Double Pendulum (One pendulum hanging from the other)

[CAF123]

Homework Statement

A body of mass m hangs by a light inelastic string of length a from a body of mass 3m which in turn hangs from a fixed point O by a string of length a. The masses are confined to motion in a vertical plane through O. Denote the horizontal displacements of the two particles from the vertical by ##x_1## (for upper mass) and ##x_2##(for lower mass). Assume that they are both small compared to a.

1) Determine the kinetic and potential energy in terms of ##x_1##, ##x_2##.
2) Rescale the coordinates ##x_i = \mu_i z_i## where ##\mu_i## are constants such that that the kinetic energy would take the form ##T = \frac{1}{2}(\dot{z_1}^2 + \dot{z_2}^2)##. Rewrite the total energy in terms of these new rescaled coordinates.

The Attempt at a Solution

I think I have made a good attempt at the above but I want to be sure of my answer before I start the next part of the question.

Define coord system with +ve y down and +ve x right. Then x1 = x1 , y1 = acosθ₁ = ##\sqrt{a^2 - x_1^2}.## For the other mass: x2 = x2, y2 = ##\sqrt{a^2 - x_1^2} + \sqrt{a^2 - x_2^2}.##
By Taylor expansion/simplification I get to $$\frac{m_1gx_1^2}{2a} + \frac{m_2gx_2^2}{2a} + \frac{m_1}{2}[\dot{x_1}^2] + \frac{m_2}{2}[\dot{x_2}^2],$$ m1 = 3m and m2 = m.

To get the rescaled version, I just subbed in what they give (after taking the derivative etc..). To get the required T, I think the condition ##\mu_1^2 m_1 = \mu_2^2m_2 = 1## must hold. Provided this is right, $$E = \underline{z}^T G \underline{z} + \frac{1}{2} \underline{\dot{z}}^T \dot{z}.$$

I can then rearrange this into the form ##\underline{\ddot{z}} + n^2 G \underline{z} = 0, ##with ##n = \pm \sqrt{2}## and ##G## a diagonal matrix being ##\frac{g}{2a} I_2##, ##I_2## 2x2 identity.

Many thanks.

 

[voko]

Note the expression for energy you got is the sum of energies of two independent simple pendula. Can this be correct?

 

[CAF123]

Note the expression for energy you got is the sum of energies of two independent simple pendula. Can this be correct?

I would agree, this does not make sense since they are clearly coupled. However, is it not the case that by rescaling the coordinates the system becomes decoupled? (Is that not the purpose of the rescaling)?

(The physical reasoning behind this is not clear, although)

 

[voko]

The original energy equation is already essentially decoupled, rescaling has nothing to do with that. So you must have made a mistake composing it.

 

[CAF123]

The original energy equation is already essentially decoupled, rescaling has nothing to do with that. So you must have made a mistake composing it.

Hmm..I'll look over it again and get back to you

EDIT: Yes, I see the error. I'll fix it and then post the new expression.

 

[CAF123]

If I didn't make any further mistakes, then I have:$$E = \frac{m_1g x_1^2}{2a} + \frac{m_2g x_2^2}{2a} + \frac{m_1}{2}[\dot{x_1}^2] + \frac{m_2}{2}[\dot{x_2}^2 + \frac{x_1^2}{a^2} + \frac{2x_1 x_2}{a^2} + \frac{x_2^2}{a^2}]$$

 

[voko]

I see that you have tried to account for the fact that the velocity of the lower mass depends on the velocity of the upper mass. But I don't understand why it is so complex. It should really be along the lines (v + u)^2.

Then, how about the potential energy of the lower mass?

 

[CAF123]

I see that you have tried to account for the fact that the velocity of the lower mass depends on the velocity of the upper mass. But I don't understand why it is so complex. It should really be along the lines (v + u)^2.

I got something of the form (u + v)^2 but then I expanded this. Perhaps I neglected too many terms. What I get is:$$\frac{m}{2}\left[\underline{\dot{x_2}^2} + \left(\frac{-x_1}{\sqrt{a^2 - x_1^2}} - \frac{x_2}{\sqrt{a^2 - x_2^2}}\right)^2\right]$$ When I expand I get: $$ \frac{m}{2}\left[\underline{\dot{x_2}}^2 + \frac{x_1^2}{a^2 - x_1^2} + \frac{2x_1 x_2}{a^2} + \frac{x_2^2}{a^2 - x_2^2}\right]$$

This is for the last kinetic energy term (due to the lower mass).

Then, how about the potential energy of the lower mass?

