this graphic: The red x marks the respective frame of reference.

In case 2 of the image the frame of reference is at the center of the wheel. From it's perspective it is simply rotating around the center, thus the given velocity distribution (in blue).

But from the frame of reference of a person standing on the floor, e.g. the point of contact (case 1 and 2) the whole wheel is also moving to the right, consisting of an overall translational velocity to the right (case 3). So the velocity distributions of case 2 and 3 (literally) add up to case 1. Now the ICOR is simply the point where there is no velocity at this instant, which turns out to be the point of contact. This is true for the point on the wheel where it touches the ground, as well as for the point on the ground.

  • $\begingroup$ Ok I got my first two doubts cleared. However I would like to discuss a bit more on the rolling cylinder's ICOR being the point of contact. $\endgroup$
    – user388824
    Nov 19, 2016 at 15:28
  • $\begingroup$ It is true that if the ground would move, slipping would occur, but even then the point of contact cant be ICOR just because it is at rest. And, BTW when we say "point of contact", exactly which point do we consider? The point on ground which touches the cyl or the point on the cyl which touches the ground? $\endgroup$
    – user388824
    Nov 19, 2016 at 15:33
  • $\begingroup$ The full velocity equation for rigid body kinematics is $v_b = v_a + \omega$ X $r_{ba}$. So for slipping the velocity of the contact point is nonzero. The ICOR is found by simply setting one of the velocities to zero (definition of ICOR) and knowing the angular velocity you can solve for the location of the ICOR. In the case of slipping it is not necessarily (and almost definitely) not the point of contact with the ground. $\endgroup$
    – nrabbit
    Nov 19, 2016 at 23:00

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