Problem 6/115  Page 478, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition

Given: A uniform rectangle of weight W, base B and height H supported in a position such that the base is horizontal.  The rectangle is pivoted about its lower right corner.  The rectangle is released from rest and allowed to rotate through an angle q.

Find: The angular speed of the rectangle after rotating through the angle q.

Parameter Values:
W = 250 lb = Weight of rectangle.
B = 4 ft = Base of rectangle.
H = 3 ft = Height of rectangle.
q = 30 degrees = Angle of rotation.

1. Mechanical System: Rectangle as it falls through the angle of interest.

2. Free Body Diagram: Drawn at an arbitrary position between release and the angle of interest.  Shows a weight force acting vertically downward through the center of the rectangle.  Shows horizontal and vertical pin forces acting at the fixed pivot point (O).  The x axis is horizontal to the right and the y axis is vertically upward.

3. Equations:
a) Work = 0 - W Dh
Note that the forces at the pivot point are applied to a point of zero velocity and do no work.  The work down by gravity can be calculated in terms of the weight (W = m g) and the change in height.  In order to evaluate the change in height we can see that the mass center is initially a distance H/2 above the pivot point.  After the rectangle has rotated through some angle we can now locate the mass center relative to the pivot point by traveling a distance B/2 along the base edge of the rectangle and then a distance H/2 parallel to the other edge of the rectangle.  As we are interested only in the vertical portion of this position, we can see that traveling along the base edge lowers us a distance equal to B/2 sin(q) (edge makes an angle of q with the horizontal).  As we travel along the other edge, we make an angle of q with the vertical.  Thus we traverse a vertically upward distance of H/2 cos(q).  We can combine these two distances to yield the fact that the height of the mass center above the pivot point after falling through the angle q is H/2 cos(q) - B/2 sin(q).  The change in height will be this vertical height minus the initial height.
Work = -W ( H/2 cos(q) - B/2 sin(q) - H/2 )
b) D Kinetic Energy  = 1/2 I_O w² - 0
As the rectangle is released from rest, its initial kinetic energy is zero.  As it has a fixed pivot point, the final kinetic energy can be evaluated in terms of the angular speed and the mass moment of inertia about the pivot point.  The mass moment of inertia about the pivot point can be evaluated using the transfer theorem.  Using the mass moment of inertia for a rectangle and transferring it to the corner:
D Kinetic Energy  = 1/2 ( 1/12 W/g (B² + H²) + W/g ( B/2² + H/2²) )  w²
   = 1/6 W/g ( B² + H²) w²

4. Solve:
-W ( H/2 cos(q) - B/2 sin(q) - H/2 ) = 1/6 W/g ( B² + H²) w²
Solving for the angular velocity:
 w²  = 3 g ( H - H cos(q) + B sin(q)) / ( B² + H²)
w = 3.05 rad/s

Note that the weight does not influence the final angular speed.