Turns out Princeton makes some really, really good problems to do if you like doing physics. So, just like when your binge-watching a good TV show, I need another one, another one, another one…

In fact… lets write some python code for it

while True:
  s = "another one, "
  print("I need " + s)

Don’t actually run it - its an infinite loop and might crash your system.

If you do run it, hit CTRL+C if you want it to stop.

Problem

This is Princeton Graduate Prelims J08M.1, entitled “Pendulum on a Sled”.

A plane pendulum consists of a bob of mass $m$ suspended by a massless rigid rod of length $\ell$ that is hinged to a sled of mass $M$. The sled slides without friction on a horizontal rail. Gravity acts with the ususal downward acceleration $g$.

  1. Taking $x$ and $\theta$ as generalized coordinates, write the Lagrangian for the system.
  2. Derive the equations of motion for the system.
  3. Find the frequency $\omega$ for small oscillations of the bob about the vertical.
  4. At time $t=0$ the bob and the sled, which had been previously at rest, are set in motion by a sharp tap delivered to the bob. The tap imparts a horizontal impulse $\Delta P = F \Delta t$ to the bob. Find expressions for the values of $\dot \theta$ and $\dot x$ just after the impulse.

Part 1

We let $x$ be the horizontal displacement of the sled and let $\theta$ be the angle made with the vertical by the pendulum. Letting the sled’s original location be the origin, the coordinates of the sled are $(x,0)$ and the coordinates of the bob are $(x + \ell \sin{\theta}, -\ell \cos \theta)$. The total kinetic energy is then

\[K = \frac{1}{2} (M+m) \dot{x}^2 + m \dot{x} \ell \dot{\theta} \cos{\theta} + \frac{1}{2} m \ell^2 \dot{\theta}^2\]

Only the bob has any changing potential energy, so up to a constant, the potential energy is

\[U = -mg \ell \cos{\theta}\]

The Lagrangian is then

\[\mathcal{L} = T-U = \frac{1}{2} (M+m) \dot{x}^2 + m \dot{x} \ell \dot{\theta} \cos{\theta} + \frac{1}{2} m \ell^2 \dot{\theta}^2 + mg \ell \cos{\theta}\]

Part 2

We use the Euler-Lagrange equations on both $\theta$ and $x$, starting with the former.

\[\ell \ddot{\theta} + \ddot{x} \cos{\theta} + g \sin{\theta} = 0\]

and

\[(M+m)\ddot{x} + m \ell \ddot{\theta} \cos{\theta} = m \ell \dot{\theta}^2 \sin{\theta}\]

Part 3

For small oscillations, $\theta \ll 1$. This means that we can use the small angle approximation, so $\sin \theta \approx \theta$ and $\cos \theta \approx 1$. Moreover, $\dot{\theta}^2 \ll \frac{g}{\ell}$, so we can ignore terms that include this factor. We can then rewrite our equations as

\[\ddot{x} + \ell \ddot{\theta} + g \theta = 0\]

and

\[(M+m) \ddot{x} + m \ell \ddot{\theta} = 0\]

Using the latter equation to solve for $\ddot{x} = - \frac{m}{M+m} \ell \ddot{\theta}$, we can substitute to find that

\[\ddot{\theta} = - \frac{g}{\ell} \frac{M+m}{M} \theta\]

This means that the frequency of oscillations is $\omega = \sqrt{\frac{g}{\ell} \frac{M+m}{M}}$.

Part 4

This impulse just gives us initial conditions. Using the effective mass of $M+m$, $\dot{x} = \frac{\Delta P}{M+m}$.

By conservation of momentum, $(M+m) \dot{x} + m \ell \dot{\theta} \cos{\theta}$ is a conserved quantity. At this moment, $\theta = 0$, so $\cos{\theta} = 1$. This means that $\dot{\theta} = - \frac{\Delta P}{m \ell}$.

Aftermath

Another nice problem by the Princeton Grad Committee. Thanks!

Also another way to use the Euler-Lagrange equation, which is a fun way of doing problems using only energy.