Question

The particle has a mass of 0.35 kg and is confined to move along the smooth vertical slot due to the rotation of the arm OA. The rod is rotating with a constant angular velocity θ = 2 rad/s. Assume the particle contacts only one side of the slot at any instant. (Figure 1)

Part A

Determine the magnitude of the force of the rod on the particle when θ=30°.

Part B

Determine the magnitude of the normal force of the slot on the particle when θ=30°.

Answer 1

r = 0.5/cos theta = 0.5 sec theta r = 0.5 sec theta tan theta. theta r = 0.5 sec theta tan theta theta + 0.5 sec^3 theta theta^2 + 0.5 sec theta tan^2 theta theta^2 At theta = 30 degree theta = 2 rad/s^2 theta = 0 r = 0.5774 m r = 0.6667 m/s r = 3.8490 m/s^2 a_r = r - r theta^2 = 3.8480 - 0.5774(2)^2 = 1.5396 m/s^2 a_theta = r theta + 2r theta middot 0 + 2(0.6667)(2) = 2.667 m/s^2 Rightarrow Sum F_r = ma_r N_p cos 30 degree -(0.35)(9.81) sin 30 degree = 0.35 (1.5396) N_P = 2.6045N + Sum F_theta = ma_theta F - 2.6045 sin30 degree - 0.35(9.81) cos30 degree = 0.35 (2.667) F=5.209N