Pendulums

Pendulums are nothing more than objects than swing back and forth around some pivot point. Of course they do this under a gravitational field. What do you think we are, space cadets?

A couple of different examples are shown here. The simplest one is just a mass dangling from a string

Let's apply $F~=~ ma$ to this situation. The acceleration in the tangential direction $a_t$ can be obtained by considering the components of the forces in those direction. The easiest one is the tension $T$. That's just zero in the tangential direction because the string is perpendicular to this direction. The only one left over is the component of the weight. Well that's just $-mg \sin\theta$. So we have

$$ma_t ~=~ -mg \sin\theta$$ (1.44)

But what is $a_t$? Well from what we figured out earlier, $a_t ~=~ L \alpha$. Here $L$ is the length of the string. So

$$mL {d^2\theta\over dt^2} ~=~ -mg \sin\theta$$ (1.45)

Or

$${d^2\theta\over dt^2} ~=~ -{g\over L}\sin\theta$$ (1.46)

This is the equation we have to solve for this pendulum. Before we solve it, let's look at a more complicated situation, a ``physical pendulum".

Here you can see that we've pivoted this rigid body close to the top and it's going to swing back and forth. Because we're dealing with a rigid body we should use all this stuff that we learned about rigid body motion. That is, $\tau ~=~ I\alpha$.

What is $\tau$? Well the force of gravity acts on the center of mass of the object, and let's say the distance between the pivot point and the center of mass is also called $L$. So we have

$$\tau ~=~ -Lmg\sin\theta$$ (1.47)

So $I\alpha ~=~ \tau$ reads

$$I{d^2\theta\over dt^2} ~=~ -mgL \sin\theta$$ (1.48)

or

$${d^2\theta\over dt^2} ~=~ -{mgL\over I} \sin\theta$$ (1.49)

Both the equation for the simple pendulum and this equation are almost the same. The constant multiplying the sine function is the only difference. So how do we solve this? One method which is a commonly used technique, is to get the answer for very small, deviations in $\theta$. You can check out on a calculator that

$$\sin \theta \approx \theta$$ (1.50)

when $\theta << 1$. So if we stick to very small amplitude oscillations, we can safely use this approximation. Then eqn. 1.49 becomes

$${d^2\theta\over dt^2} ~=~ -{mgL\over I} \theta$$ (1.51)

This looks just like the equation for simple harmonic motion eqn. 1.33, except that we're using $\theta$ as a variable rather than $x$ and our constant $k/m$ has now change to $mgL\over I$. Previously we got that $\omega^2 ~=~ k/m$ so the only difference now is that we have a different constant, so

$$\omega^2 ~=~ {mgL\over I}$$ (1.52)

Let's check out what this says for a simple pendulum. In that case $I ~=~ mL^2$ so

$$\omega^2 ~=~ {mgL\over mL^2} ~=~ {g \over L}$$ (1.53)

As usual the masses cancel, and the answer depends only on gravity and the length of the string. If you quadruple the length of the string, the oscillation frequency goes down by a factor of two, which means the period doubles.

Notice this is only true for small oscillations. Suppose we increase the angle $\theta$, what happens to the period? It actually gets longer and longer. Eventually something catastrophic happens when you start the pendulum from $\theta ~=~ \pi$. Assuming we have a thin rod and not string, you can start off the pendulum from this point. That is, pointing directly up. If you start it pointing exactly up, it'll remain that way indefinitely. Of course that's impossible to do because this is an example of an unstable equilibrium. The point is that, as you approach this point, the ball will stay at the top for a very long time, and it stays there longer the closer you can start it to $\theta ~=~ \pi$. So the period actually diverges in this limit. So we have to keep in mind that our above formula for the oscillation frequency is only true in the limit of small angles.