Inverted pendulum balance dynamics equations

[aerograce]

Homework Statement

Hi guys! I am reading a paper on inverted pendulum. However, I can't understand the dynamics equations of this inverted pendulum. You may look at the images below. Basically the thruster is placed at top plate, while the middle plate is a base plate where pivot is placed. And then bottom plate is where you place your counterweight. Upon thruster firing, the pendulum arm will rotate.

The equations are attached in the pictures below. I(T) and I(CW) represent the moments of inertia of the thruster and counterweight respectively.

My question is, can anyone help me explain these equations? I understand that the left side is the net force on horizontal direction based on Newton's second law, and the left side has the thrust caused by pivots stiffness, weight and thrust. However, I don't understand where the momentum of inertia * angular acceleration / arm length comes in.

Homework Equations

Relevant equations for this question will be Newton's second law, and also rotational dynamics.

The Attempt at a Solution

My attempt at understanding the equations are stated in the question part.

 

[TSny]

As you say, the right hand side of equation (3.6) must be the sum of the horizontal forces acting on $m_{T}$. So, the complicated fraction must represent the horizontal component of a force acting on $m_{T}$. Conceptually, what are the forces that act on $m_{T}$?

 

[aerograce]

As you say, the right hand side of equation (3.6) must be the sum of the horizontal forces acting on $m_{T}$. So, the complicated fraction must represent the horizontal component of a force acting on $m_{T}$. Conceptually, what are the forces that act on $m_{T}$?

Hello! Sorry for my late reply. I can't seem to get the forces acting on mT. The force acting on mT will be gravity, and then force caused by pivot stiffness, and then thrust. However, I can't seem to catch the meaning of that moment inertia * angular acceleration over arm length, isn't it denoting the total external force instead of one component?

 

[TSny]

I think of the system as consisting of two point masses ( $m_T$ and $m_{CW}$), an upper rod of length $l_T$ and a lower rod of length $l_{CW}$. I think of the couples $k_{tp}$, $k_b$, and $k_{bp}$ as acting on the rods (not the point masses).

The upper mass $m_T$ is connected to the upper rod. So, the upper rod will exert a force on $m_T$. This force is unknown, but we can break it up into an unknown horizontal force $F_x$ and an unknown vertical force $F_y$. So, you can draw a free body diagram for the upper mass and set up $ΣF = ma$ for horizontal and vertical components.

You can also draw a free body diagram for the upper rod with the intention of setting up $\sum{\tau} = I \ddot{\theta}$. Apparently, you are to assume that half the couple $k_{b}$ acts on the upper rod and half acts on the lower rod. (I don't see why that would necessarily have to be the case unless the upper section of the system is identical to the lower section.)

 

[aerograce]

I think of the system as consisting of two point masses ( $m_T$ and $m_{CW}$), an upper rod of length $l_T$ and a lower rod of length $l_{CW}$. I think of the couples $k_{tp}$, $k_b$, and $k_{bp}$ as acting on the rods (not the point masses).

The upper mass $m_T$ is connected to the upper rod. So, the upper rod will exert a force on $m_T$. This force is unknown, but we can break it up into an unknown horizontal force $F_x$ and an unknown vertical force $F_y$. So, you can draw a free body diagram for the upper mass and set up $ΣF = ma$ for horizontal and vertical components.

You can also draw a free body diagram for the upper rod with the intention of setting up $\sum{\tau} = I \ddot{\theta}$. Apparently, you are to assume that half the couple $k_{b}$ acts on the upper rod and half acts on the lower rod. (I don't see why that would necessarily have to be the case unless the upper section of the system is identical to the lower section.)

Thank you so much!:)

 

[alaspina]

I know this is a little late, but I've found myself working on the same problem. I just wanted to correct TSny as mT and mcw are definitely not treated as point masses in this system. The components with the angular acceleration (theta double dot), are due to the rotational inertia of the thruster and the counterweight. I don't think he mentions it in the paper, but that could be the only explanation as why we see those two terms appear for the equations you posted in figure 3. If they were point masses, I believe we would only get one of the components, which is m*(distance to rotation point)^2 * theta_doubledot.

What I don't understand however is the way he treated the signs for the gravity force terms of the thruster and counterweight.

