| Topic: |
Science > Physics |
| User: |
"bit188" |
| Date: |
14 Oct 2007 04:46:33 PM |
| Object: |
Help with Lagrangian mechanics |
I'm stuck on a problem involving the Lagrangian formulation of
classical mechanics. Here is the problem and what I've done so far;
help is greatly appreciated.
(Problem: A wedge of mass M and angle [alpha] slides freely on a
horizontal plane. A particle of mass m slides freely on the wedge.
Determine the motion of the particle as well as that of the wedge.)
Solution (so far): Let l denote the length of the hypotenuse of the
wedge. Our kinetic and potential energies are
T_m = (m/2)[(dl/dt)^2]
V_m = -mglsin[alpha]
I'm having trouble specifying the speed of the wedge, so I can't come
up with the kinetic energy. Obviously, however, the wedge's potential
energy is zero.
Thanks again.
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| User: "Bruce Scott TOK" |
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| Title: Re: Help with Lagrangian mechanics |
15 Oct 2007 08:13:13 AM |
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I'm having trouble specifying the speed of the wedge, so I can't come
up with the kinetic energy. Obviously, however, the wedge's potential
energy is zero.
Use the centers of gravity of the wedge and the particle for your two
position coordinates. Express the Lagrangian in terms of those
coordinates. If there is no rotation involved, that's all you need.
As an exercise, you should be able to prove that the momentum and
kinetic energy of a distributed object is given by those of its C/M.
Rotation is slightly trickier, but only slightly (if you're good at
doing integrals).
--
ciao,
Bruce
drift wave turbulence: http://www.rzg.mpg.de/~bds/
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| User: "tadchem" |
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| Title: Re: Help with Lagrangian mechanics |
15 Oct 2007 07:40:21 PM |
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On Oct 15, 9:13 am, Bruce Scott TOK <Use-Author-Supplied-Address-
Header@[127.1]> wrote:
I'm having trouble specifying the speed of the wedge, so I can't come
up with the kinetic energy. Obviously, however, the wedge's potential
energy is zero.
Use the centers of gravity of the wedge and the particle for your two
position coordinates. Express the Lagrangian in terms of those
coordinates. If there is no rotation involved, that's all you need.
As an exercise, you should be able to prove that the momentum and
kinetic energy of a distributed object is given by those of its C/M.
Rotation is slightly trickier, but only slightly (if you're good at
doing integrals).
--
ciao,
Bruce
drift wave turbulence: http://www.rzg.mpg.de/~bds/
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
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| User: "Bruce Scott TOK" |
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| Title: Re: Help with Lagrangian mechanics |
17 Oct 2007 04:38:20 AM |
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Tom Davidson misunderstood me...
On Oct 15, 9:13 am, Bruce Scott TOK <Use-Author-Supplied-Address-
Header@[127.1]> wrote:
I'm having trouble specifying the speed of the wedge, so I can't come
up with the kinetic energy. Obviously, however, the wedge's potential
energy is zero.
Use the centers of gravity of the wedge and the particle for your two
position coordinates. Express the Lagrangian in terms of those
coordinates. If there is no rotation involved, that's all you need.
I didn't say use the CG of the system, but to use the CG of each object
as each of two coordinates in the Lagrangian. The formulation of the
problem suggests the wedge will not rotate (otherwise, you would use an
angle of its rotation as an additional coordinate).
The directions of motion of the two CG's may suggest using a _vector_
coordinate for each of them. In 2D motion you have 4 dependent
variables (etc).
--
ciao,
Bruce
drift wave turbulence: http://www.rzg.mpg.de/~bds/
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| User: "tadchem" |
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| Title: Re: Help with Lagrangian mechanics |
14 Oct 2007 07:12:51 PM |
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On Oct 14, 5:46 pm, bit188 <mari...@webclique.net> wrote:
I'm stuck on a problem involving the Lagrangian formulation of
classical mechanics. Here is the problem and what I've done so far;
help is greatly appreciated.
(Problem: A wedge of mass M and angle [alpha] slides freely on a
horizontal plane. A particle of mass m slides freely on the wedge.
