Lagrangian function for beetle on paper

In summary, the conversation is about a circle with a radius of ##a##, diameter ##AB##, and moment of inertia ##4ma^2## being drawn on a sheet of paper that is pivoted at ##A##. An insect of mass ##m## walks around the circle at a uniform speed of ##2a\Omega## relative to the paper. The angle turned through by the paper, measured from an initial instant at which the insect passes through ##B## and at which the paper is momentarily at rest, is taken as a generalized coordinate, denoted by ##q##. The corresponding generalized momentum is denoted by ##p##. The Lagrangian function for the system is given by $$L = 2
  • #1
CAF123
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Homework Statement


[/B]
A circle of radius ##a##, with diameter ##AB##, is drawn on a sheet of paper which lies on a smooth horizontal table. The paper is pivoted with a pin at ##A## and has moment of inertia ##4ma^2## about a vertical axis through ##A##. An insect of mass ##m## walks around the circle with uniform speed ##2a\Omega## relative to the paper. The angle ##q## turned through by the paper, measured from an initial instant at which the insect passes through ##B## and at which the paper is momentarily at rest, is taken as a generalised co-ordinate; ##p## denotes the corresponding generalised momentum. Show that the lagrangian function for the system is
$$L = 2ma^2 [(1 + \cos^2 \Omega t)\dot{q}^2 + 2\Omega \dot{q} \cos^2 \Omega t]$$

2. Homework Equations

L = T - V

The Attempt at a Solution


My set up is shown in a picture below. ##q## is measured from ##AB## to the vertical axis and ##\Omega## is measured from ##AB## to the line connecting the centre of the circle to the mass ##m##. With this, $$x = a \cos \left(\frac{\pi}{2} - q\right) + a \cos\left(\frac{\pi}{2} - (q-\Omega)\right)$$ and similar expression for ##y## are the coordinates of the beetle in this frame. I just wanted to check if my set up is fine and that I have defined the angles ##q## and ##\Omega## correctly as given in the question.
 

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  • #2
CAF123 said:
$$x = a \cos \left(\frac{\pi}{2} - q\right) + a \cos\left(\frac{\pi}{2} - (q-\Omega)\right)$$ and similar expression for ##y## are the coordinates of the beetle in this frame.

Note that q and Ω do not have the same dimensions. So, you can't have (q - Ω) in your last term. Also, did you drop a factor of 2? Note that the speed of the bug relative to the paper is 2aΩ.

I just wanted to check if my set up is fine and that I have defined the angles ##q## and ##\Omega## correctly as given in the question.

I think you are setting it up OK. When I crank through to get the Lagrangian, I get an opposite sign for the last term in L. But, that might be due to a choice of which direction is positive for defining the angle q.
 
  • #3
TSny said:
Note that q and Ω do not have the same dimensions. So, you can't have (q - Ω) in your last term. Also, did you drop a factor of 2? Note that the speed of the bug relative to the paper is 2aΩ.
Ah yes, so define ##2\phi## as the angle from AB to the line connecting to the mass from the centre of the circle. Then ##v_t = a (2\dot{\phi}) = 2a \Omega##, ##v_t## is the tangential velocity of the bug relative to the circle.

Then the equations for the position of the beetle are $$x = a \sin q + a(\sin q \cos 2\phi - \cos q \sin 2\phi) \,\,;\,\,y = a \cos q + a(\cos q \cos 2\phi - \sin q \sin 2\phi)$$
 
  • #4
OK, but I don't agree with the sign of your last term in y.
 
  • #5
TSny said:
OK, but I don't agree with the sign of your last term in y.
Ok I have the lagrangian function now although I have a different sign in the last term which I think you said you also had. The next part of the problem was to derive the Hamiltonian function, which I did (using the Lagrangian given in the question with '+' in last term) and this is $$H(q,p,t) = \frac{(p-4ma^2\Omega \cos^2 \Omega t)^2}{8ma^2(1+\cos^2 \Omega t)} + \text{const}$$ It is a show that, so this is the given answer on sheet too.

Then write down Hamilton's equations of motion and deduce that by the time the bug reaches A the paper will have turned through an angle ##\frac{1}{2}(\sqrt{2}-1)\pi##.

Attempt: Hamilton's eqns of motion are $$\dot{q} = \frac{\partial H}{\partial p} = \frac{(p - 4ma^2 \Omega \cos^2 \Omega t)}{4ma^2(1+\cos^2 \Omega t)}$$ and $$\dot{p} = -\frac{\partial H}{\partial q} = 0 \Rightarrow p = \text{const}$$

So using the first equation I get, $$4ma^2 \int_{q_i}^{q_f} dq = p \int_0^T \frac{1}{1+\cos^2 \Omega t}dt - 4ma^2 \Omega \int_0^T \frac{\cos^2\Omega t }{1+\cos^2\Omega t} dt$$ Now ##\phi = \Omega t## and ##2\phi \in [0,\pi] \Rightarrow \phi \in [0,\pi/2]##. So we have $$4ma^2 \Delta q = \frac{p}{\Omega}\int_0^{\pi/2} \frac{1}{1+\cos^2 \phi} d\phi - 4ma^2\int_0^{\pi/2} \frac{\cos^2 \phi}{1+\cos^2\phi} d\phi$$ My question is how do I eliminate the ##p##?

Many thanks.
Edit: I should also point out that in the question we are given the integral $$\int_{0}^{\pi/2} \frac{d\theta}{1+\cos^2 \theta} = \frac{\pi}{2\sqrt{2}}$$
 
  • #6
CAF123 said:
$$H(q,p,t) = \frac{(p-4ma^2\Omega \cos^2 \Omega t)^2}{8ma^2(1+\cos^2 \Omega t)} + \text{const}$$ It is a show that, so this is the given answer on sheet too.