Is there something wrong with it?

 

[voko]

I got something of the form (u + v)^2 but then I expanded this.

Before you go expanding: what is the velocity of the lower mass?

For potential energy: what is the vertical position of the lower mass?

 

[CAF123]

Before you go expanding: what is the velocity of the lower mass?

All of the expression in the squared brackets in the previous post.
Edit: I forgot the chain rule - the expression i wrote intially is dimensionally ridiculous.

For potential energy: what is the vertical position of the lower mass?

##\sqrt{a^2 - x_1^2} + \sqrt{a^2-x_2^2}##

 

[voko]

All of the expression in the squared brackets in the previous post.
Edit: I forgot the chain rule - the expression i wrote intially is dimensionally ridiculous.

The square of the velocity of the upper mass is ##\dot{x}_1^2 + \dot{y}_1^2 ##. The lower mass: ## (\dot{x}_1 + \dot{x}_2)^2 + (\dot{y}_1 + \dot{y}_2)^2##. Are these what you have been using before expanding and linearizng?

##\sqrt{a^2 - x_1^2} + \sqrt{a^2-x_2^2}##

Correct. But then I don't understand why the potential energy term for the lower mass does not have ##x_1##. Or was that combined with its kinetic energy?

 

[CAF123]

The square of the velocity of the upper mass is ##\dot{x}_1^2 + \dot{y}_1^2 ##. The lower mass: ## (\dot{x}_1 + \dot{x}_2)^2 + (\dot{y}_1 + \dot{y}_2)^2##. Are these what you have been using before expanding and linearizng?

Yes for the upper mass but no to the lower mass. I was doing for the lower mass: $$\frac{m}{2}(\dot{x_2}^2 + \dot{y_2}^2)$$ and expressing y2 in terms of x1 and it's derivative.

Correct. But then I don't understand why the potential energy term for the lower mass does not have ##x_1##. Or was that combined with its kinetic energy?

Because it is yet another silly error. Full potential term: $$V = V_1 + V_2 = -m_1gy_1 - m_2gy_2 = -m_1g\sqrt{a^2 - x_1^2} - m_2g(\sqrt{a^2 - x_1^2} + \sqrt{a^2 - x_2^2})$$

 

[CAF123]

My final expression for T: $$\frac{m_1}{2}[\underline{\dot{x_1}}^2 + \frac{x_1^2 \dot{x_1}^2}{a^2 - x_1^2}] + \frac{m_2}{2}[\underline{\dot{x_2}^2} + (\frac{-x_1 \dot{x_1}}{\sqrt{a^2 - x_1^2}} - \frac{x_2 \dot{x_2}}{\sqrt{a^2 - x_2^2}})^2 ]$$

before I expand.

 

[voko]

Yes for the upper mass but no to the lower mass. I was doing for the lower mass: $$\frac{m}{2}(\dot{x_2}^2 + \dot{y_2}^2)$$ and expressing y2 in terms of x1 and it's derivative.

I just noticed that you defined ## y_2 ## as the sum of ## y_1 ## and second mass's vertical displacement relative the first. So my expression is indeed incorrect in this notation. Yours is too, anyway, because the horizontal component of the velocity is ## \dot{x}_1 + \dot{x}_2 ##.

I would still suggest that you redefine ## y_2 ## to be the second mass's vertical displacement relative the first, that should simplify the expression.

Because it is yet another silly error. Full potential term: $$V = V_1 + V_2 = -m_1gy_1 - m_2gy_2 = -m_1g\sqrt{a^2 - x_1^2} - m_2g(\sqrt{a^2 - x_1^2} + \sqrt{a^2 - x_2^2})$$

This I agree with.

 

[CAF123]

Yes, because I defined my coordinates wrt ceiling. So my kinetic term is: $$\frac{m_1}{2}\left[\frac{\dot{x_1}^2a^2}{a^2 - x_1^2}\right] + \frac{m_2}{2}\left[(\dot{x_1}+ \dot{x_2})^2 + \left(\frac{-2x_1\dot{x_1}}{\sqrt{a^2 - x_1^2}} + \frac{x_2\dot{x_2}}{\sqrt{a^2 - x_2^2}}\right)^2\right]$$,
where $$\dot{y_1} + \dot{y_2} = \frac{-2x_1\dot{x_1}}{\sqrt{a^2 - x_1^2}} + \frac{x_2\dot{x_2}}{\sqrt{a^2 - x_2^2}}$$

 

[voko]

Why -2?

 

[CAF123]

Why -2?

