Question

In a four bar chain ABCD, AD is fixed and is 150 mm long. The crank AB is 40 mm long and rotates at 120 r.p.m. clockwise, while the link CD = 80 mm oscillates about D. BC and AD are of equal length. Find the angular acceleration of link CD when angle BAD = 60°

Answer 1

Given

$N_{BA}=120 r.p.m \implies \omega_{BA}=2 \pi *\frac{120}{60}=12.568 rad/s$

Since the length of the crank AB = 40 mm = 0.04 m , therefore velocity of B with respect to A or even velocity of B is given by

$v_{BA}=v_B = \omega _{BA}*AB=12.568 * 0.04 = 0.503 m/s$

By the method of measurement , we find that

$v_{CD}= v_C= 0.385 m/s$

Angular velocity of link CD , $\omega _{CD}=\frac{v_{CD}}{CD}=\frac{0.385}{0.08}=4.8 rad/s$ . Clockwise about point D