Question

The bar shown remains in contact with the vertical wal surface A and with edge B. At the instant shown, A is moving upward with a constant velocity of 6 ft/s. Determine (a) the angular velocity in rad/s and (b) the angular acceleration in rad/s/s of the bar at the instant y 8 ft. Assume w 11 ft.

Answer 1

Now, from figure tan theta = y/w Now, differentiate the above equation w.r to time sec^2 theta d theta/dt = 1/w dy/dt sec^2 theta (w) = 1/w V_A equation (1) sec theta = squareroot y^2 + w^2/w w = 1/w V_A times 1/sec^2 theta w = 1/w V_A times w^2/(w^2 + y^2) w = w(V_A)/w^2 + y^2 equation (2) \r\n w = 11 ft times 6 ft/s/(11 ft)^2 + (8 ft)^2 = 0.357 rad/s b) Angular acceleration alpha = -w(V_A)^2 2y/(w^2 + y^2)^2 = -11(6 ft/s)^2 2 times 8 ft/(11^2 + 8^2)^2 = -0.185 rad/s^2