Figure 1 -- Weights and ramp:
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This
is a very simple example of the use of the Lagrangian formulation of
Newtonian mechanics to solve a problem. We have two weights,
m1 and
m2, attached together with a string of length
l (
figure 1).
One is on a ramp; the other is hanging from the string. The
string goes over a pully. Both weights are initially stationary.
We'll start by neglecting friction, then do it again with
friction
later in the page.
With
x units of string between the pulley and weight
m1, it drops
units below the level of the pulley. So, the (gravitational) potential energy of weight
m1 is
(1)
When weight
m1 slides
x units down the ramp, weight
m2 must
rise x units. So, the (gravitational) potential energy of weight
m2 is
(2)
The kinetic energy of the system is, of course,
(3)
So, following equation (
lagrange.4), the Lagrangian for the system is:
(
4)
The partial derivatives we need are:
(
5)
As soon as we start to write the equation of motion, we realize this substitution helps a lot:
(
5b)
And with substitions (
5b), the equation of motion, following (
lagrange.3), is:
(
6)
We can see immediately from (6) that there are three cases. In case (7a), weight
m1 slides down the ramp, and weight
m2 rises.
(
7a)
In case (7b), the weights balance, and nothing happens.
(
7b)
And in case (7c), weight
m1 slides up the ramp, and weight
m2 falls.
(
7c)
We can integrate (
6) twice to obtain a closed-form solution. Assuming the weights are stationary to start with, and
m1 starts at
x0:
(8) |
Once Again, with Friction
We'll assume there's friction at work when
m1 slides along the ramp, and we'll use the friction model given in (
lagrange.9). In this case there's just one dimension, so it's very simple:
(9)
We will assume
f1 is positive for case (
7a), and negative for case (
7c). (In case (
7b)
nothing moves and friction doesn't matter!) The "cos θ" term is
due to the fact that the force of gravity is straight down, which is at
angle θ to a line perpendicular to the ramp's surface. Note that
the
dimensions of
f2 are time/distance, while
f1 is dimensionless.
Plugging equations (
5) and friction force (9) into the equation of motion (
lagrange.8), we obtain
(10)
or, slightly rearranged, and again substituting (
5b):
(
11)
We
can see a couple of things immediately by looking at (11). First,
if the "constant friction" is too large, the weights won't move at all;
the initial acceleration will be zero. In order for them to move,
we must have:
(12)
As
the velocity increases, the viscous friction term becomes larger, and
eventually the acceleration again goes to zero; the terminal velocity,
when acceleration is zero, must be:
(
13)
Finally, we can solve for the velocity and position. Let's rearrange (11) a bit:
(
14)
That looks like an exponential. The coefficient of
is positive; the right hand side may be either positive or negative depending on which way the system slides (case (
7a) or case (
7c)). Let's substitute
a and
b for the constants, to reduce clutter, with
a>0:
(15)
A solution to the homogeneous equation (with b=0) is:
(16)
and with a little fiddling we obtain the particular solution (for b≠0):
(17)
Adjusting this to match the initial conditions at t=0 (which were x=x
0,
=0), we obtain:
(18)
Taking a firm grip on our pencil and looking back at (
14), we substitute the definitions for
a and
b back into (18), differentiate, and divide out some common factors to obtain the equations for the position and velocity:
(19a)
(19b) |
At time t=0, (19b) is certainly zero, and as t->
, we see that (19b) approaches the terminal velocity we found in (
13),
which encourages us to think that these rather ugly equations might
actually be correct. As is often the case, introducing a general
friction function made the problem a lot messier. Keep in mind
that, in this case, the sign we assumed for
f1 depends on the ratio of the masses and the angle of the ramp.
Page
created on 12/1/06. Minor corrections, 12/2/06