Problem 6/97  Page 464, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition

Given: A slender rod released from rest in a vertical position.  The upper end of the rod is supported by a light roller on an inclined plane that makes an angle of q with the horizontal.  The rod has length L and mass M.

Find:  The initial acceleration of the upper end of the rod (A) just after release.

1. Mechanical System: Rod at the instant just after release.

2. Free Body Diagram: Shows the rod at the instant just after release (rod is still vertical).  A weight force acts vertically downward through the center of the rod.  A normal force acts on the roller at the upper end of the rod in a direction that is perpendicular to the inclined plane and hence at an angle of q from the vertical.  The x axis is horizontal to the right and the y axis is vertically upward.

3. Equations:
a) S F = -N sin q i + ( N cos q - M g ) j
S M_G = L/2 j X ( -N sin q i + N cos q j)
    = L/2 N sin q k

b) m a_G = M (a_A + w X ( w X r_G/A ) + a X r_G/A)
The roller (A) is moving along the inclined plane in a straight line.  Thus any acceleration it has is along the incline.  As the rod is released from rest the angular velocity at this instant is zero.  The rod rotates in the xy plane so the angular acceleration is in the z direction.  The position vector from the roller to the mass center is of length L/2 vertically downward.
m a_G = M ( N cos q - M g( -cos q i - sin q j ) + a k X -L/2 j )
      = M ( a_A ( -cos q i - sin q j ) + a L/2 i )
I_G a = 1/12 M L² a k
(Using the mass moment of inertia for a slender rod).

4. Solve:
-N sin q = M ( -a_A cos q  + a L/2 )
 N cos q - M g = -M  a_A sin q
 L/2 N sin q =  1/12 M L² a

From the third equation one can solve for the angular acceleration in terms of the normal force:
a = 6 N sin q / ( M L)
From the first equation one can solve for the normal force in terms of the acceleration of the roller.
N = M  a_A cos q / ( sin q + 3 sin q )
   = M  a_A cos q / 4 sin q
From the second equation, one can solve for the acceleration of the roller.
a_A = g / ( sin q + cos² q / 4 sin q )
Multiplying the top and bottom of the above equation by the sine of q, and noting that sin²(q) is equal to 1-cos²(q):
a_A = g sin q  / ( 1 -  cos² q + cos² q / 4 )
a_A = g sin q  / ( 1 -  [3/4] cos² q )