Conservation Laws in Lagrangian Mechanics

[Lengalicious]

Homework Statement

(i) A particle of mass m moves in the x - y plane. Its coordinates are x(t) and y(t).
What is the kinetic energy of this particle?

(ii) The potential energy of this particle is V (y). The actual form of V will remain
unspecied, except that it depends only on the y coordinate. Write down the Lagrangian
for this particle. Obtain the equation of motion.

(iii) Apart from the energy, what other physical quantity is conserved, and why?

Homework Equations

N/a 3. The Attempt at a Solution [/b
Ok so I'm new to Lagrangian, anyway from what i understand would the kinetic energy just be? T(v)=1/2mv^2 ; where v = ydot, but i don't understand whether it is y dependent or x dependent, would it be y dependent because the potential energy given in part 2 is y dependent so they act in same plane?
In part 2, L(y,v) = T(y,v)-V(y,v) ; where v=ydot, so
after using the lagrangians equation d(mv)/dt=ydot since y is not specified?
and for part 3 I am taking a complete guess at momentum because F=ydot=dp/dt. No idea very stuck =/ if someone could explain would be much appreciated thanks.

 

[BruceW]

starting with the first question: It tells you the coordinates are x and y. Don't think too much about this, what would be the speed of the particle? (don't think ahead to the next question)

 

[Lengalicious]

speed of the particle is xdot? if its displacement is x. or ydot if its displacement is y =/

 

[BruceW]

and what about if it has some general displacement in x and/or y?

 

[Lengalicious]

T(xdot,ydot)=1/2mv(xdot,ydot)^2? Not sure

 

[BruceW]

you just write the velocity as a vector, then do the dot product with itself, what will this get?

 

[Lengalicious]

ohh so EDIT: (xdot,ydot)*(xdot,xydot) = xdot^2+ydot^2? So would it be T(xdot,ydot)=1/2m(xdot^2+ydot^2)^2

 

[BruceW]

exactly. So you've got question 1, now for question 2. What is the Lagrangian? And then, what are the equations of motion? (I'm guessing they are asking for the Euler-Lagrange equations here). Have you learned about them?

 

[Lengalicious]

Erm what I've learned is that when L(x,xdot)=T-V where T = KE V = PE, L is the lagrangian, where the equation of motion is d(∂L/∂xdot)/dt=∂L/∂x, is that Euler lagrange?

 

[BruceW]

Yes, that is the Euler-Lagrange equation for x. And in this problem, we also have a y coordinate. So what is the Euler-Lagrange equation for y? (I don't know if you have used Lagrangian mechanics with multiple coordinates before, but you can probably just guess what the equation for y will be).

 

[Lengalicious]

ok so ∂L/∂xdot = 1/2m(4xdot^3+4xdotydot^2), and ∂L/∂x = 0 since T is velocity dependent and V has only y dependance
then: ∂L/∂ydot = 1/2m(4ydotxdot^2+4ydot^3), and ∂L/∂y = V(y)prime?
So d(1/2*m*(4xdot^3+4xdot*ydot^2))/dt=0 and d(1/2m(4ydotxdot^2+4ydot^3))/dt=V(y)prime, but what is V(y)prime? and once i have come this far, then what do i do?

 

[BruceW]

I'm not sure about your value for ∂L/∂xdot ... remember that you're holding ydot constant when you do this partial differentiation. (and you've got a similar problem for ∂L/∂ydot)

 

[Lengalicious]

L = 1/2*m*(xdot^2+ydot^2)^2 = 1/2*m*(xdot^4+2xdot^2*ydot^2+ydot^4) so when holding y constant ∂L/∂xdot = 1/2*m*(4xdot^3+4xdot*ydot^2) no? I really can't see where I'm going wrong

 

[BruceW]

Ah, whoops. Sorry, I should have spotted this from earlier. you've put T(xdot,ydot)=1/2m(xdot^2+ydot^2)^2 But this isn't quite right, because you don't need to square the brackets. You got the speed squared: (xdot,ydot)*(xdot,xydot) = xdot^2+ydot^2 so you don't need to square again to get T.

 

[Lengalicious]

Ahhhh ok taking the dot product of the velocity makes so much more sense to me now lol, aha, ok so once i take the partial derivative and substitute into lagrangian will that be end of part (ii)? I will have motion of equation for x coord and equation for y coord? Although the question just asks for singular motion equation not plural, so do i have to combine the 2 some how?

 

[BruceW]

yep, that's right.

 

[Lengalicious]

Ok thanks very much, for the last part will it become obvious what the other conserved quantity is when i have the lagrangian? If so how do i spot it? EDIT: So i get d(1/2*m*2xdot)/dt=0 and d(1/2*m*2ydot)/dt=∂V(y)/∂y for the lagrangians, from this i don't understand how to spot what is conserved =/

EDIT: So i get m*xdotdot = F = 0 and m*ydotdot=∂V(y)/∂y

 

[BruceW]

you're looking for something complicated, but its nice and simple. You've got d(m xdot)/dt = 0, so what is conserved with time in this equation?

 

[Lengalicious]

The Force?, and what about the other equation? I take it i disreguard that since its not = to 0

EDIT: The question also asks why? What would i answer to that bit?

EDIT: Ok actually is it the mass and velocity that are conserved or force, bit confused. Ohhhh m*v=momentum right? So its momentum

 

[BruceW]

In the equation for y, you're right, its not clear that anything is conserved, since we don't know the form of the potential yet. Also, your equation for y is not quite right. (remember that L=T-V).

In the equation for x, it's not the force which is conserved. For something to be conserved, we don't need it to equal zero, we just require that it doesn't change with time. (I guess you could also say the force in the x direction is conserved, since it is always zero, but there is another thing which is conserved that people more often talk about).

 

[Lengalicious]

oh yeh bit of a mistake i disreguarded the V, so ∂L/∂ydot = 1/2*m*2ydot-∂V(y)/∂ydot? And yeh d(mv)/dt=0 so its the momentum that is conserved as it does not vary with time :D

 

[Lengalicious]

Anyway, thanks ALOOOOT, greatly appreciate all the help you've given me. Btw i notice your an undergrad right? What year if you don't mind me asking? Bit of a nooblet myself in the 1st year.

 

[BruceW]

yeah, it's the momentum which is conserved (technically, the x component of momentum, since the y component isn't necessarily conserved, since we don't know the form of the potential energy yet).

∂L/∂ydot = 1/2*m*2ydot-∂V(y)/∂ydot Yes this is true, and the ∂V(y)/∂ydot disappears, since the potential does not depend on speed (we hope). But on the other side of the equation, we've got ∂L/∂y which does not equal ∂V(y)/∂y because we know L=T-V, so what should the equation for y be?

 

[Lengalicious]

I missed the minus sign right? Anyway I am shattered, going to go bed thanks for all the help, appreciate it beyond comprehension, night.

 

[BruceW]

yep, no worries man. I'm going to sleep now too!