Determine the motion of the particle as well as that of the wedge.)
Solution (so far): Let l denote the length of the hypotenuse of the
wedge. Our kinetic and potential energies are
T_m = (m/2)[(dl/dt)^2]
V_m = -mglsin[alpha]
I'm having trouble specifying the speed of the wedge, so I can't come
up with the kinetic energy. Obviously, however, the wedge's potential
energy is zero.
Thanks again.
Back to basics:
The Lagrangian function L is defined as the difference of the kinetic
T and potential V energies:
L = T - V
T is expressed in terms of the general position coordinates (q) and
general velocities (q-dot), while V depends solely on position
coordinates (q).
The Lagranges equations for _each_ of the particles in the system
takes the form:
d[ d(L) / d(q-dot) ] / dt = d(L) / d(q)
These equations are separate equations for each particle in the system
and for each orthogonal coordinate.
They serve to guide you in the selection of a coordinate system in
which momenta - the d(L) / d(q-dot) term - is conserved and the
equations of motion are separable.
Hints:
Both particles will move laterally in opposite directions to conserve
the horizontal position of the center of mass.
The energy to move *both* the wedge and the mass will come solely from
the potential energy change of the sliding mass as it moves downward
(vertically).
Each object has its own mass and motion.
This should suggest a stationary frame of reference passing through
the initial center of mass.
Set up the Lagrange equations for the initial equation using the
suggested frame of reference and integrate.
Tom Davidson
Richmond, VA
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| User: "Thomas" |
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| Title: Re: Help with Lagrangian mechanics |
17 Oct 2007 10:49:02 AM |
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bit188 wrote:
I'm stuck on a problem involving the Lagrangian formulation of
classical mechanics. Here is the problem and what I've done so far;
help is greatly appreciated.
(Problem: A wedge of mass M and angle [alpha] slides freely on a
horizontal plane. A particle of mass m slides freely on the wedge.
Determine the motion of the particle as well as that of the wedge.)
Solution (so far): Let l denote the length of the hypotenuse of the
wedge. Our kinetic and potential energies are
T_m = (m/2)[(dl/dt)^2]
V_m = -mglsin[alpha]
I'm having trouble specifying the speed of the wedge, so I can't come
up with the kinetic energy. Obviously, however, the wedge's potential
energy is zero.
Thanks again.
Do you necessarily need the Lagrangian formulation? The point is that
it (like the Hamiltonian formulation) often tends to make things
unnecessarily complicated in comparison to the Newtonian form:
here the gravitational force on the particle along the slope of the
wedge is obviously m*g*sin(alpha), which you can decompose into the
vertical and horizontal components m*g*sin^2(alpha) and
m*g*sin(alpha)*cos(alpha) respectively (that is the horizontal
acceleration of the particle is g*sin(alpha)*cos(alpha)). So according
to Newton's third law, the horizontal force on the wedge will be equal
and opposite to the horizontal force on the particle and thus its
acceleration is -m/M*g*sin(alpha)*cos(alpha).
Thomas
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| User: "tadchem" |
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| Title: Re: Help with Lagrangian mechanics |
18 Oct 2007 02:39:25 AM |
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On Oct 17, 11:49 am, Thomas <thomas.s...@gmail.com> wrote:
bit188 wrote:
I'm stuck on a problem involving the Lagrangian formulation of
classical mechanics. Here is the problem and what I've done so far;
help is greatly appreciated.
(Problem: A wedge of mass M and angle [alpha] slides freely on a
horizontal plane. A particle of mass m slides freely on the wedge.
Determine the motion of the particle as well as that of the wedge.)
Solution (so far): Let l denote the length of the hypotenuse of the
wedge. Our kinetic and potential energies are
T_m = (m/2)[(dl/dt)^2]
V_m = -mglsin[alpha]
I'm having trouble specifying the speed of the wedge, so I can't come
up with the kinetic energy. Obviously, however, the wedge's potential
energy is zero.
Thanks again.
