Relative Acceleration for Links with 2 points in the same link

Example: The planar mechanism shown below consists of two rigid links, AB of length L and BC

of length 2L, pinned at A, B, and C. The slider ,C, moves along the guide track at an angle θ. At the instant shown, link AB rotates counterclockwise at a constant angular velocity given by ω = ω_o k rad/sec . Find the angular acceleration of link BC, the acceleration of the slider, C. All lengths are

in feet.

Strategy: Apply the relative velocity and relative acceleration equations between points A and B

on link AB and between B and C on link BC.

Solution: a_B = a_A + a_B/A |_{n +} a_B/A|t = ˗ ω_o² L i + α_AB x L i = ˗ ω_o² L i

a_C = a_B + a_C/B |_{n +} a_C/B|t = ˗ ω_o² L i ˗ ω_BC² (2L) i + α_BC k x 2L i

a_C = ( ˗ ω_o² L ˗ 2L ω_BC²) i + ( 2L α_BC ) j

Since the slider is constrained to move along the guide a_C = a_C ( cos θ i + sin θ j )

So a_C ( cos θ i + sin θ j ) = ( ˗ ω_o² L ˗ 2L ω_BC²) i + ( 2L α_BC ) j

Equate scalar components: a_C cos θ = ˗ ω_o² L ˗ 2L ω_BC²

and a_C sin θ = 2L α_BC = a_Cy

From the relative velocity equation (for this example previously calculated) ω_BC = ˗ (1/2) ω_o

giving a_C = ˗ 3Lω_o² sec θ / 2 and a_Cx = ˗ 3Lω_o² / 2

Now from a_C sin θ = 2L α_BC , α_BC = (˗ 3Lω_o² / 2 ) tan θ / 2L = ˗ (3/4) ω_o² tan θ

α_BC = ˗ (3/4) ω_o² tan θ k rad/sec² (result)

Therefore a_Cy = ˗ (3/2) Lω_o² tan θ

a_C = ˗ 3Lω_o² / 2 i + ˗ (3/2) Lω_o² tan θ j ft/sec² (result)

Return to Notes on Dynamics

Copyright © 2019 Richard C. Coddington
All rights reserved.