# Pendulum but not quite

1. Feb 26, 2009

### mangaluve

I've successfully been able to simulate e pendulum (wow!). Now I want something slighly more complicated (at least I think it is). Suppose I take some 2d-object, for instance a square. Somewhere on the square I make a hole, and then I hang it on the wall on a nail. Then I move it a bit and let it fall (like a pendulum, but there is no string, instead the pivot is connected to the object itself). What is the physics behind the movements of this square?

2. Feb 26, 2009

### CompuChip

I suppose that it will rotate around the pivot, such that at the end of the motion the center of gravity will be vertically aligned with the pivot.

3. Feb 26, 2009

### Dr.D

This is what is called a physical pendulum, and the motion is governed by Newton's law in the form
sum of torques = I * alpha

4. Feb 26, 2009

### csprof2000

In general, all you have to do is calculate the moment of inertia of the body, and use that for I. Other than that, it behaves just like a regular pendulum. In fact, your other simulation should work for the new value of I.

5. Mar 2, 2009

### Bob S

The total energy of the body (physical square) is the rotational energy about its center of mass plus the energy of the linear motion of its center of mass. This in turn is equal to mgh, where h is the vertical distance the center of mass dropped from its release point.

Last edited: Mar 2, 2009
6. Mar 2, 2009

### Dr.D

What Bob S says is true, but it is awkward. The easier way to express the total kinetic energy of the body is to use the MMOI with respect to the fixed point, in which case the kinetic energy is simply
T = (1/2) Io w^2
where Io is the MMOI with respect to the fixed point.

7. Mar 2, 2009

### Bob S

OK. But doing the moment of inertia (MOI) integral about an arbitrary point in the square is messy. It is much easier to do it about the center of mass. Consider a circular disk with the pivot point off center. The MOI about the center is simple to calculate, as is the translational energy of the center of mass. I recall in class doing this problem as Dr. D. proposes, and finding that the solution separates into two terms, like the ones I describe.

8. Mar 2, 2009

### Dr.D

But the MMOI with respect to an arbitrary point is easily obtained from the centroidal MMOI by the parallel axis theorem.

9. Mar 3, 2009

### Bob S

I agree with Dr.D that the most elegant way of solving the above problem is with the parallel axis theorem. However, a similar problem, that of a "pendulum but not quite" where a solid spherical ball of mass M and radius b is rolling but not slipping on the inside of a spherical bowl of radius R. One could use the approach that the point of contact of the ball with the bowl is instantaneously at rest, and use parallel axis theorem, or add the sum of translational energy and rotational energy to get the total energy, which is equal to Mgh. This latter approach is more general.

10. Mar 3, 2009

### Dr.D

But you can get into trouble there because you are relying on the fact that the point of contact is an instant center of rotation, but it is also an accelerated point most of the time. That can get you into big problems real fast! At that point, it is usually better to go back to the mass center for moment sums.