Pendulum in a stopping train Can a pendulum reach the vertical ($\theta = 180^°$) in a train that brakes hard? I think the answer is yes but I don't see how to show that. I used the frame of reference of the train but I found that $W(\vec{f_{ie}})=0$ where $\vec{f_{ie}}=m\vec{a}$ (initially the train is at constant speed, so the force is perpendicular to displacement). This is the diagram I did:
 A: Suppose that the pendulum is initially vertical when suddenly a constant acceleration $\vec{A} = -A\hat x$ is applied to the train. In the reference frame of the train, this will cause an inertial force of $-m\vec A = mA\hat x$ to be applied on the mass. Putting aside the tension in the string for a moment, the mass moves under the influence of the force $m(\vec g + \vec A) = m(-g\hat y + A \hat x) $.
This problem is equivalent to that of an ordinary pendulum moving under uniform gravity, with $\vec g$ replaced by $\vec g + \vec{A}$. The direction of $\vec{g} + \vec{A}$ defines the equilibrium direction about which the pendulum swings (and where it points if it is not swinging at all).
When $\vec{g} + \vec{A}$ makes an angle $\theta$ with the vertical ($-\hat y$), the pendulum will swing another $\theta$ after passing that direction before it comes to a stop. Therefore the highest the pendulum will get is at $2\theta$ from the downward direction. The condition is therefore
$$2\theta \ge 180^\circ$$
$$\theta \ge 90^\circ$$
meaning $\vec g + \vec A$ would need to have an upward component, or at least be horizontal. For no finite value of $\vec A$ is $\vec g + \vec A$ horizontal, so the pendulum never points upward.
If you remove the restriction that the starting position be vertical or the acceleration be uniform, the pendulum can get to the point where it points upward. An easy example is where $A \ge g$ and the pendulum initially points along $- \hat x$ (or $\theta = -90^\circ$).
For an example where the pendulum initially points downward but the acceleration is non-uniform, suppose the train initially moves with a speed of $V$ but comes to an instantaneous halt. In the train's frame, the pendulum starts moving right with a speed of $V$. The pendulum will get vertical if
$$V \ge \sqrt{5gR}$$
where $R$ is the length of the string. You can show this using energy conservation, as follows. The centrifugal force when the pendulum reaches the top  is provided by the string tension $T$ and the weight $mg$:
$$mg + T = mv^2/R.$$
In order for the string to remain taut, we require
$$T = m(v^2/R - g) \ge 0$$
$$v \ge \sqrt{gR}. $$
Conservation of mechanical energy implies
$$ \frac{1}{2}mV^2 = \frac{1}{2}mv^2 + 2mgR $$
$$ V \ge \sqrt{5gR}. $$
If the pendulum uses a rigid rod instead of a string, the requirements are $v \ge 0$ and $V \ge 2\sqrt{gR}$ instead.
