The drive link in the mechanism shown runs in a linear beari

The drive link in the mechanism shown runs in a linear bearing with a velocity v_c directed to the left. Determine the angular velocity of link AB and BC as a function of v_c, x and L.

Solution

Initial angle BAC, cos theta = (x^2 + L^2 - (L/2)^2) / (2Lx) = x / (2L) + (3L/8)/x

Differentiating it with time,

-w*sin theta = 1/ (2L) * (dx/dt) + (3L/8) * (-1 / x^2) * (dx/dt)

Putting dx/dt = -Vc

-w*sin theta = -Vc / (2L) + 3L*Vc/(8x^2)

w = [Vc / (2L) - 3L*Vc/(8x^2)] / sin theta

Now sin theta = sqrt (1 – cos^2 theta)

Sin theta = sqrt (1 – (x^2 / 4L^2 ) – (9L^2 / 64x^2) – 3/8)

Sin theta = sqrt (5/8 – (x^2 / 4L^2) - (9L^2 / 64x^2))

So, w = [Vc / (2L) - 3L*Vc/(8x^2)] / sqrt (5/8 – (x^2 / 4L^2) - (9L^2 / 64x^2))