In
the figure, two bars are hinged in the vertical position. The lower bar is pinned to a fixed support
and its rotation impeded by a torsional spring. The spring is in a relaxed
position when .
The upper bar is pinned to a collar which is constrained to move along a
vertical shaft. A force, F, is applied at the point β, where a pin connects the two bars. The force
remains parallel to the upper bar as the two bars rotate. Write the equations
of motion using Lagrange’s equations.
The first objective is to compute the virtual work performed by the force F.
Here, F does work through the point β as the bars rotate through the rotation angle .
(In this problem
will be our generalized coordinate). By
definition, the virtual work is
Where F is the generalized force and is the change in the position vector of the
point β. For
we have
(1.1)
To compute ,
we allow the bars to rotate by an incremental amount
.
The incremental change in the position of the pin β is then
The force F in vector form is .
Applying equation (1.2)
we’re now able compute the virtual work
(1.3)
Having the virtual work now allows us to compute the
generalized force. Remember, the generalized force is what turns up in the
equations of motion not the virtual work. The generalized force
for the coordinate
is simply
To compute the kinetic energy we will need to calculate the translational velocity of the upper bar at its center of mass. We’ll first start by calculating the velocity of the pin at β.
(1.5)
The velocity of the center of mass for the upper bar is then
(1.6)
We should note that we do not need to calculate the velocity of the lower bar since the lower bar is rotating about the fixed point α. The bar is in pure rotation and only the rotational term shows up in its kinetic energy. The upper bar has both translational and rotational motion and is therefore in general motion. Subsequently, the upper bar’s kinetic energy will have two terms. Writing the total kinetic energy for the system we have
(1.7)
The moments of inertia and
are moments of the lower bar about the pin α and the upper bar its center of mass.
and
Making these substitutions we have the final form for the kinetic energy
The
next step is to form the potential energy which is a sum of potentials in both
upper and lower bars and the torsional energy stored in the spring. Using the
pin at α as the zero reference line we have for the potential energy for
each bar
and
The torsional energy stored in the spring is the change in from its rest position
.
Summing these energies we have for the total potential energy for the system
Having both the total kinetic and total potential energies
we are ready to write the Lagrangian function .
Using equations (1.8)
and (1.9)
gives us the following
(1.10)
The equations of motion for the generalized coordinate will be
Starting with the first term,
The second term in (1.11) will be
Substituting the results from equations (1.12), (1.13) and (1.4) into (1.11), we have the following for the final equations of motion for the system.
(1.14)