Bar AD has the angular motions shown Suppose that omegaAB 4

Bar AD has the angular motions shown Suppose that omega_AB = 4.5 rad/s and alpha_AB = 6.3 rad/s^2. Part A Determine the velocity of the slider block C at this instant Part B Determine the acceleration of the slider block C at this instant.

Solution

a) The initial distance of the slide is,

OC = 1.5 x cos theta where theta = arc sin OA / AC

AC = 1.5m and OA = sin 60 - sin 45 = 0.15m

theta = 5.7 degree

OC = 1.5 x cos 5.7 = 1.49 m (initial length OC)

Final length OC = cos 45 + cos 60 = 1.2m

The displacement per second of the slider is diff of initial length and final length of OC,

1.49 - 1.2 = 0.29 m / s

b) The total time for the acceleration is omega over alpha which is equal to 0.7 sec

Now the rate of displacement of OC in 0.7 sec interval is the acceleration, 0.29 / 0.7 = 0.4 m/s2