Do you necessarily need the Lagrangian formulation? The point is that
it (like the Hamiltonian formulation) often tends to make things
unnecessarily complicated in comparison to the Newtonian form:
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.
here the gravitational force on the particle along the slope of the
wedge is obviously m*g*sin(alpha), which you can decompose into the
vertical and horizontal components
AH! The value of selecting a convenient frame of reference and system
of coordinates appears!
m*g*sin^2(alpha) and
m*g*sin(alpha)*cos(alpha) respectively (that is the horizontal
acceleration of the particle is g*sin(alpha)*cos(alpha)). So according
to Newton's third law, the horizontal force on the wedge will be equal
and opposite to the horizontal force on the particle and thus its
acceleration is -m/M*g*sin(alpha)*cos(alpha).
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.
Thomas
Tom Davidson
Richmond, VA
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| User: "Puppet_Sock" |
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| Title: Re: Help with Lagrangian mechanics |
17 Oct 2007 05:02:55 PM |
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On Oct 17, 11:49 am, Thomas <thomas.s...@gmail.com> wrote:
[snip]
Do you necessarily need the Lagrangian formulation?
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
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| User: "mL" |
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| Title: Re: Help with Lagrangian mechanics |
18 Oct 2007 05:42:06 PM |
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Thomas wrote:
Do you necessarily need the Lagrangian formulation?
No, but in this case (with 2 degrees of freedom) the
Lagrangian method is more straightforward and easier
to apply.
The point is that it (like the Hamiltonian formulation)
often tends to make things unnecessarily complicated in
comparison to the Newtonian form: here the gravitational
force on the particle along the slope of the wedge is
obviously m*g*sin(alpha), which you can decompose into the
vertical and horizontal components m*g*sin^2(alpha) and
m*g*sin(alpha)*cos(alpha) respectively (that is the horizontal
acceleration of the particle is g*sin(alpha)*cos(alpha)).
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.
So accordingto Newton's third law, the horizontal force on the
wedge will be equal and opposite to the horizontal force on the
particle and thus its acceleration is -m/M*g*sin(alpha)*cos(alpha).
No.
/mL
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| User: "Thomas" |
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| Title: Re: Help with Lagrangian mechanics |
19 Oct 2007 12:01:26 PM |
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On 18 Oct, 23:42, mL <mL.bey...@elsewhere.xxx> wrote:
Thomas wrote:
Do you necessarily need theLagrangianformulation?
No, but in this case (with 2 degrees of freedom) theLagrangianmethod is more straightforward and easier
to apply.
The point is that it (like the Hamiltonian formulation)
often tends to make things unnecessarily complicated in
comparison to the Newtonian form: here the gravitational
force on the particle along the slope of the wedge is
obviously m*g*sin(alpha), which you can decompose into the
vertical and horizontal components m*g*sin^2(alpha) and
m*g*sin(alpha)*cos(alpha) respectively (that is the horizontal
acceleration of the particle is g*sin(alpha)*cos(alpha)).
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.
The horizontal component of the contact force IS exactly
m*g*sin(alpha)*cos(alpha) , and since we have a closed system, this
must be identical for both the particle and the wedge but with a
different sign. So this should result then in the horizontal
accelerations g*sin(alpha)*cos(alpha) for the particle and -m/
M*g*sin(alpha)*cos(alpha) for the wedge.
But if you arrive at a different solution, why don't you post it here?
Thomas
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| User: "mL" |
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| Title: Re: Help with Lagrangian mechanics |
19 Oct 2007 04:56:19 PM |
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Thomas wrote:
On 18 Oct, 23:42, mL <mL.bey...@elsewhere.xxx> wrote:
Thomas wrote:
Do you necessarily need theLagrangianformulation?
No, but in this case (with 2 degrees of freedom) the Lagrangian
method is more straightforward and easier to apply.
The point is that it (like the Hamiltonian formulation)
often tends to make things unnecessarily complicated in
comparison to the Newtonian form: here the gravitational
force on the particle along the slope of the wedge is
obviously m*g*sin(alpha), which you can decompose into the
vertical and horizontal components m*g*sin^2(alpha) and
m*g*sin(alpha)*cos(alpha) respectively (that is the horizontal
acceleration of the particle is g*sin(alpha)*cos(alpha)).
