First of all, if $\theta_0=0$, then obviously the particle will just sit there forever at the bottom of the hill. This is called a stable equilibrium point. Likewise, if you set the pendulum at rest at the bottom of its arc, it will stay there forever.

Now what if you make $\theta_0$ a little bigger? The particle will oscillate back and forth on the hill between $\pm \theta_0$. Try dragging the initial angle slider again to some small value in the above animation and then press start. The horizontal dashed line represents the total energy of the particle. If $\theta_0$ is still fairly small, the particle just gently rocks back and forth between the turning points—the points where the potential energy curve intersects the total energy line. When the total energy equals the potential, there's nothing left over for the kinetic energy, meaning that the particle has stopped and is about to turn around. This gentle rocking at small angles corresponds to the simple case where the pendulum oscillates side-to-side in a sinusoidal pattern. It's called simple harmonic motion, and it crops up (almost) any time you perturb a system slightly away from a stable equilibrium point.

But what if we release the pendulum from a big angle, like $\theta_0$ just shy of $\pi$? This is the hard case, where we can't write down a simple solution for $\theta(t)$. We saw that the pendulum will again oscillate back and forth, but it spends a long time turning around at the biggest angles $\pm \theta_0$. But we can see why very easily just by glancing at the shape of the potential! Notice how the hill flattens out near the peaks at $\theta = \pm \pi.$ Then as the particle climbs up the hill, it slowly comes to a stop because it's crawling up a very gradual incline. And likewise as it falls back down, it slowly picks up speed because it's such a gentle slope. Try dragging the angle to a large initial value now, press reset, and then press start again.

To be clear, the motion of the pendulum is not literally the same as the motion of a particle on a hill—the two setups have their own $F = ma$ equations, and they're not interchangeable. But the qualitative features of the two motions are very similar, and that's why the particle on a hill is a useful analogy.

So even if you don't have a computer to numerically solve the $F=ma$ equation like I've done in the animation, you could predict for yourself what the motion of the pendulum would look like with almost no work just by drawing the picture of the potential!

I want to show you one more example to prove to you just how powerful this idea is. This time we're going to look at the orbit of a planet like the Earth around a star like the Sun, or even something like a comet passing through the solar system. You probably know that the orbit of a planet around a star is an ellipse, but is it possible to understand why without going through all the work of solving the $F=ma$ equations for the planet? And that is quite a bit of work—as I'll show you in an upcoming video where we'll derive the orbit of a planet around a star. Make sure you're subscribed on YouTube to see that video when it comes out.

Say that we have a planet or a comet of mass $m$ that's moving around a star of mass $M$. The star is going to be much more massive than the planet, so it's a very good approximation to assume that the star remains fixed in one place, and that's where we'll put our origin:

According to Newton's law of gravity, the force on the planet will then be

$$F = - \frac{GMm}{r^2}$$

pointing toward the star at the origin—that's what the minus sign means here. Now you can write down the $F=ma$ equations and attempt to solve them, either analytically or numerically. They're tough equations, but with a bit of work you can derive, say, the elliptical trajectory of a planet caught in the star's orbit. I'll show you how to do that in that upcoming video. But again, it takes quite a bit work; can we understand the shape of the orbit more quickly by thinking about the shape of the potential again?

Remember that the force is minus the slope of the potential, $F(r) = -U'(r)$, so that the gravitational potential energy is

$$U = -\frac{GMm}{r}.$$

Here's what it looks like:

According to what we've learned, the particle will slide right on down the hill, and the planet will likewise plunge straight into the starry fireball at $r = 0$, right? Fortunately not! We need to remember that this isn't just a one-dimensional problem. The planet will move around in a plane, with coordinates $x$ and $y$ or, equivalently, polar coordinates $r$ and $\theta$.

Let's write down the total energy. Like the pendulum, the planet will have an angular component of the velocity $\omega r$. But because $r$ isn't fixed here as it was for the pendulum, the planet can also have a radial component of velocity $v_r.$ Then the kinetic energy gets contributions from both pieces, $\frac{1}{2} m v^2 = \frac{1}{2} mv_r{}^2 + \frac{1}{2} m r^2 \omega^2$, and the total energy is

$$E = \frac{1}{2} m v_r{}^2 + \frac{1}{2} m r^2 \omega^2 - \frac{GMm}{r}.$$

Once again, you can show that $E$ is a constant of the motion. And while we're on the subject, there's another constant of the motion—the angular momentum $L$:

$$L = mr^2 \omega.$$

You can again prove that $\frac{\mathrm{d}L}{\mathrm{d}t} = 0$ by taking the derivative and applying the $F = ma$ equations. I'll show you how that works in that upcoming video. For now just note that the fact that $L = mr^2\omega$ is constant means that if we know $r$ at any given time, we automatically know $\omega = \frac{L}{mr^2}.$ We can use that to eliminate $\omega$ from our expression for the energy:

$$E = \frac{1}{2} m v_r{}^2 + \frac{L^2}{2mr^2} - \frac{GMm}{r}.$$

Something remarkable has happened: we've eliminated all references to the other coordinate $\theta$, and written the energy entirely in terms of one coordinate $r$. It looks just like the standard energy $E = \frac{1}{2} m v_r{}^2 + U_\mathrm{eff}(r)$ for a particle moving in one dimension, but instead of the potential being $-\frac{GMm}{r}$ alone, an additional term gets added to it:

$$U_\mathrm{eff}(r) = -\frac{GMm}{r} + \frac{L^2}{2mr^2}.$$

$U_\mathrm{eff}$ is called the effective potential, and it's what will enable us to understand the orbits at a glance by drawing a picture. Whereas the $\propto - \frac{1}{r}$ potential hill had catastrophic consequences that made it seem like a planet will fall to its doom, now we can see that if the planet has angular momentum $L$, the story is fortunately quite different. As $r\to 0$, $\frac{1}{r^2}$ becomes big (and positive) much faster than $-\frac{1}{r}$ becomes super negative because of that power of 2. So near $r = 0$, the new term is what dominates the effective potential, and it drastically changes the shape:

This is very good news if you're hoping not to plummet to your death in a giant ball of fire! A particle moving along the effective potential hill won't fall down to $r = 0$, because the hill climbs higher and higher as you approach.

Now we can understand the shape of the planet's motion by thinking about how a particle would slide along the effective potential hill. Let's draw one more animation to see what's going on.