Question

Drawing Problem 5 The arm is rotating at an instantaneous angular velocity 6 -5 rad/s, and an insta angular acceleration of 6 -2 rad/s when 0 90 degrees. Determine the normal force (undirection) the arm must exert on the 0.5 kg particle to keep it on a horizonal path defined by r-0.2/0 meter.(40 points) 0.2 筚亨 ! Notes

Answer 1

theta = 90 degree = pi/2 rad, theta = 5 rad/sec, theta = 2 rad/s^2 r theta = 0.2, r(pi/2) = 0.2 => r = 0.1273 m Differentialting Therefore r theta + r theta = 0 => r (pi/2) + 0.1273 (5) = 0 => r = -0.405 m/s again Differntiating r theta + r theat + r theta + r theta = 0 => r(pi/2) + 2(- 0.405 times 5) + (0.1273 times 2) = 0 => r = 2.418 m/s62 v = v_r e_r + v_theta e_theta, v_r = r = -0.405 m/s, v_theta = r theta = 0.1273 (5) = 0.637 m/s Therefore v^- = -0.405 e_r + 0.637 e_theta m/s accelaration a = a_re_r + a_theta e_theta rightarrow a_r = r - r theta^2 = 2.418 - (0.1273)(25) = -0.765 m/s^2 rightarrow a_theta = 2 r theta + r theta = 2(- .405)(5) + (0.1273 times 2) a_theta = -3798 m/s^2 Therefore sigma F_r = ma_r => N coa 32.49 = 0.5(- 0.768) => N = -0.453 N sigma F_theta = Ma_theta => R - N sin 32.49 = 0.5 (- 3.798) => R = -16.55 Ns