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r = 600 mm

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  1. B r = 600 mm O C q 200 g A Problem 12.127 A small 200-g collar C can slide on a semicircular rod which is made to rotate about the vertical AB at the constant rate of 6 rad/s. Determine the minimum required value of the coefficient of static friction between the collar and the rod if the collar is not to slide when (a) q = 90o, (b) q = 75o, (c) q = 45o. Indicate in each case the direction of the impending motion.

  2. Problem 12.127 B r = 600 mm O C q 200 g A Solving Problems on Your Own A small 200-g collar C can slide on a semicircular rod which is made to rotate about the vertical AB at the constant rate of 6 rad/s. Determine the minimum required value of the coefficient of static friction between the collar and the rod if the collar is not to slide when (a) q = 90o, (b) q = 75o, (c) q = 45o. Indicate in each case the direction of the impending motion. 1. Kinematics: Determine the acceleration of the particle. 2. Kinetics: Draw a free body diagram showing the applied forces and an equivalent force diagram showing the vector ma or its components.

  3. Problem 12.127 B r = 600 mm O C q 200 g A Solving Problems on Your Own A small 200-g collar C can slide on a semicircular rod which is made to rotate about the vertical AB at the constant rate of 6 rad/s. Determine the minimum required value of the coefficient of static friction between the collar and the rod if the collar is not to slide when (a) q = 90o, (b) q = 75o, (c) q = 45o. Indicate in each case the direction of the impending motion. 3. Apply Newton’s second law: The relationship between the forces acting on the particle, its mass and acceleration is given by SF = ma . The vectors F and a can be expressed in terms of either their rectangular components or their tangential and normal components. Absolute acceleration (measured with respect to a newtonian frame of reference) should be used.

  4. Problem 12.127 Solution B r = 600 mm O C q 200 g A Kinematics. w B r = 600 mm O C q an A an = (r sinq) w2 an = (0.6 m) sinq ( 6 rad/s )2 r sinq an = 21.6 sinqm/s2

  5. Problem 12.127 Solution B r = 600 mm O C q 200 g A F O N q man = (0.2) 21.6 sinq = 4.32 sinq N (0.2 kg)(9.81 m/s2) Kinetics; draw a free body diagram. =

  6. Problem 12.127 Solution man = (0.2) 21.6 sinq = 4.32 sinq N + SFt = 0: F - 0.2 (9.81) sin q = - 4.32 sinq cos q F = 0.2 (9.81) sin q - 4.32 sinq cos q + SFn = man: N - 0.2 (9.81) cos q = 4.32 sinq sin q N = 0.2 (9.81) cos q + 4.32 sin2q F = mN F Apply Newton’s second law. O = N q (0.2 kg)(9.81 m/s2) For a given q, the values of F , N , and m can be determined

  7. Problem 12.127 Solution F O = N q man = (0.2) 21.6 sinq = 4.32 sinq N (0.2 kg)(9.81 m/s2) Solution: (a) q = 90o, F = 1.962 N, N = 4.32 N, m = 0.454 (down) (b) q = 75o, F = 0.815 N, N = 4.54 N, m = 0.1796 (down) (c) q = 45o, F = -0.773 N, N = 3.55 N, m = 0.218 (up)