The upper mass m_T is connected to the upper rod. So, the upper rod will exert a force on m_T. This force is unknown, but we can break it up into an unknown horizontal force F_x and an unknown vertical force F_y. So, you can draw a free body diagram for the upper mass and set up ΣF=ma for horizontal and vertical components.

If he considers the summation in the x direction, I don't see why we would have to account for the gravity term since it is in the y direction. I understand that the rod would exert a horizontal force and vertical force, but if you take the free body diagram of the top mass (thruster), shouldn't you be left with only the reaction force in the y direction equal to the weight and in the x direction equal to the thrust of the thruster?

 

[TSny]

I know this is a little late, but I've found myself working on the same problem. I just wanted to correct TSny as mT and mcw are definitely not treated as point masses in this system. The components with the angular acceleration (theta double dot), are due to the rotational inertia of the thruster and the counterweight. I don't think he mentions it in the paper, but that could be the only explanation as why we see those two terms appear for the equations you posted in figure 3. If they were point masses, I believe we would only get one of the components, which is m*(distance to rotation point)^2 * theta_doubledot.

OK. I’m probably not interpreting the problem correctly. I’m able to get the result expressed in equation (3.6) in the OP, but my method is probably bogus. The quickest way I’ve found to get (3.6) is to analyze the top part of diagram 3.5 in the frame of reference moving with the base plate. Thus, consider the diagram below:

Since the base plate has acceleration $\ddot x_b$ relative to the lab, this is a noninertial reference frame. This is taken into account by adding the fictitious force $-m_T \ddot x_b$, as shown in red. I took the lower end of the rod of length $l_T$ as the origin for torques. There is a reaction force on the lower end of the rod (the origin) which I didn’t bother to indicate since it does not produce any torque about the origin.

I took the torque at the lower end to be described by the torsional constant $k_b/2$, rather than by $k_b$. This assumes that half the torque produced by this joint goes to the top part of the system while the other half goes to the bottom part (not shown).

Setting up $\sum \tau = I \ddot \theta$, I obtain $$F_T l_T \cos \theta + m_T g l_T \sin \theta – m_T \ddot x_b l_T \cos \theta – \frac{k_b}{2} \theta – k_{tp} \theta = I_{\rm sys} \ddot \theta$$
Assume $\theta$ is small so that $\cos \theta$ may be replaced by 1 and $\sin \theta$ may be replaced by $\theta$.

Also, I took the moment of inertia of this system to be $I_{\rm sys} = I_T + m_T l_T^2$. This might be one of the places where I’m misinterpreting the system. I treated $m_T$ as a point mass revolving about the origin. This gives a contribution of $m_T l_T^2$ to the moment of inertia. I then treated $I_T$ as an additional moment of inertia due to other parts of the system such as the rod.

With these assumptions, the torque equation becomes $$F_T l_T + m_T g l_T \theta – m_T \ddot x_b l_T – \frac{k_b}{2} \theta – k_{tp} \theta = \left( I_T + m_T l_T^2 \right) \ddot \theta$$.
Rearrangement of this equation gives (3.6)

Again, I’m not claiming that this is a correct analysis. I included a fictitious force only for the localized mass $m_T$, but I did not include a fictitious force for any other part of the system such as the rod. This amounts to neglecting the mass of the other parts of the system, even though I included a contribution $I_T$ from other parts of the system. Likewise, I did not include any weight other than the weight of $m_T$. Thus, this derivation is shaky, to say the least.

What I don't understand however is the way he treated the signs for the gravity force terms of the thruster and counterweight.
If he considers the summation in the x direction, I don't see why we would have to account for the gravity term since it is in the y direction. I understand that the rod would exert a horizontal force and vertical force, but if you take the free body diagram of the top mass (thruster), shouldn't you be left with only the reaction force in the y direction equal to the weight and in the x direction equal to the thrust of the thruster?

The original way that I worked the problem was to work in the lab frame and draw separate free-body diagrams for the mass $m_T$ and the rod of length $l_T$. There will be a force of interaction between the rod and the mass $m_T$. Setting up Newton's laws for the rod and $m_T$ separately, you can solve for the interaction force. The horizontal component of the force of interaction includes a term involving the weight $m_T g$. This is how the vertical weight ends up contributing to the net horizontal force on $m_ T$. This method also led to the result in equation (3.6).