Pendulum in a braking train

Problem

A pendulum hangs from the ceiling of a railway carriage. The carriage decelerates uniformly. State in which direction the pendulum tilts at steady state and calculate the angle if the deceleration is $2{,}0\ \text{m/s}^2$.

Solution

We analyse the pendulum from the point of view of an observer at rest on the ground (inertial frame). The pendulum, attached to the carriage, decelerates together with it: its acceleration is directed opposite to the motion (backward relative to the direction of travel).

For the pendulum to have this acceleration, the resultant force must point backward. The forces acting are the weight $\vv{P}$ (vertical) and the string tension $\vv{T}$ (along the string). For their sum to be horizontal and directed backward, the string must tilt forward, i.e. in the direction of motion: the mass “carries on” forward by inertia while the carriage slows down.

We resolve the tension into vertical and horizontal directions, with $\theta$ the angle from the vertical.

Vertical direction (no acceleration): $T\cos\theta = mg$

Horizontal direction (the horizontal component of the tension provides the acceleration, of magnitude $a$): $T\sin\theta = ma$

Dividing the two equations term by term, the tension and the mass cancel out: $\tan\theta = \frac{a}{g}$

Substituting $a = 2{,}0\ \text{m/s}^2$ and $g = 9{,}81\ \text{m/s}^2$: $\tan\theta = \frac{2{,}0}{9{,}81} = 0{,}2039 \quad\Rightarrow\quad \theta = \arctan(0{,}2039)$

$\ev{\theta \approx 11{,}5^\circ \quad \text{tilted forward, in the direction of motion}}$

It is the same effect felt on a bus when it brakes: the body tends to be thrown forward, exactly as the pendulum mass does.

Links

Topics: Dynamics Concepts: Newton’s second law · Tension Skills: Vector resolution