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 r = 600 mm C A B 200 g O 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.

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Presentation on theme: " r = 600 mm C A B 200 g O 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."— Presentation transcript:

1  r = 600 mm C A B 200 g O Problem 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)  = 90 o, (b)  = 75 o, (c)  = 45 o. Indicate in each case the direction of the impending motion.

2 Solving Problems on Your Own Problem  r = 600 mm C A B 200 g O 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)  = 90 o, (b)  = 75 o, (c)  = 45 o. 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 Solving Problems on Your Own Problem  r = 600 mm C A B 200 g O 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)  = 90 o, (b)  = 75 o, (c)  = 45 o. 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  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.

4 Problem Solution  r = 600 mm C A B 200 g O Kinematics.  r = 600 mm C A B O anan  r sin  a n = (r sin  )  2 a n = (0.6 m) sin  ( 6 rad/s ) 2 a n = 21.6 sin  m/s 2

5 Kinetics; draw a free body diagram. Problem Solution  r = 600 mm C A B 200 g O  (0.2 kg)(9.81 m/s 2 ) O N F ma n = (0.2) 21.6 sin  = 4.32 sin  N =

6 Problem Solution Apply Newton’s second law.  (0.2 kg)(9.81 m/s 2 ) O N F ma n = (0.2) 21.6 sin  = 4.32 sin  N = +  F t = 0: F (9.81) sin  = sin  cos  F = 0.2 (9.81) sin  sin  cos  +  F n = ma n : N (9.81) cos  = 4.32 sin  sin  N = 0.2 (9.81) cos  sin 2  F =  N For a given , the values of F, N, and  can be determined

7 Problem Solution  (0.2 kg)(9.81 m/s 2 ) O N F ma n = (0.2) 21.6 sin  = 4.32 sin  N = Solution: (a)  = 90 o, F = N, N = 4.32 N,  = (down) (b)  = 75 o, F = N, N = 4.54 N,  = (down) (c)  = 45 o, F = N, N = 3.55 N,  = (up)


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