Horizontal Circular Motion With Lagrange

In summary, the stone is moving on a horizontal circular path with constant velocity, but the force required to turn it at a given radius and speed is not the same. The Lagrange equation gives 0 = 0, which is not what I'm looking for.
  • #1
Fascheue
17
3
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.
 
Physics news on Phys.org
  • #2
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.
 
  • Like
Likes Fascheue
  • #3
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.
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.
 

What is horizontal circular motion with Lagrange?

Horizontal circular motion with Lagrange refers to the study of objects moving in circular paths that are parallel to the ground, using the Lagrangian mechanics approach. This approach is based on the principle of least action and allows for the analysis of systems with multiple degrees of freedom.

What is the Lagrangian mechanics approach?

The Lagrangian mechanics approach is a mathematical method for solving problems in classical mechanics. It is based on the principle of least action, which states that the motion of a system is determined by minimizing the action, a quantity that combines the kinetic and potential energies of the system.

What is the significance of horizontal circular motion with Lagrange?

Horizontal circular motion with Lagrange is significant because it allows for the analysis of complex systems with multiple degrees of freedom, such as objects moving in circular paths on a horizontal plane. It also provides a more elegant and efficient way of solving problems in classical mechanics compared to traditional methods.

What are some real-world applications of horizontal circular motion with Lagrange?

Horizontal circular motion with Lagrange has various real-world applications, such as in the design of roller coasters, satellite orbits, and gyroscopes. It is also used in the analysis of planetary motion and the behavior of charged particles in magnetic fields.

What are the limitations of horizontal circular motion with Lagrange?

Although horizontal circular motion with Lagrange is a powerful tool for solving problems in classical mechanics, it has some limitations. It is not applicable to systems with non-conservative forces, such as friction or air resistance. It also does not take into account the effects of relativity and quantum mechanics, which are necessary for certain systems.

Similar threads

  • Introductory Physics Homework Help
2
Replies
55
Views
645
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
31
Views
1K
  • Introductory Physics Homework Help
Replies
20
Views
987
  • Introductory Physics Homework Help
2
Replies
40
Views
2K
  • Introductory Physics Homework Help
Replies
26
Views
407
  • Introductory Physics Homework Help
Replies
6
Views
959
  • Introductory Physics Homework Help
Replies
30
Views
496
  • Introductory Physics Homework Help
Replies
10
Views
2K
Back
Top