With the sign conventions that you chose initially for ##q## and ##\Omega##, I get $$H(q,p,t) = \frac{(p+4ma^2\Omega \cos^2 \Omega t)^2}{8ma^2(1+\cos^2 \Omega t)}$$ but I think the sign difference has to do with your choosing q positive in the clockwise direction and ##\Omega## positive in the counterclockwise direction.

Then write down Hamilton's equations of motion and deduce that by the time the bug reaches A the paper will have turned through an angle ##\frac{1}{2}(\sqrt{2}-1)\pi##.

I'm probably making a mistake somewhere, but I'm getting an answer of ##\frac{1}{2\sqrt{2}}(\sqrt{2}-1)\pi## [EDIT: I get this answer if I assume that the bug and paper are both initially at rest and then the bug suddenly starts walking. I get the answer given in the problem if I assume that at t = 0, the paper is at rest but the bug is already in motion. After re-reading the problem I see that the latter interpretation is intended. So, I'm now agreeing with the answer stated in the problem.]
Attempt: Hamilton's eqns of motion are $$\dot{q} = \frac{\partial H}{\partial p} = \frac{(p - 4ma^2 \Omega \cos^2 \Omega t)}{4ma^2(1+\cos^2 \Omega t)}$$ and $$\dot{p} = -\frac{\partial H}{\partial q} = 0 \Rightarrow p = \text{const}$$

So using the first equation I get, $$4ma^2 \int_{q_i}^{q_f} dq = p \int_0^T \frac{1}{1+\cos^2 \Omega t}dt - 4ma^2 \Omega \int_0^T \frac{\cos^2\Omega t }{1+\cos^2\Omega t} dt$$ Now ##\phi = \Omega t## and ##2\phi \in [0,\pi] \Rightarrow \phi \in [0,\pi/2]##. So we have $$4ma^2 \Delta q = \frac{p}{\Omega}\int_0^{\pi/2} \frac{1}{1+\cos^2 \phi} d\phi - 4ma^2\int_0^{\pi/2} \frac{\cos^2 \phi}{1+\cos^2\phi} d\phi$$ My question is how do I eliminate the ##p##?

OK, but due to the sign difference, I get a + in front of the last integral in ##\Delta q##.

Can you see the physical interpretation of the conserved conjugate momentum ##p##? If so, you should be able to deduce its value. [EDIT: Actually, with the initial conditions as stated in the problem, you can deduce ##p## directly from the relation between ##p## and ##\dot{q}##.]
 
Last edited:
  • #7
TSny said:
With the sign conventions that you chose initially for ##q## and ##\Omega##, I get $$H(q,p,t) = \frac{(p+4ma^2\Omega \cos^2 \Omega t)^2}{8ma^2(1+\cos^2 \Omega t)}$$ but I think the sign difference has to do with your choosing q positive in the clockwise direction and ##\Omega## positive in the counterclockwise direction.
Yes, I think I would have got the same, I just used the expression given on the sheet with the sign of the last term flipped.

OK, but due to the sign difference, I get a + in front of the last integral in ##\Delta q##.

I see why the sign changes here given the differences in the sign of the last term in the lagrangian, but wouldn't this give a different numerical answer for ##\Delta q##?

Can you see the physical interpretation of the conserved conjugate momentum ##p##? If so, you should be able to deduce its value. [EDIT: Actually, with the initial conditions as stated in the problem, you can deduce ##p## directly from the relation between ##p## and ##\dot{q}##.]
Yep, since ##p## is constant throughout the motion, evaluate at ##t=0##. This is a point mass moving with angular speed ##\Omega## so that ##I = m(2a)^2\Omega = 4ma^2 \Omega##. Is there any physical insight into why this remains constant though?
 
  • #8
CAF123 said:
I see why the sign changes here given the differences in the sign of the last term in the lagrangian, but wouldn't this give a different numerical answer for ##\Delta q##?
I think the sign changes won't make a difference in the final magnitude of ##\Delta q##, just a difference in overall sign. I believe there will be a sign difference in the value of ##p## that will make up for the sign difference in the Lagrangian.

Yep, since ##p## is constant throughout the motion, evaluate at ##t=0##. This is a point mass moving with angular speed ##\Omega## so that ##I = m(2a)^2\Omega = 4ma^2 \Omega##. Is there any physical insight into why this remains constant though?

You can show that ##p## represents the total angular momentum of the system about point A. The net external torque is zero about A.
 

What is a Lagrangian function?

A Lagrangian function is a mathematical tool used in mechanics to describe the motion of a system. It is based on the concept of energy, and it takes into account the kinetic and potential energies of a system.

How is a Lagrangian function used to describe the motion of a beetle on paper?

A Lagrangian function can be used to describe the motion of a beetle on paper by considering the beetle as a point mass and taking into account the forces acting on it, such as gravity and friction. This allows for the prediction of the beetle's trajectory and behavior.

What variables are used in a Lagrangian function for a beetle on paper?

The variables used in a Lagrangian function for a beetle on paper are typically the position and velocity of the beetle, as well as any external forces acting on it, such as the weight of the beetle and the resistance from the paper surface.

How is a Lagrangian function different from a Newtonian function?

A Newtonian function is based on the concept of force, while a Lagrangian function is based on energy. This means that a Lagrangian function takes into account the total energy of a system, while a Newtonian function only considers the forces acting on it.

What are the benefits of using a Lagrangian function for studying the motion of a beetle on paper?

Using a Lagrangian function allows for a more comprehensive understanding of the beetle's motion, as it takes into account both energy and forces. It also allows for the prediction of the beetle's behavior in different scenarios, such as varying paper surfaces or external forces.

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