Problem 6/55  Page 445, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition

Given: Uniform slender bar attached to a frictionless pin at one end.  At the instant of interest the bar is at an angle q below the horizontal.  The bar has mass M and length L.  At the instant of interest the bar is swinging in a vertical plane with an angular speed of w.

Find: At the instant of interest the forces exerted by the pin on the bar.

Parameter Values:
q = 30 degrees = Bar angle below the horizontal at the instant of interest.
L = 0.9 m = Bar length.
M = 8 kg = Bar mass.
w = 2 rad/s = Bar angular speed at the instant of interest.
 
1. Mechanical System:  Bar at the instant of interest.

2. Free Body Diagram: Shows the bar at the instant of interest.  Includes the weight force acting vertically downward.  Includes horizontal and vertical forces at the pin.  The x axis is horizontal to the right and the y axis is vertically upward.

3. Equations:
a) S F = Ax i + ( Ay - M g ) j
S M_A = L/2 ( cos q i - sin q j ) X - M g j
(Note that we have a fixed pivot point so we can sum moments about that pivot point).
 = -M g L/2 cos q  k

b) m a_G = M ( a_A + w X (w X r_G/A) + a X r_G/A )
A is a fixed pivot point and hence has zero acceleration.  The angular speed is given and the direction is clockwise (-z).  The angular acceleration is unknown but is parallel to the z axis.  The position vector from A to G is of length L/2 along the bar.
m a_G = M ( -w k X ( -w k X L/2 ( cos q i - sin q j ) ) + a k X L/2 ( cos q i - sin q j )
    = M w² L/2 ( -cos q i + sin q j ) + M a L/2 ( cos q j + sin q i )
I_A a = ( 1/12 M L² + M (L/2)² ) a k
Using the transfer theorem to evaluate the mass moment of inertia about the pivot point.
I_A a = 1/3 M L² a k

4. Solve:
Ax = M ( a L/2 sin q - w² L/2 cos q )
Ay - M g = M ( a L/2 cos q + w² L/2 sin q )
-M g L/2 cos q = 1/3 M L² a

Solving the third equation for the angular acceleration:
a = -3/2 g/L  cos q
The first equation can now be solved for Ax:
Ax = M ( -3/4 g  cos q sin q   - w² L/2 cos q )
    = -38.0 N
Ay = M ( g - 3/4 g cos q cos q   + w² L/2 sin q )
   = 41.5 N
A = (Ax² + Ay²)^1/2 = 56.3 N