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.
The horizontal component of the contact force IS exactly
m*g*sin(alpha)*cos(alpha) ,
No - the particle accelerates! By Newton's 2nd law, F = ma,
components of the two acting forces don't sum up to zero.
... and since we have a closed system, this
must be identical for both the particle and the wedge but with a
different sign. So this should result then in the horizontal
accelerations g*sin(alpha)*cos(alpha) for the particle and -m/
M*g*sin(alpha)*cos(alpha) for the wedge.
But if you arrive at a different solution, why don't you post it here?
OK - some steps. Axes, and a crude FBD of particle m :
. N
. /
./
+----x m
| |
|y |
mg
F = ma (a = alpha; x' = dx/dt, x" = dx'/dt, etc):
(1) N sin(a) = mx"
(2) mg - N cos(a) = my"
Similarly for the wedge (coordinate X):
(3) - N sin(a) = M X"
To solve for X" you need, besides eq (1)-(3), a kimematic
equation that relates X", x" and y". Try it!
/mL
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| User: "Thomas" |
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| Title: Re: Help with Lagrangian mechanics |
20 Oct 2007 11:24:52 AM |
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mL wrote:
Thomas wrote:
On 18 Oct, 23:42, mL <mL.bey...@elsewhere.xxx> wrote:
Thomas wrote:
Do you necessarily need theLagrangianformulation?
No, but in this case (with 2 degrees of freedom) the Lagrangian
method is more straightforward and easier to apply.
The point is that it (like the Hamiltonian formulation)
often tends to make things unnecessarily complicated in
comparison to the Newtonian form: here the gravitational
force on the particle along the slope of the wedge is
obviously m*g*sin(alpha), which you can decompose into the
vertical and horizontal components m*g*sin^2(alpha) and
m*g*sin(alpha)*cos(alpha) respectively (that is the horizontal
acceleration of the particle is g*sin(alpha)*cos(alpha)).
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.
The horizontal component of the contact force IS exactly
m*g*sin(alpha)*cos(alpha) ,
No - the particle accelerates! By Newton's 2nd law, F = ma,
components of the two acting forces don't sum up to zero.
You still seem to be misunderstanding what I am saying:
it is the overall horizontal force that is zero, not the force on the
particle or wedge individually. Since there is no external horizontal
force, we must have according to Newton's third law F12=-F21 and thus
F=F12+F21=0.
As an example, assume you push a person on roller skates with a force
F12 whilst being yourself on roller skates. If the other person has
mass m1 and you have mass m2, this will accelerate the other person
forwards with F12/m1 and you backwards with -F12/m2.
... and since we have a closed system, this
must be identical for both the particle and the wedge but with a
different sign. So this should result then in the horizontal
accelerations g*sin(alpha)*cos(alpha) for the particle and -m/
M*g*sin(alpha)*cos(alpha) for the wedge.
But if you arrive at a different solution, why don't you post it here?
OK - some steps. Axes, and a crude FBD of particle m :
. N
. /
./
+----x m
| |
|y |
mg
F = ma (a = alpha; x' = dx/dt, x" = dx'/dt, etc):
(1) N sin(a) = mx"
(2) mg - N cos(a) = my"
Similarly for the wedge (coordinate X):
(3) - N sin(a) = M X"
To solve for X" you need, besides eq (1)-(3), a kimematic
equation that relates X", x" and y". Try it!
There are no further equations needed. You just need to specify N,
which is N=m*g*cos(alpha) (i.e. the gravitational force acting on
the particle projected on the slope of the wedge).
So according to your equations, this gives then
(1) mx'' = m*g*sin(alpha)*cos(alpha)
(2) my''= mg*(1-cos^2(alpha)) = m*g*sin^2(alpha)
(3) MX'' = -m*g*sin(alpha)*cos(alpha)
which is exactly the result I gave above already.
