Locating the instantaneous centre

Instantaneous centre (IC) of a body in plane motion is the point (on the body or its extension) whose velocity is zero at that instant. At that instant the body appears to rotate about it with angular velocity $\omega$, so the velocity of any point P is $v_P = \omega\,(IP)$, perpendicular to $IP$.

Methods of locating the IC

1. Directions of velocities of two points known (not parallel). Draw perpendiculars to $v_A$ at A and to $v_B$ at B. Their intersection is the IC (example: ends of a sliding ladder, or crank pin and slider of an engine).

      A ->v_A     perpendicular lines meet at I
      |
      |       B
      |      /|
      +-----I-+ 

2. Velocities parallel at A and B and both perpendicular to AB. Join the tips of the velocity vectors with a straight line; its intersection with the line AB gives the IC. If the vectors point the same way the IC lies outside AB (on the side of the smaller velocity); if they are opposite it lies between A and B. Because $v_A/IA = v_B/IB$, this follows from similar triangles.

3. Velocities parallel but not perpendicular to AB, equal in magnitude. The IC is at infinity: the body is in instantaneous translation ($\omega = 0$).

4. Rolling without slipping. The point of contact C with the fixed surface has zero velocity, so the IC is at the contact point. For a wheel of radius $r$: $\omega = v_O/r$, and the top point moves at $2v_O$.

Is the acceleration of the IC zero?

No. The IC has zero velocity but in general it has acceleration, so it must not be used to find accelerations as if it were a fixed point. For a rolling wheel, for example, the contact point has an acceleration $\omega^2 r$ directed toward the wheel centre.