A simple pendulum hangs in a wagon moving with constant…

A simple pendulum hangs in a wagon moving with constant acceleration $a$. How does its equilibrium angle $\theta$ from the vertical depend on $a$?

Explanation of Method

A pendulum in an accelerating wagon.

Step 1: Forces on the bob

In the wagon's frame the bob is at rest under gravity $mg$ (down), the inertial force $ma$ (backward, opposite to $\vec a$) and the string tension $T$. The string must point along the resultant of the first two.

Step 2: Angle

\[ T\sin\theta = ma, \]

\[ T\cos\theta = mg. \]

\[\begin{gathered} \theta = \arctan\frac{a}{g} \\ \text{(bob deflected backward)} \end{gathered}\]

The string tension is $T = m\sqrt{g^2 + a^2}$.

$a$ $mg$ $ma \ \text{(inertial)}$ $T$ $\theta$ $\tan\theta = \frac{a}{g}$

Drawn for $a = 0.4g$. In the wagon's frame the bob feels gravity and a backward inertial force $ma$; the string lines up with their resultant, so the pendulum leans backward, opposite to the acceleration.

Result

\[\boxed{\,\begin{gathered} \theta = \arctan\frac{a}{g} \\ \text{(bob deflected backward)} \end{gathered}\,}\]

Remark

About the original solution. The formula $\tan\theta = a/g$ is correct, but it says the pendulum "leans forward". It leans backward, opposite to the acceleration, like a passenger pressed into the seat of an accelerating car. (Seen from the ground: the string must pull the bob forward, so it has to tilt forward from the bob to the suspension point, i.e. the bob hangs behind.)