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Relative Velocity and Acceleration in a Plane Click here for discussion of relative acceleration.

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 velocity of link BC, the velocity of the slider, C, and the instantaneous center of link BC. All lengths are in feet.

Strategy: Apply the relative velocity equation between points A and B on link AB and between

B and C on link BC. Locate the instantaneous center of link BC by the intersection of lines

perpendicular to the velocity of points B and C.

Solution: v_B = v_A + v_B_/A= v_A + ω_o k x L i = L ω_o j

v_C = v_B + v_C_/B= L ω_o j + ω_BC k x 2L i = ( L ω_o + 2L ω_BC ) j

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

So v_C ( cos θ i + sin θ j ) = ( L ω_o + 2L ω_BC ) j

Equate scalar components: v_C cos θ = 0 and v_C sin θ = ( L ω_o + 2L ω_BC )

From the i-component: v_C = 0 and from the j-component L ω_o + 2L ω_BC = 0

Therefore v_C = 0 i + 0 j ft/sec and ω_BC = ˗ (1/2) ω_o k (results)

Next locate the instantaneous center for link BC.

Perpendiculars intersect at C. Therefore C is the instantaneous center for link BC. (result)

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