I just skipped a step and brought two terms together:
$$\dot{y_1} + \dot{y_2} = \frac{-x_1 \dot{x_1}}{\sqrt{a^2 - x_1^2}} -\frac{x_1\dot{x_1}}{\sqrt{a^2 - x_1^2}} - \frac{x_2 \dot{x_2}}{\sqrt{a^2 - x_2^2}}$$

 

[voko]

I do not understand why the vertical velocity of the lower mass has two x1-based terms. Physically, it does not seem to make sense.

 

[CAF123]

I do not understand why the vertical velocity of the lower mass has two x1-based terms. Physically, it does not seem to make sense.

For the upper mass, ##y_1 = \sqrt{a^2 - x_1^2}##. For the lower mass, ##y_2 = \sqrt{a^2 - x_1^2} + \sqrt{a^2 - x_2^2}##

Then ##\dot{y_1} + \dot{y_2} ## gives the result, no?

 

[voko]

For the upper mass, ##y_1 = \sqrt{a^2 - x_1^2}##. For the lower mass, ##y_2 = \sqrt{a^2 - x_1^2} + \sqrt{a^2 - x_2^2}##

If you insist that ## y_2 ## is defined this way, let me denote ##y_2' = y_2 - y_1##, which is the vertical velocity of the lower mass relative the upper mass. The total vertical velocity of the lower mass is either ## \dot{y}_1 + \dot{y}_2' ## or simply ## \dot{y}_2 ## in your notation, but not ##\dot{y_1} + \dot{y_2} ##

 

[CAF123]

If you insist that ## y_2 ## is defined this way, let me denote ##y_2' = y_2 - y_1##, which is the vertical velocity of the lower mass relative the upper mass. The total vertical velocity of the lower mass is either ## \dot{y}_1 + \dot{y}_2' ## or simply ## \dot{y}_2 ## in your notation, but not ##\dot{y_1} + \dot{y_2} ##

Yes, so sorry voko. When I draw the sketch it all becomes clear.
So I think it would be best if I summarized my x1,x2,y1,y2 for clarity:
x1: From ceiling going right until in line with m1
y1: From ceiling going down until in line with m1, m1 the upper mass

x2: From ceiling going right all the way to m2.
y2: From ceiling going down all the way to m2:

the potential function is agreed upon and now the kinetic term:
$$T = \frac{m_1}{2}[ \dot{x_1}^2 + \dot{y_1}^2] + \frac{m_2}{2}[ \dot{x_2}^2 + \dot{y_2}^2],$$

x1 = x1, x2 = x2, y1 = ##\sqrt{a^2 - x_1^2}##, ##y_2 = \sqrt{a^2 - x_2^2} + \sqrt{a^2 - x_1^2}##

 

[voko]

Do I understand correctly that both x1 and x2 are measured from the fixed vertical line (presumably the line of equilibrium)? That's the only way I can justify the second term in the latest expression for kinetic energy.

There is nothing wrong with this definition, but you should keep in mind that this changes the formula for y2.

 

[CAF123]

Do I understand correctly that both x1 and x2 are measured from the fixed vertical line (presumably the line of equilibrium)? That's the only way I can justify the second term in the latest expression for kinetic energy.

There is nothing wrong with this definition, but you should keep in mind that this changes the formula for y2.

Indeed it does. Is it possible to express y2 in terms of solely x1 and x2 using my definition?

 

[CAF123]

Actually on second thought, I'll go back to my original definition:
So my final expression for T (hope it is right this time!) is $$T = \frac{m_1}{2}\left[\dot{x_1}^2 + \frac{x_1^2 \dot{x_1}^2}{a^2 - x_1^2} \right] + \frac{m_2}{2}\left[(\dot{x_1} + \dot{x_2})^2 + (\frac{-x_1 \dot{x_1}}{\sqrt{a^2 - x_1^2}} - \frac{x_2 \dot{x_2}}{\sqrt{a^2 - x_2^2}})^2\right]$$

 

[voko]

It looks good, but you might want to lose the unnecessary minus signs in the second term :)

 

[CAF123]

Since I have nothing to aim for, what terms should I ignore when I expand it all out?
There are a lot of terms containing say x1 and it's derivative so these are crossed terms which I don't think I can get into the matrix eqn (unless I have for example a term ##\underline{x}^T \underline{\dot{x}}##

 

[voko]

I would say you should apply the techniques you mastered in the previous problems, and if that becomes problematic, bring it up here.

 

[CAF123]

I would say you should apply the techniques you mastered in the previous problems, and if that becomes problematic, bring it up here.

