Horizontal Circular Motion With Lagrange

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SUMMARY

The discussion focuses on applying Lagrangian mechanics to analyze horizontal circular motion. Participants clarify that while the kinetic energy (T) remains constant due to constant speed, the potential energy (U) is also constant since elevation does not change. The Lagrangian is expressed as L = T - U, leading to a constraint equation for circular motion, 0 = (x² + y²) - R², which must be incorporated into the Lagrangian using a Lagrange multiplier. This approach allows for the derivation of equations of motion for both x and y coordinates, along with an additional equation for the multiplier.

PREREQUISITES
  • Understanding of Lagrangian mechanics and the Lagrange equation
  • Familiarity with kinetic energy (T) and potential energy (U) concepts
  • Knowledge of circular motion dynamics and constraints
  • Ability to perform calculus operations, including derivatives
NEXT STEPS
  • Study the derivation of the Lagrangian for constrained systems
  • Learn about Lagrange multipliers and their application in mechanics
  • Explore the relationship between circular motion and centripetal force
  • Investigate the implications of two-dimensional motion in Lagrangian mechanics
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in advanced mechanics, particularly those studying Lagrangian dynamics and circular motion analysis.

Fascheue
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Homework Statement
A 12 kg stone is tied to a 2.7 m “massless” rope, which can deal with tensions of up to 210 N. Forcing the stone on a circular trajectory in a horizontal plane using the rope, how fast can it be rotated uniformly before the rope likely gets torn apart? What happens for a rope that is twice as long? Neglect air resistance effects.
Take into account that the rotating rope holding the stone will make an angle α < π/2 with the vertical direction (i.e. the axis of rotation). The faster the stone goes around, the closer the angle α gets to this limit.
If you solved the problem using Newton’s formulation of classical mechanics, try using the Lagrangian approach next!
Relevant Equations
L = T - U
pd(L)/pd(x) = d/dt(pd(L)/pd(x’))
Where pd represents a partial derivative.
In the situation described in the problem, the mass is moving on a horizontal circular path with constant velocity. Wouldn’t this make L and U both constant? Then the Lagrange equation would give 0 = 0, which isn’t what I’m looking for. Any help would be appreciated.
 
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You can approach this problem in various ways. Maybe you should do it the very easiest way you can think of first so you have something to put on your bulletin board to cheer you up.

Velocity has both magnitude and direction. Is circular motion really a constant velocity? Can you write down a formula giving the position, x and y, for the stone as it undergoes circular motion? Can you take the first and second derivatives of this? Can you then get the acceleration? What about the magnitude of the acceleration?

Circular motion: What is the force required to turn the stone at a given radius and speed? You might start with the back-of-the-text equation for this. Should be a quick solve. Do it with variables first. Then solve for the speed that breaks the rope. Then substitute the known values for case 1 then for case 2.

Not instantly obvious what they mean by "Newton's formulation." Probably F=ma. That means you need to write down the equation for circular motion, and solve for the acceleration. Hey, you just did that if you followed the previous paragraphs.

As to the Lagrange formulation: Maybe you should actually write down the T and U terms for a particle undergoing acceleration, but not changing elevation, and see what you think from there. Then see if you can derive an equation involving, for example, position and momentum. Then see if you can relate that to circular motion.
 
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DEvens said:
You can approach this problem in various ways. Maybe you should do it the very easiest way you can think of first so you have something to put on your bulletin board to cheer you up.

Velocity has both magnitude and direction. Is circular motion really a constant velocity? Can you write down a formula giving the position, x and y, for the stone as it undergoes circular motion? Can you take the first and second derivatives of this? Can you then get the acceleration? What about the magnitude of the acceleration?

Circular motion: What is the force required to turn the stone at a given radius and speed? You might start with the back-of-the-text equation for this. Should be a quick solve. Do it with variables first. Then solve for the speed that breaks the rope. Then substitute the known values for case 1 then for case 2.

Not instantly obvious what they mean by "Newton's formulation." Probably F=ma. That means you need to write down the equation for circular motion, and solve for the acceleration. Hey, you just did that if you followed the previous paragraphs.

As to the Lagrange formulation: Maybe you should actually write down the T and U terms for a particle undergoing acceleration, but not changing elevation, and see what you think from there. Then see if you can derive an equation involving, for example, position and momentum. Then see if you can relate that to circular motion.
I meant to write constant speed, not constant velocity. Also, I forgot to mention that I’ve already solved the problem with Newtonian mechanics. I’m struggling with the Lagrangian part of the problem.

Wouldn’t the Lagrangian just be a constant in this case? If the speed is constant, T should be constant. If the elevation is unchanging U seems like it should be constant as well.
 
Fascheue said:
Wouldn’t the Lagrangian just be a constant in this case? If the speed is constant, T should be constant. If the elevation is unchanging U seems like it should be constant as well.

I said to write it down. What's the formula for the Lagrangian? And how does the acceleration enter? If you just ignore me when I make suggestions, I will wander away.
 
DEvens said:
I said to write it down. What's the formula for the Lagrangian? And how does the acceleration enter? If you just ignore me when I make suggestions, I will wander away.
L = T - U

T = (1/2)Mx’ ^2
U = mgh

If I plug these values into the Lagrange equation the right-hand side becomes ma and the left-hand side becomes 0.
 
One last try. How do you introduce the fact that the stone is forced to go in a circle into the Lagrangian? And what happened to the y coordinate? The stone is not only moving in the x direction.

There's a reason this was at the end of the problem. You've got a 2-D problem that is constrained. And that can be dealt with in the Lagrange formalism. It's usual for the Lagrange formalism, but the Lagrange formalism is often shied away from in constrained problems, because people don't know the appropriate method.

What is the constraint? The stone must go in a circle. So you have something like this.

##0 = (x^2 + y^2) - R^2 ##

Then you add that to the Lagrangian with a Lagrange undetermined multiplier.

## L = \frac{1}{2} M [ (x^\prime )^2 + (y^\prime )^2 ] +\Lambda (x^2 + y^2 - R^2) ##

Then you read up on how to deal with that. Note that you have an extra coordinate in your system now, the Λ. So you wind up with an equation of motion for x, another for y, and a third for Λ. But there's something pretty special about the Λ equation. Fun wow!

I guess I was assuming that, because your homework assignment included "try it with the Lagrange formalism" that you had been taught this.
 

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