Restoring Force

Question

Consider a pendulum hanging straight down. Say I pull it to the $\textbf{right}$ some angle $\phi$. Using Newton's laws in polar coordinates, I find that in the that force in the $\hat\phi$ direction is simply the weight in that direction pulling it back down to equilibrium.

$$F_{\phi} = -mg\sin\phi$$

I get that it is negative because the force acts to pull the pendulum bob back down to $\phi = 0$. Also, usually when Newton's laws are derived in polar coordinates, $\hat\phi$ is make to point counterclockwise.

My question is what if the pendulum was pulled to the $\textbf{left}$ and let go? Would $F_{\phi}$ still have that negative sign? In polar coordinates, $\phi$ is positive going counterclockwise, which is the way the pendulum will go now that I pulled it to the left. However, I feel like it would still be negative. So does that mean $\phi$ is positive only when the pendulum is swinging away from equilibrium (it swings away from equilibrium in two directions, one counterclockwise and the other clockwise) and negative when it is swinging towards it? How can this be?

Answer 1

$F_\phi = -mg\sin\phi$ is valid for any direction. But remember $\sin\phi$ can itself be negative, changing the overall sign of the quantity $F_\phi$.

If positive $\phi$ is counterclockwise, then when the bob is to the left (clockwise) of equilibrium, $\phi<0$. This in turn means $\sin\phi<0$ so that $F_\phi=-mg\sin\phi>0$. So even with the negative sign in front of that expression, it can be positive.

The end result is $F=-mg\sin\phi \hat \phi$ points toward the equilibrium point (positive direction) when the bob is clockwise (left) of the equilibrium, just as you would expect.