1. 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

2. 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.

3. 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.

4. 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

5. = 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

6. 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

7. = 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)