Instantaneous Center of Zero Velocity

v_{A} = ω \cdot d_{I A}, v_{B} = ω \times r_{B / I}

At any instant a planar rigid body with rotation has a unique point I of zero velocity. Then every point moves as if the body were rotating about I: v=ω d. Locate I by drawing perpendiculars to velocity vectors.

$I$: Instantaneous center
$d_{I} A$: Distance from I to A

Step-by-step solved example

Bar 2 m, end A has v_A=4 m/s perpendicular to the bar, B is on a horizontal slot (v_B horizontal). ω?

IC is the point of zero velocity; v = ω d from I.

1. Locate I
Perp to v_A along the bar; perp to v_B vertical through B. I at the intersection.
2. Simple case
If B is instantaneously at rest it is the IC, so ω = v_A / L = 4/2 = 2 rad/s.
$ω = 2 rad/s$

Answer: ω = v/d to the IC (e.g. 2 rad/s if B is instantaneously at rest)

10 practice questions

0/10 correct

1.The IC has velocity

2.Rolling wheel IC is at

3.Translation (ω=0) puts I at

4.v=6 m/s at 1.5 m from I → ω =

5.Acceleration of I is generally

6.Two non-parallel v’s locate I at

7.Pin support that is at rest is

8.Relative-velocity formula v_B=v_A+ω×r_{B/A} is

9.If ω=2 rad/s and d=0.5 m, v =

10.The IC can lie outside the body