16.90 (Fixed Axis Rotation, but of a composite body)
Comments/Hints:
1. This is a composite body, consisting of a disk and a slender rod.
The mass center of the disk is at the pin at C, and the mass center of the
slender rod is at the midpoint between A and B. At the position shown, the
assembly has an angular velocity of w = 10 rad/sec
clockwise.
2. Draw a free body diagram of the composite body. It will show pin
reactions Cx and Cy, the mg of disk C acting downward at pin C, and the mg of
bar AB acting down at the midpoint of AB.
3. Draw a kinetic diagram (KD) of the composite body. There are two
ways to do this:
(a) Continue to treat the disk and the bar AB separately: Draw both
ICa and IABa
couples, plus man and mat terms at the mass center of AB.
(b) Combine the disk and the rod into a single composite body with a
single mass center and a single I. (But this requires finding the mass
center, etc.). It's probably easier to do (a), that is, keep the bodies
separate.
4. IC = 0.016 kg-m2 (about the pin at C), and IAB
= 0.0018 kg-m2 (about the mass center of AB). The man
term for the rod AB is mw2r = (1.5
kg)(102)(0.08 + 0.06) = 21.
5. Write the equations of motion: (a) Sum forces in the
horizontal (normal) direction = man; (b) Sum forces in
the vertical direction = mat;
(c) Sum moments about C = Sum of the kinetic moments about C = ICa
+ IABa + mat(0.14).
6. Solve for a = 43.6 rad/sec2 CW, at
= 0.14a = 6.11 m/s2 downward; and Cy
= 54.6 N upward.