I'm struggling to model the movement of a complex object based on an external force.

Let's start with a simple example of what I'm looking for.

We have a block $b$ of mass $M_b$, moving friction-less along a surface. It's influenced by a varying external force $F_b(t)$.


How does the block move? We have $a_b(t) = \frac{F_b(t)}{M_b}$, so assuming we start with $x_b(0) = 0$ and $v_b(0) = 0$, we have $x_b(t) = \frac{1}{2} \frac{F_b(t)}{M_b} t^2$.

I need the same calculation for a more complex object, which is basically an inverted pendulum.


On the block, we mount an axis, and connect a long rod with mass $M_{rod}$ to it. It has a large mass $M_{head}$ on its top.

The two main points I'm struggling with:

  1. The inertia of the object changed. As long as the rod is almost upright, it's somewhat intuitive that the block will accelerate slower than before (the lower parts of the rod have to be accelerated as well), and faster than if the rod was fixed (making this a fixed object with mass $M_b + M_{rod} + M_{head}$). I'm pretty sure that this has to be factored in as a backwards force the rod exerts to the block, but I have no idea how to calculate its strength.
  2. We get additional forces when the rod is no longer upright.


    Gravity pulls on the rod and head with the force $F_g$ in the shared center of mass. Using $\alpha$, this can be split into a force $F_t$ that acts as a torque on the rod and head, and a force $F_p$ that pushes down the rod onto the block.

Are these considerations right so far? How could I continue?

The ultimate goal is a function $x(t)$ depending on the properties of the objects and $F_b(t)$ – but I'm more interested in the way to get there than the actual result. (This is why I tagged this homework-and-exercises, although this is not homework.)

  • 3
    $\begingroup$ Love the way you drew your diagrams! May I ask what you used for that? $\endgroup$
    – Floris
    May 2, 2015 at 12:35
  • 2
    $\begingroup$ Thanks! They were drawn in OneNote on a laptop with a digitizer built in - so they were literally hand-drawn with a pen on-screen. $\endgroup$
    – fefrei
    May 2, 2015 at 12:47

1 Answer 1


You are really asking about the reaction forces felt on a rod when you push on its end. For a rod, you can work with a quantity called the reduced mass (see for example this excellent answer for the derivation). As long as the rod is balanced on its end, the reduced mass tells you exactly how much greater the inertia of the cart appears to be:

$$m_r = \frac{1}{\frac{1}{m} + \frac{h^2}{I}}$$

where $I$ is the moment of inertia of the "rod" about its center of mass, and $h$ is the distance from the center of mass to the point where you apply the force. This reduced mass would have to be added to the mass of the cart to understand the instantaneous acceleration when a force is applied while the rod is balanced.

In this equation, you can see readily that if you apply force at the center of mass, $h=0$ and the reduced mass is equal to the actual mass; for any other displacement, the rod will rotate and give less "resistance".

Now that is just the situation when the rod is vertical. Once the rod is not vertical, you will have a component of horizontal force that appears (from the rod falling over); and the distance $h$ becomes $h\cos\alpha$.

An actual equation of motion will be a complex thing - I am not even sure it has an analytical solution. The motion of the inverted pendulum by itself has an interesting solution (which I touched upon in my earlier answer about balancing a pencil on its tip) - but the derivation I gave there is only valid when the bottom of the pencil is fixed.

Let me know if this is enough of a starting point for you, or if you would like me to go deeper. You stated you wanted "the way to get there, not the actual result".

  • $\begingroup$ Thanks so far! I will take a deeper look into that next week. $\endgroup$
    – fefrei
    May 2, 2015 at 17:15
  • $\begingroup$ I think I now understand all I need, especially due to your earlier answer. I won't continue solving this - as I mentioned, I was doing this to learn (for a slightly more complicated, but similar, problem). Thanks a lot! $\endgroup$
    – fefrei
    May 7, 2015 at 8:50

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