Moment of inertia - inverted pendulum

[truongpm]

Hi,

I have been trying to understand the dynamics behind an inverted pendulum from this webpage:

http://www.library.cmu.edu/ctms/ctms/simulink/examples/pend/pendsim.htm

I don't get what the moment of inertia (I) of the rod is. I was thinking it should be about the end of the rod: m*(2L)^2, since L is defined as half the length of the rod, but the listed I value seems to be solved for a rod spinning about its center of gravity: 1/12*m*(2L)^2. Can someone explain this issue to me? Why should I be about the rod's cg?

Thanks,

- Paul

 

[electricpete]

let mu = mass per distance along radial axis

Find the moment of inertia about the center of the rod. It is twice the moment of inertia of one half the rod about the center:
I1=2*int(mu*r^2*dr,r=0..L)

D
I1=2*[r^3/3], evaluated at r=L minus same thing at r=0:
I1 = 2/3 L^3*mu

substitute: mu=m/(2*L) into I1:
I1 := 1/3 L^2 m


substitute L=0.3,m=0.2 into I1:
I1 = 0.006






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(2B)+(2B)' ?

 

[electricpete]

Why should I be about the rod's cg?

You can choose it anywhere you want, and you should of course end up with the same numerical solution for a given problem.

The particular choice of 2 coordinates x and theta as shown results in two equations (F = M*d^2x/dt^3 and T = I * d^2theta/dt^2) which are already decoupled. i.e. they are the modal variables. So it is very easy to solve for derivative of state variables. If you chose x and theta as angle from the bottom, they would not end up being decoupled, but with a very small amount of additional algebra you can still solve for derivatives of state variables to feed it into Matlab to get the same solution. (But you'd still get same solution)

The best choice of state variables to give the simplest, most direct and most intuitive path to solution can be a tricky thing when you have angles and displacements. For this simple problem, I think there is some intuition that suggests that a rigid body has 2 independent modes: one for translation of center of mass and one for rotation about center of mass, which leads to this selection without much fuss. I think maybe Lagrange can help provide insight about selection of variables and developing equation for more complicated systems. I studied it a few years ago and remember I was impressed by what you could do to develop equations of motion more simply. But I’ve forgotten most of it.


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(2B)+(2B)' ?

 

[electricpete]

Whoops, sorry... I was wrong to say these are the decoupled / modal variables. I was thinking about pendulum alone with force applied at bottom, not he additional mass at he bottom. With mass at he bottom the COM of entire system is not same as COM of the bar, so these are not the modal variables.

I'm not sure there is a big benefit to selecting center of bar for summing moments as opposed to bottom of bar for summing moments (maybe someone can comment?). Again you will come up with same solution either way (as long as you apply equations correctly.)

=====================================
(2B)+(2B)' ?

 

[truongpm]

Hmm.. I guess they are treating the rod and cart separately and doing the equations of motion based on that, so you are right about translation and rotation about the rod's cg. As for the coupling, that is handled by defining what those interaction forces (N and P) are?

 

[electricpete]

It looks like four: inertia equations.

Fnetx on M = M*d^2(x)/dt^2
Fnetx on m = m*d^2(xp)/dt^2
Fnety on m = m*d^2*(yp)/dt^2
Tnet on I = I*d^2*(theta)/dt^2

On first glance it looks like a 4DOF system. However there is one constraint which is that the positions of pendulum COM (xp,yp) must always remains a distance L from the position of mass M (x,0). Once you consider that constraint, I believe it turns out o be a 3DOF problem as we can see from the fact that if we know one variable theta, we can compute the two variables (xp,yp) and we still know the complete condition of the system (including the kinetic energy of each of the three inertia's m,M, I).

=====================================
(2B)+(2B)' ?

 

[electricpete]

As for the coupling, that is handled by defining what those interaction forces (N and P) are?

I guess you could say that. I would say the coupling remains in the model, rather than being removed anaytically or thru selection of variables. Simulink is obviously capable of solving coupled ordinary differential equations.

=====================================
(2B)+(2B)' ?

