Problem 6-33 Revisited, the Angular Momentum of the Plate About a Pivot Point: H_A

Problem 6-33 on page 440 of our text book involved a plate hinged along one edge.  The figure depicts the plate (green) and three coordinate axes (light blue).  The coordinate axes are to be taken as attached to the plate and moving with it.  The Y axis runs through the center of the plate and along its longest dimension.  The X axis runs through the midpoint of one edge of the plate and is perpendicular to the largest surface of the plate.  The Z axis is in the plane of the plate and is along the shortest edge in the plate plane.  The only permitted movement of the plate is a rotation about this Z axis.  The plate has a length of 0.4 m along the Y axis and a length of 0.25 m along the Z axis.  Its thickness in the X direction is small and will be taken to be zero.  One end of the plate (parallel to the Z axis) is welded to a light axle.  The axle is shown in yellow in the figure.  The axle passes through and is supported by two fixed bearings, A and B.  The bearings are shown in gray in the figure.  The bearings are each a distance of 0.08 m from the edge of the plate in the -Z and +Z directions respectively.  The bearings permit the axle to rotate freely about the Z axis but prevent any other movement.  The plate is uniform and has a mass of 20 kg.  We can use the formula for the inertia tensor of a rectangular parallelepiped to evaluate the angular mass matrix about the mass center for the plate.  Earlier we developed an expression for the angular mass matrix of the plate about point A.  The components of this angular mass matrix were given relative to the X, Y, Z axes shown in the figure.  Recall that these axes are attached to the plate.  This matrix is repeated here for convenience:
 
           |   2.011       0          0        |
I_A =    |      0       0.945    -0.82    |  kg m²
           |      0      -0.82      1.067   |

Observing that point A is a fixed hinge point for the plate, then the above angular mass matrix (inertia tensor) should be most useful in evaluating the angular momentum of the plate.  In particular:

H_A = I_A w_P/I
P = Plate
A = Point fixed in plate at hinge location A.
I = Fixed coordinate system that never moves (not shown in the figure).
w_P/I = Angular velocity of the plate relative to the fixed coordinate system I.
 I_A = Angular mass matrix of the plate about the point A.
H_A  = Angular momentum of the plate about the point A.

In order to evaluate the angular momentum of the plate about the point A we need an expression for the angular velocity of the plate.  Noting that the Z axis of the plate never changes direction, then the direction of the angular velocity of the plate must always be parallel to the Z axis.  Thus while the magnitude of that angular velocity may change as the plate rotates about the fixed axis, the direction can never change.  Thus:

w_P/I = w K
K = Direction of angular velocity vector of the rigid body P relative to the fixed inertial coordinate system I.  Note that this direction, the Z axis attached to the plate, happens to be fixed both in the plate and in absolute space.
w = Magnitude of the angular velocity of the rigid body P relative to the fixed inertial coordinate system I.  Note that this magnitude may well change as the plate rotates.

We now have expressions for all of the terms involved in evaluating the angular momentum of the plate about the fixed pivot point A.  We have both the angular mass matrix and the angular velocity.  So we are ready to multiply them together and get the angular momentum, right?  WRONG!  Before we dare even think about multiplying them together we must make sure that the components of the angular mass matrix and the components of the angular velocity vector have been expressed relative to the same set of axes.  Good fortune is smiling on us because in this case both entities have been expressed relative to the XYZ axes fixed in the plate.  The angular mass matrix was developed using those axes.  The angular velocity vector has only one component and it is along one of those axes fixed in the plate (Z axis).  Thus we are indeed set to perform the multiplication and evaluate the angular momentum of the plate.  Expressing the angular velocity vector as a three by one column matrix with the only non-zero component being in the third or Z slot yields the following expression for the angular momentum:

            |   2.011       0          0        |                      0
H_A =    |      0       0.945    -0.82    |  kg m²           0
             |      0      -0.82      1.067   |                     w

Performing the required matrix product yields the following expression for the angular momentum of the plate about point A:

                 0
H_A =   -0.82 kg m² w
            1.067 kg m² w

It is crucial to recall that the components of this expression for the angular momentum are expressed relative to the same coordinate system used in expressing the angular mass matrix (inertia tensor) and the angular velocity vector.  Thus the three components are relative to the XYZ axes attached to the plate.  Stating this explicitly:

H_A =   -0.82 kg m² w J + 1.067 kg m² w  K

It is of critical importance to observe that the above formula for the angular momentum of the plate is valid throughout the entire motion history of the plate.  This is because both of the entities entering into the calculation of the angular momentum vector were always valid.  No matter how long you watch the plate, any spin it might have is about the Z axis.  The magnitude of this spin might change but that has been accounted for by denoting that magnitude as a variable w.  This might be further emphasized by noting that this w is some, as yet unknown, function of time, call it w(t).  The inertial tensor components were evaluated relative to axes fixed in the plate, thus as long as the plate doesn't change size, these components never change either.  Thus the angular momentum formula is always appropriate.  What price have we paid for this generality?  The angular momentum vector has been expressed in terms of a coordinate system (XYZ) that is itself rotating.  We will need to keep an eye on this as we proceed further with this problem.

Additionally it is important to note that as prophesied earlier, the angular momentum vector is not parallel to the angular velocity vector. The role of the off-diagonal terms of the mass matrix in producing coupling effects is now clear.  An angular velocity about the Z axis has produced an angular momentum term in the Y direction.  Amazing!

Note also that as this angular momentum vector is expressed in the XYZ system, it can change in two ways.  First of all the magnitude of the angular velocity, w, can change.  Less obvious is the observation that as the plate rotates, the Y axis which is fixed in the plate along the long edge of the plate, also changes direction.  Thus the direction of the unit vector J may change as the plate rotates.  This observation is crucial to correctly evaluating the forces acting on the plate.

Several homework problems involve the evaluation of angular momentum so make sure you can follow the above example!

So we can calculate an angular momentum of the plate about some weird point.  So what?  The problem involved finding the bearing forces.  How does this help us get the bearing forces?  That brings us back to our fundamental formulas.  The angular momentum is crucial to the calculation of the forces and moments acting on the plate.  Want to know how?  Go back to the fundamental formulas and find out!
 
Copyright (1998) by Nels Madsen.  Last Updated: February 28, 1998