Thomas
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| User: "PD" |
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| Title: Re: Help with Lagrangian mechanics |
20 Oct 2007 11:50:11 AM |
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On Oct 20, 11:24 am, Thomas <thomas.s...@gmail.com> wrote:
mL wrote:
Thomas wrote:
On 18 Oct, 23:42, mL <mL.bey...@elsewhere.xxx> wrote:
Thomas wrote:
Do you necessarily need theLagrangianformulation?
No, but in this case (with 2 degrees of freedom) the Lagrangian
method is more straightforward and easier to apply.
The point is that it (like the Hamiltonian formulation)
often tends to make things unnecessarily complicated in
comparison to the Newtonian form: here the gravitational
force on the particle along the slope of the wedge is
obviously m*g*sin(alpha), which you can decompose into the
vertical and horizontal components m*g*sin^2(alpha) and
m*g*sin(alpha)*cos(alpha) respectively (that is the horizontal
acceleration of the particle is g*sin(alpha)*cos(alpha)).
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.
The horizontal component of the contact force IS exactly
m*g*sin(alpha)*cos(alpha) ,
No - the particle accelerates! By Newton's 2nd law, F = ma,
components of the two acting forces don't sum up to zero.
You still seem to be misunderstanding what I am saying:
it is the overall horizontal force that is zero, not the force on the
particle or wedge individually. Since there is no external horizontal
force, we must have according to Newton's third law F12=-F21 and thus
F=F12+F21=0.
This is a common mistake. Misusing Newton's 3rd law gives the
misimpression that the sum of forces is always zero. If that were
true, then the 2nd law would always have zero on both sides.
Returning back to free-body diagrams, it's crucial to remember that
behavior of that body (and that's the motion you're trying to solve
for, recall) depends on the forces acting ON THAT BODY. You do NOT
include in that list the forces that body exerts on other bodies.
The task at hand is to find the motions of the wedge and the particle
-- those individual bodies -- not the motion of the combined system.
As an example, assume you push a person on roller skates with a force
F12 whilst being yourself on roller skates. If the other person has
mass m1 and you have mass m2, this will accelerate the other person
forwards with F12/m1 and you backwards with -F12/m2.
... and since we have a closed system, this
must be identical for both the particle and the wedge but with a
different sign. So this should result then in the horizontal
accelerations g*sin(alpha)*cos(alpha) for the particle and -m/
M*g*sin(alpha)*cos(alpha) for the wedge.
But if you arrive at a different solution, why don't you post it here?
OK - some steps. Axes, and a crude FBD of particle m :
. N
. /
./
+----x m
| |
|y |
mg
F = ma (a = alpha; x' = dx/dt, x" = dx'/dt, etc):
(1) N sin(a) = mx"
(2) mg - N cos(a) = my"
Similarly for the wedge (coordinate X):
(3) - N sin(a) = M X"
To solve for X" you need, besides eq (1)-(3), a kimematic
equation that relates X", x" and y". Try it!
There are no further equations needed. You just need to specify N,
which is N=m*g*cos(alpha) (i.e. the gravitational force acting on
the particle projected on the slope of the wedge).
So according to your equations, this gives then
(1) mx'' = m*g*sin(alpha)*cos(alpha)
(2) my''= mg*(1-cos^2(alpha)) = m*g*sin^2(alpha)
(3) MX'' = -m*g*sin(alpha)*cos(alpha)
which is exactly the result I gave above already.
Thomas
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| User: "mL" |
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| Title: Re: Help with Lagrangian mechanics |
20 Oct 2007 01:54:24 PM |
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Thomas wrote:
mL wrote:
Thomas wrote:
On 18 Oct, 23:42, mL <mL.bey...@elsewhere.xxx> wrote:
Thomas wrote:
Do you necessarily need theLagrangianformulation?
No, but in this case (with 2 degrees of freedom) the Lagrangian
method is more straightforward and easier to apply.
