Natural Science Forum / Physics / General Physics / October 2007

On Oct 15, 9:13 am, Bruce Scott TOK <Use-Author-Supplied-Address-
Header@[127.1]> wrote:

The wedge will move only horizontally.  The sliding mass will move
both horizontally and vertically (with the horizontal component
possessing an equal and opposite moment to that of the wedge).  A line
drawn between the Centers of Mass (CoM) of the two will rotate unless
the angle of the incline is 0.

Draw a diagram.  It may help.  Look at the 'before' (sliding mass at
the top of the wedge) and 'after' (sliding mass at the bottom of the
wedge just touching the tabletop) and remember that horizontal
position of the CoM of the SYSTEM (the weighted mean of the CoMs of
the two masses) is conserved.  Drawn so that the wedge moves to the
left, the sliding mass will move to the right AND will drop.

The OP is misleading himself by selecting displacement along the
incline of the wedge as a coordinate.  This is oblique to the motion
of the wedge itself as well as subject to acceleration as the wedge
itself moves.

The CoM of a system is the reference frame of choice only if there is
not an "absolute" frame like the table in billiard ball problems (as
is often the case in simple mechanics/dynamics) with respect to which
all other motions can be measured.

A better choice of frame of reference would be the 'ground' surface,
with respect to which the wedge moves ONLY in a horizontal direction,
and with respect to which only the sliding mass has a vertical
component to its motion.  The motions become thus separable, and the
equations of motion become much simpler to solve.

In general conservation of energy and conservation of momentum are
very useful tools, as long as *all* relevant energies are considered.
In elastic collisions and frictionless motion simple kinetic and
potential energy are sufficient.  When inelasticity or friction arise,
they too must be considered as sinks for energy.

For example, when considering the transfer of momentum and energy
between two non-spherical 3-D molecules in space, the redistribution
of energy to vibrational, rotational, and translational modes
(inelastic interactions) must be handled with consideration of the
maintenance of thermal equilibrium in all available modes of molecular
motion.  The problem becomes a statistical one because of the chaotic
sensitivity of the outcome of any individual collision to the exact
orientations and velocities of the colliding particles.  Individual
outcomes cannot be well predicted, but statistical averages over LARGE
numbers of such trials can be determined.

Tom Davidson
Richmond, VA

[snip]

I suspect that the OP had the problem of setting up the
Lagrangian formulation of the problem, not just solving
the equations of motion any old way he could.
Socks

No.  The point of Lagrangian mechanics is that relationships between
momentum and energy exist that are invariant with respect to changes
in coordinate systems.  It allows the investigator the freedom to
choose whatever coordinate system is most convenient for solving the
problem.

AH! The value of selecting a convenient frame of reference and system
of coordinates appears!

Meanwhile the vertical acceleration affects only the sliding mass.
Given the Lagrangian function of each mass and the *indicated*
coordinate system (recall that the OP had originally selected one axis
as parallel to the slope of the wedge) the problem of the complex
motion of the system is easily decomposed into separate problems of
the motion of each coordinate.

While the problem seems almost trivial for two bodies in two
dimensions, it becomes more significant in multi-body systems in 3
dimensions such as the problem of the internal vibrations of a
molecule.  It is *really* helpful in molecular spectroscopy to be able
to identify the orthonormal vibration modes of a molecule or complex
ion.

Benzene (to take a very simple example) includes 12 atoms, each with
three position coordinates and three velocity coordinates.  Rather
than try to represent these 72 dimensions in X-Y-Z coordinates, it is
far more convenient to identify the spatial coordinates of the CoM of
the molecule and its translation motion and bulk rotation first.  Then
the various other motions become internal motions involving the
bending and stretching of molecular bonds with varying orders of
symmetry. These internal motions, according to the Lagrange equations,
are all completely independent of each other.

If an internal motion, for example, changes the electrostatic dipole
moment of the molecule, then it will do so with a characteristic force
constant, associated with a resonant frequency (in perfect analogy to
simple harmonic motion of a Hooke's Law restoring force) which is
generally independent of the resonant frequency of other modes of
motion.  Then an applied electromagnetic field which varies with the
same frequency will energize that motion distinctively from all
others.

The Lagrangian allows us to identify these vibrations as orthogonal
coordinates (in terms of linear combinations of individual atomic
motions) with the help of some Debye-Hueckel theory and applied linear
algebra.

Tom Davidson
Richmond, VA

No, but in this case (with 2 degrees of freedom) the
Lagrangian method is more straightforward and easier
to apply.

No, that's wrong. Besides mg, there's a contact force from
the wedge, N say, acting on the particle. That means that you
need more equations.  It's a good practice to start by drawing
the appropriate free body diagrams.

No.

/mL