Question

The answer was provided by my Dynamics professor.

The 4 kg slender bar shown is 2 m in length. The bar started from rest in an initial position relative to an inertial reference frame. For the position shown, v,-21+6) m/s and ???- 12k rad/s. Treating the bar as a rigid body, find the work done on the bar as it moved from its initial position to its present position. 30° 65.9 Nm

Answer 1

Given data, the 4 - kg slender bar shown is 2m in length the bar started from rest in an initial position relative to an inertial reference frame for position shown v_sim A = 2 i + 6 j m/s w_sim AB = 12 k rad/s Treating the bar as a rigid body. the work done on the bar as it moved from is initial position to its present position. mass m = 4 ib = 1.81 kg radius r = 2ft = 0.61 m velocity v = [2^2 + 6^2]^1/2 = 6.325 ft = 1.93 m/s angular velocity w = 12 rad/s. step 2 the work done w = [w translation + w rotation] sin = [1/2 mv^2 + 1]/2 mr^2 2 w^2] sin (30) = [1/2 times 1.81 times 1.93^2 + 1/2 times 1.81 times (12 times 0.61)^2] 0.5 = [3.4 + 48.5] 0.5 = [51.9] times 0.5 = 25.953.