Typo in Landau mechanics pendulum problem?

[Peeter]

Homework Statement

Attempting a mechanics problem from Landau's mechanics (3rd edition) I get a different answer, as shown below. Error by me, or typo in the textbook? I can't find any errata page for the text, but since it's an older book, perhaps no such page is maintained.

Chapter 1 problem 3a is to calculate the Lagrangian of a pendulum where the point of support is moving in a circle (figure in this google books url).

Homework Equations

See below.

The Attempt at a Solution

The coordinates of the mass are

$$p = a e^{i \gamma t} + i l e^{i\phi},$$

or in coordinates

$$p = (a \cos\gamma t + l \sin\phi, -a \sin\gamma t + l \cos\phi).$$

The velocity is

$$\dot{p} = (-a \gamma \sin\gamma t + l \dot{\phi} \cos\phi, -a \gamma \cos\gamma t - l \dot{\phi} \sin\phi),$$

and in the square
$$\dot{p}^2 = a^2 \gamma^2 + l^2 \dot{\phi}^2 - 2 a \gamma \dot{\phi} \sin\gamma t \cos\phi + 2 a \gamma l \dot{\phi} \cos \gamma t \sin\phi=a^2 \gamma^2 + l^2 \dot{\phi}^2 + 2 a \gamma l \dot{\phi} \sin (\gamma t - \phi).$$

For the potential our height above the minimum is

$$h = 2a + l - a (1 -\cos\gamma t) - l \cos\phi = a ( 1 + \cos\gamma t) + l (1 - \cos\phi).$$

In the potential the total derivative $\cos\gamma t$ can be dropped, as can all the constant terms, leaving

$$U = - m g l \cos\phi,$$

so by the above the Lagrangian should be (after also dropping the constant term $m a^2 \gamma^2/2$
$$\mathcal{L} = \frac{1}{{2}} m \left( l^2 \dot{\phi}^2 + 2 a \gamma l \dot{\phi} \sin (\gamma t - \phi) \right) + m g l \cos\phi.$$

This is almost the stated value in the text
$$\mathcal{L} = \frac{1}{{2}} m \left( l^2 \dot{\phi}^2 + 2 a \gamma^2 l \sin (\gamma t - \phi) \right) + m g l \cos\phi.$$

It looks like an innocent enough typo (text putting in a $\gamma$ instead of a $\dot{\phi}$). Also oddly, there's a second reference after that point that also doesn't make sense where they refer to the omission of the total derivative $m l a \gamma \cos( \phi - \gamma t)$ ... a term that I didn't have when multiplying out my velocity?

Is there consensus that there are a pair of typos here, and if not, can somebody spot the error in my calculation?

 

[Dickfore]

Your calculation looks correct.

However, notice that the second term in the kinetic energy is:
$$\begin{align} & m a l \, \gamma \, \dot{\phi} \, \sin(\gamma t - \phi) \\ = & -m a l \, \gamma \, (\gamma - \dot{\phi}) \, \sin(\gamma t - \phi) + m a l \, \gamma^2 \, \sin(\gamma t - \phi) \\ = & \frac{d}{dt} \,\left(m a l \, \gamma \, \cos(\gamma t - \phi) \right) + m a l \, \gamma^2 \, \sin(\gamma t - \phi) \end{align}$$
After dropping the total time derivative, you get the master's result.

 

[Peeter]

After dropping the total time derivative, you get the master's result.

Very clever (of both you and the master;). Thanks!