 

[electricpete]

Correction in bold:
Once you consider that constraint, I believe it turns out to be a 3DOF problem as we can see from the fact that if we know one variable theta, we can compute the two variables (xp,yp) and we still know the complete condition of the system (including the kinetic energy of each of the three inertia's m,M, I).
should have ben:
Once you consider that constraint, I believe it turns out to be a 2DOF problem as we can see from the fact that if we know two variables: X and theta (along with their derivatives), we can still compute (xp,yp) and we still know the complete condition of the system (including the kinetic energy of each of the three inertia's m,M, I).
In other words, it is a 2DOF system since 2 variables (and their derivatives) completely describe the system energy.

=====================================
(2B)+(2B)' ?

 

[FeX32]

They are considering each of the DOF's separately as this makes it easier to use the Newton-E method to derive the differential equations. There are a few other methods to derive the equations of motion. For example, in this situation the Langrange method is more systematic. There are also some other uncommon methods like the principle of virtual work (usually for forward and reverse dynamics) and the very uncommon connection method (can be used for more complicated systems like robots).
Mass moments of inertia are important for a problem like this.
The inertia should be taken about the axis of rotation, always. In this case it's the pin joint at the bottom attaching the rod to the cart. This joint constrains the rod to rotate only about this axis. It of course also translates due to the motion of the cart, which comprises the 2nd DOF.
The moment of inertia in this case is 1/3*mL^2 as seen here

http://en.wikipedia.org/wiki/List_of_moments_of_inertia

For more information about moments of inertia and their relations to dynamic systems there are a few dynamics texts books I would refer you to.


Fe

 

[FeX32]

BTW,
The inverted pendulum system is a very good model to learn how to solve dynamical systems and control them. I remember it was the model I chose to learn nonlinear control theory on at uni.




Fe

 

[FeX32]

Inverted pendulum by Langrange.
I looked it over as it is wiki so there is a possibility of mistakes. But it looks fine to me.

http://en.wikipedia.org/wiki/Inverted_pendulum

Fe

 

[electricpete]

The inertia should be taken about the axis of rotation, always. In this case it's the pin joint at the bottom attaching the rod to the cart.

I would respectfully disagree. There is not just one way to skin the cat. You can take moment about whichever point you want. We have in front of us one example where it is solved taken moment of inertia about center of beam (link in original post) and one where it is solved by taking moment of inertia about the bottom "hinge" (your link). Both will surely give the same answer.

=====================================
(2B)+(2B)' ?

 

[zekeman]

In these problems where translation forces and motion are involved you ALWAYS use the cg as he origin for I. The reason is that the total KE is energy of the rotating translating rod is
1/2Icgw^2 + 1/2mx'^2 +1/2my'^2

where x' and y' are the orthogonal coordinates of velocity at the cg

while 1/2 Iw^2 is the KE only when there is no translation of the pivot point.

 

[FeX32]

electricpete,

Thanks. You are correct. I guess my wording was not 100% correct. It does come out easier if you choose a pin joint when deriving the equation of motion as some of the forces applied to the rigid body don't create a moment there. This goes for most mechanisms.
Thanks again for the correction.

zekeman,

I agree totally when we are deriving the DE's using an energy method. Lagrange or virtual work ect.
But using the Newton-Euler method you can choose any point (and inertia), where the joint is the most efficient choice in many cases.
In the pendulum example we end up with the inertial part of the DE as torque=(angular accel)*(inertia about joint) in the DE which is intuitive and correct. Then the inertia is just Icg+m*l^2 by parallel axis theorem.

http://www.library.cmu.edu/ctms/ctms/examples/pend/invpen.htm

I'm sure we all agree, we just have different ways of thinking about it.




Fe

 

[truongpm]

Thanks for the responses guys!

FeX32, if I analyzed about the joint, would the DE simply be thetaddot = 1/(Icg + mL^2) * mgLsin(theta)? And everything else would remain the same in the link I showed?

 

[FeX32]

Yes, you are just missing the term associated with xdd.
Summing moments about the joint you directly get

(Icg+mL^2)*thetadd+mgL*sin(theta)+mLxdd*cos(theta)=0

You then sum forces in the most principle direction. So yes the same as that link.

To solve the DE's I sugest using both Simulink and ODE45.




Fe

 

[FeX32]

correction:
(Icg+mL^2)*thetadd-mgL*sin(theta)+mLxdd*cos(theta)=0


Fe

 

[zekeman]

FeX32
I agree with your comment,(18 Jul 11 0:07); equations both ways are equivalent. Guess my personal preference got in the way.

 

[FeX32]

. Same thing happen to me with electricpete



Fe