# Homework Help: Lagrangian Mechanics question

1. Mar 23, 2012

### Lengalicious

1. The problem statement, all variables and given/known data
(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?

2. Relevant 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 dont understand whether it is y dependant or x dependant, would it be y dependant because the potential energy given in part 2 is y dependant 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 im taking a complete guess at momentum because F=ydot=dp/dt. No idea very stuck =/ if someone could explain would be much appreciated thanks.

Last edited: Mar 23, 2012
2. Mar 23, 2012

### 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)

3. Mar 23, 2012

### Lengalicious

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

4. Mar 23, 2012

### BruceW

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

5. Mar 23, 2012

### Lengalicious

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

6. Mar 23, 2012

### BruceW

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

7. Mar 23, 2012

### 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

Last edited: Mar 23, 2012
8. Mar 23, 2012

### 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?

9. Mar 23, 2012

### Lengalicious

Erm what i've learnt 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?

Last edited: Mar 23, 2012
10. Mar 23, 2012

### 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).

11. Mar 23, 2012

### Lengalicious

ok so ∂L/∂xdot = 1/2m(4xdot^3+4xdotydot^2), and ∂L/∂x = 0 since T is velocity dependant 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?

12. Mar 23, 2012

### 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)

13. Mar 23, 2012

### 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

Last edited: Mar 23, 2012
14. Mar 23, 2012

### 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.

15. Mar 23, 2012

### 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?

16. Mar 23, 2012

### BruceW

yep, that's right.

17. Mar 23, 2012

### 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 dont understand how to spot what is conserved =/

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

Last edited: Mar 23, 2012
18. Mar 23, 2012

### 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?

19. Mar 23, 2012

### 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

20. Mar 23, 2012

### 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).