# Pendulum with a massive rod

1. Apr 21, 2015

### ellipsis

Consider a realistic pendulum with a bob of mass $M$ and a rod of mass $m$. For the purposes of torque, is it correct to treat the bob and rod as two separate point masses, one at length $L$, and the other at length $\frac{L}{2}$?

When I implemented this idea, I found that centripetal acceleration increased by a factor of $\frac{3}{2}$

The only change I detected was that the center of mass of the bob/rod combo shifted closer to the pivot, which increased the frequency of motion. In this case, the total mass had no effect.

This is the force analysis I used for a massless pendulum. For the massive pendulum, I treated the rod as a second point mass at length ##\frac{L}{2}, with its own set of forces. I then added the forces together. The acceleration due to gravity did not change (the bottom blue line), but the centripetal acceleration increased (the top blue line).

I did not specify any density information, and the pendulum swings faster, but independently of the added mass, which leads me to conclude that this is a spurious result.

If so, how should you model a pendulum with a massive rod? I'd prefer a purely force-based model (rather than Lagrangians, which I've never been introduced to)

2. Apr 21, 2015

### ianchristie

Find the centre of mass of the system of rod and bob. Then treat your pendulum as a massless rod with a point mass of M+m at the centre of mass. For small perturbations of a pendulum the period is related only to the length between the pivot and the centre of mass, not the mass. The pendulum clock in Westminster (colloquially known as Big Ben), was once (is still ?) regulated by adding or subtracting pennies from the top of the pendulum bob. It was not the extra mass that made a difference but the raising or lowering of the centre of mass.

3. Apr 22, 2015

### ellipsis

Thanks, that was very informative - and elegantly simple in retrospect. It turns out the behavior of a pendulum with a rod with mass is simply isomorphic to a pendulum of slightly shorter length, and the only added complexity is finding the center of mass.

4. Apr 22, 2015

### HomogenousCow

If you regard the whole thing as a rigid body, you can simply treat each piece as an infitesimal mass and integrate for the lagrangian.

5. Apr 22, 2015

### ellipsis

Like I said above, I don't really understand what a Lagrangian is, and, more importantly, how it can be practically used in i.e. a numerical simulation. Could you demonstrate?

As it is, I'm modeling the pendulum as two second-order differential equations of motion in Cartesian coordinates - which was derived using basic trigonometry. The origin is at the pivot point, in this model.

$$\frac{d^2x}{dt^2} = \frac{gxy - xv^2}{L^2}$$
$$\frac{d^2y}{dt^2} = \frac{-gx^2 - yv^2}{L^2}$$

Last edited: Apr 22, 2015
6. Apr 22, 2015

### HomogenousCow

Thing is the y and x coordinates are coupled, thus you only need one second order differential equation.
Basically, treat the infitesimal pieces of the rod as individual particles (all posessing the shared degree of freedom, the angle), and then integrate through the lengths of the rod. It's rather straightforward to do this with the lagrangian, but it's also do-able directly with the equations of motion.

7. Apr 22, 2015

### ellipsis

Frankly, I still don't get it. I'm interested in modeling the system with just one second-order differential equation, though. I realize I could alternatively model everything in polar coordinates, but that isn't the Lagrangian (which, I understand, is supposed to be in generalized coordinates).

8. Apr 22, 2015

### Staff: Mentor

To calculate the torque, yes. But you need to treat it as an extended body to calculate the moment of inertia.

9. Apr 22, 2015

### ellipsis

Thanks for the response. What's the difference between torque and the moment of inertia?

10. Apr 22, 2015

### Staff: Mentor

It's the rotational equivalent to the difference between force and mass: Torque is to force as moment of inertia is to mass.

11. Apr 22, 2015

### ellipsis

Ah, that clears it up a great deal. Nice analogy. I see now: I calculated the torque correctly using that method, but I neglected to take into account the moment of inertia - which itself is proportional to the distance of the center of mass from the pivot point.

Thanks - you have a way of simplifying topics, Doc Al.