Instantaneous Center for Bodies in Plane Motion

Example: Link, BC, is supported by a crank, AB, and a connecting link, CD as shown below

in the Figure on the left. The following data apply:

AB = 0.3 m, CD = 0.4 m, and ω_ab = 20.0 k rad/sec.

Find the instantaneous center for the link. Determine the angular velocities of link BC and

of the connecting link CD.

To locate the instantaneous center for link BC, construct a diagram showing perpendicular

lines to the velocities of B and of C as shown above in the figure on the right.

Intersection of the perpendiculars is at the instantaneous center, P. Note the direction of the

velocity of C does not matter. It is either in the positive or negative y-direction.

Now A is the instantaneous center for the crank . So v_B = AB (ω_AB) = (0.3) (20.0)( ˗ i )

v_B = ˗ 6.0 i m/sec and since P is the instantaneous center for link BC, PB x ω_BC = v_B

Solve for the angular speed of link BC . ω_BC = v_B / PB = 6.0 / 0.7 = 8.57 rad/sec

And ω_BC = ˗ 8.57 k rad/sec since link BC rotates clockwise about P (result)

Now v_C = v_B + ω_BC x r_CB = ˗ 6 i + (˗ 8.57 k) x ( ˗ 0.4 i ˗ 0.7 j ) = ˗ 6 i ˗ 3.43 j + 6 i

So v_C = ˗ 3.43 j m/sec and v_C = CD ω_CD Therefore ω_CD = v_C / CD = 3.43/0.4 = 8.57

Therefore ω_CD = 8.57 k rad/sec since connecting link rotates CCW. (result)

Click here for another example.

Return to Notes on Dynamics

Copyright © 2019 Richard C. Coddington
All rights reserved.