# Lagrangian function for beetle on paper

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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.

#### Attachments

• Lagrangian.jpg
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TSny
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$$x = a \cos \left(\frac{\pi}{2} - q\right) + a \cos\left(\frac{\pi}{2} - (q-\Omega)\right)$$ 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.

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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 $$x = a \sin q + a(\sin q \cos 2\phi - \cos q \sin 2\phi) \,\,;\,\,y = a \cos q + a(\cos q \cos 2\phi - \sin q \sin 2\phi)$$

TSny
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OK, but I don't agree with the sign of your last term in y.

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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 $$\int_{0}^{\pi/2} \frac{d\theta}{1+\cos^2 \theta} = \frac{\pi}{2\sqrt{2}}$$

TSny
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$$H(q,p,t) = \frac{(p-4ma^2\Omega \cos^2 \Omega t)^2}{8ma^2(1+\cos^2 \Omega t)} + \text{const}$$ 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:
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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?

TSny
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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.