# Problem A small 200-g collar C can slide on a

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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. q r = 600 mm C A B 200 g O

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. q r = 600 mm C A B 200 g O 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.

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. q r = 600 mm C A B 200 g O 3. Apply Newton’s second law: The relationship between the forces acting on the particle, its mass and acceleration is given by S F = m a . 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.

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

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

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

= man = (0.2) 21.6 sinq = 4.32 sinq N Problem 12.127 Solution F O N q
(0.2 kg)(9.81 m/s2) O N F man = (0.2) 21.6 sinq = 4.32 sinq N = Solution: (a) q = 90o, F = N, N = 4.32 N, m = (down) (b) q = 75o, F = N, N = 4.54 N, m = (down) (c) q = 45o, F = N, N = 3.55 N, m = (up)

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