The point is that it (like the Hamiltonian formulation)
often tends to make things unnecessarily complicated in
comparison to the Newtonian form: here the gravitational
force on the particle along the slope of the wedge is
obviously m*g*sin(alpha), which you can decompose into the
vertical and horizontal components m*g*sin^2(alpha) and
m*g*sin(alpha)*cos(alpha) respectively (that is the horizontal
acceleration of the particle is g*sin(alpha)*cos(alpha)).
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.
The horizontal component of the contact force IS exactly
m*g*sin(alpha)*cos(alpha) ,
No - the particle accelerates! By Newton's 2nd law, F = ma,
components of the two acting forces don't sum up to zero.
You still seem to be misunderstanding what I am saying:
it is the overall horizontal force that is zero, not the force on the
particle or wedge individually. Since there is no external horizontal
force, we must have according to Newton's third law F12=-F21 and thus
F=F12+F21=0.
Yes, the particle and the wedge interact via a force pair
N /-N [which is accounted for in eq (1) and (3)].
[...]
... and since we have a closed system, this
must be identical for both the particle and the wedge but with a
different sign. So this should result then in the horizontal
accelerations g*sin(alpha)*cos(alpha) for the particle and -m/
M*g*sin(alpha)*cos(alpha) for the wedge.
But if you arrive at a different solution, why don't you post it here?
OK - some steps. Axes, and a crude FBD of particle m :
. N
. /
./
+----x m
| |
|y |
mg
F = ma (a = alpha; x' = dx/dt, x" = dx'/dt, etc):
(1) N sin(a) = mx"
(2) mg - N cos(a) = my"
Similarly for the wedge (coordinate X):
(3) - N sin(a) = M X"
To solve for X" you need, besides eq (1)-(3), a kimematic
equation that relates X", x" and y". Try it!
There are no further equations needed. You just need to specify N,
which is N=m*g*cos(alpha) (i.e. the gravitational force acting on
the particle projected on the slope of the wedge).
No, Newton's 3rd law does _not_ give you N; i.e. you need one
more equation.
So according to your equations, this gives then
(1) mx'' = m*g*sin(alpha)*cos(alpha)
(2) my''= mg*(1-cos^2(alpha)) = m*g*sin^2(alpha)
(3) MX'' = -m*g*sin(alpha)*cos(alpha)
which is exactly the result I gave above already.
Yes, but this result is wrong.
/mL
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| User: "bit188" |
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| Title: Re: Help with Lagrangian mechanics |
30 Oct 2007 01:05:33 AM |
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Sorry for not getting back sooner. I figured it out -- here are my
results. If there are any errors, please let me know.
Generalized coordinates are X and x, the position of the wedge and the
distance along the hypotenuse from the thin edge of the wedge to m,
respectively. I got
x" = [(M+m)gsin(alpha)cos(alpha)]/M+msin^2(alpha)
X" = [mgsin(alpha)cos(alpha)]/M+msin^2(alpha)
Thanks.
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| User: "mL" |
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| Title: Re: Help with Lagrangian mechanics |
31 Oct 2007 05:36:55 AM |
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bit188 skrev:
Sorry for not getting back sooner. I figured it out -- here are my
results. If there are any errors, please let me know.
Generalized coordinates are X and x, the position of the wedge and the
distance along the hypotenuse from the thin edge of the wedge to m,
respectively. I got
x" = [(M+m)gsin(alpha)cos(alpha)]/M+msin^2(alpha)
X" = [mgsin(alpha)cos(alpha)]/M+msin^2(alpha)
The result for X" looks OK (for a proper choice of
the X direction), but the expression for x" can't
be right. Check it by invoking momentum conservation
in the X-direction for particle + wedge :
MX' + mx'cos(alpha) = const,
so MX" + mx"cos(alpha) = 0,
and x" = ...
/mel
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| User: "mL" |
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| Title: Re: Help with Lagrangian mechanics |
19 Oct 2007 05:25:34 PM |
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mL wrote:
[...]
..> F = ma (a = alpha; x' = dx/dt, x" = dx'/dt, etc):
Oops, used the symbol 'a' with two different
meanings there.
[...]
/mL
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