I think the first term can go to ##\frac{m_1}{2} (\dot{x_1^2})##. On the second term, expand the first binomial and don't cancel any terms. For the second binomial, I expanded but there are a lot of x1 and it's derivatives coupled together.

So then taylor expand $$\frac{x_1^2 \dot{x_1}^2}{a^2 - x_1^2}$$ to get ##(\frac{a^2}{x_1^2} - 1)\dot{x_1}^2##. The first term is very large so I can't neglect that. Since I can't neglect it I will have to have a coupled term, yes?

EDIT: I'll try again because I forgot about the condition for binomial exp to hold.
EDIT2: Doing the expansion properly, I get the term ##\frac{x_1^2}{a^2} \dot{x_1}^2## which still has derivative and x1.

 

[CAF123]

Is there a way to eliminate such terms or should I just proceed and put it in some form containing a term ##x^T \dot{x}##?

Actually ,I think the terms $$\frac{x_1^2\dot{x_1}^2}{a^2}\,,\frac{2 x_1 x_2 \dot{x_1} \dot{x_2}}{a^2}\,\,,\frac{x_2^2 \dot{x_2}^2}{a^2}$$ would all vanish since we have the form ##x_i^2/a^2 ## which is <<1 since we know ##x_1 << a => x_1^2 << a^2 ##so ##x_1^2/a^2 <<1?##

 

[CAF123]

Neglecting the above terms and then bringing in the potential term, I do indeed get a form: $$\frac{1}{2} \underline{x}^T A \underline{x} + \frac{1}{2}\underline{\dot{x}}^T B \underline{\dot{x}}$$.

My matrices are A and B, where A is diagonal with entries 4mg/a in top left, 0 top right, 0 bottom left and mg/a bottom right. Also for B I get the very peculiar matrix with m in three out of 4 entries and 4m in the last.

When I input the rescaling conditions, I end up with a contradiction. (mu must satisfy two eqns that is impossible simultaneously).

 

[voko]

I think you made a reasonable assumption in #29. Yet the result is clearly incompatible with the requirements of the problem. This may be because your x1 and x2 are not exactly what the problem requires. It may be that both x1 and x2 should be measured from the single fixed vertical line (the equilibrium line). Then I think all coupling in kinetic energy will vanish (by an argument similar to #29), so you get a diagonal matrix just like the problem requires.

 

[CAF123]

I think you made a reasonable assumption in #29. Yet the result is clearly incompatible with the requirements of the problem. This may be because your x1 and x2 are not exactly what the problem requires. It may be that both x1 and x2 should be measured from the single fixed vertical line (the equilibrium line). Then I think all coupling in kinetic energy will vanish (by an argument similar to #29), so you get a diagonal matrix just like the problem requires.

This is exactly the case. Defining x1 and x2 both from the equilibrium line, I get the required form with the potential no longer decoupled as was the case earlier. Is there a physical reasoning why simply defining my x2 differently yields a decoupled potential!

It seems to be that this question can only be done via the 'right' method. I would have thought given this the question actually state where x1 and x2 are defined from because I did spend a lot of time yesterday trying to figure out why I couldn't proceed.

 

[voko]

This is exactly the case. Defining x1 and x2 both from the equilibrium line, I get the required form with the potential no longer decoupled as was the case earlier. Is there a physical reasoning why simply defining my x2 differently yields a decoupled potential!

I am not exactly sure what you mean here. What "earlier" potential are you referring to?

It seems to be that this question can only be done via the 'right' method. I would have thought given this the question actually state where x1 and x2 are defined from because I did spend a lot of time yesterday trying to figure out why I couldn't proceed.

Frankly, I think this problem should have been done using angles rather than displacements.

 

[CAF123]

I am not exactly sure what you mean here. What "earlier" potential are you referring to?

The potential written in post #12. The potential redefining x2 from the vertical line will have the first term the same but the second will be ##\sqrt{a^2 - (x_2 -x_1)^2}## and hence coupled. The potential in #12 is not at all coupled. So depending on my definitions, I decouple the potential term. Does this seem to make physical sense?

Frankly, I think this problem should have been done using angles rather than displacements.

This is exactly how we did it in lectures and we are told to do it in terms of x1 and x2 for this question.

 

[voko]

When the coordinates of the two masses are defined in such a way that that the motion of one mass can be described by one set of coordinates, and another mass by another, disjoint, set of coordinates, kinetic energy becomes decoupled. But it has to be somewhere, so it appears in potential energy.