I am looking at this problem in Morin's Introduction to classical mechanics:
6.1. Moving plane A block of mass $m$ is held motionless on a frictionless plane of mass $M$ and angle of
inclination $θ$ (see Fig. 6.8). The plane rests on a frictionless horizontal surface. The
block is released. What is the horizontal acceleration of the plane?
Here is the more detailed diagram associated with the solution:
The quantities $x_1$ and $x_2$ represent the horizontal displacement of the plane and the block in their respective directions. Later on I switch to $q_1$ and $q_2$ to keep my sanity as I go through the Lagrangian and Hamiltonian.
The proposed solution sets out the Lagrangian in terms of the kinetic energy
$$
K=\frac12M\dot q_1^2 + \frac12 m ((\dot q_1 + \dot q_2)^2\tan^2(\theta) + \dot q_2^2)
$$
and gravitational potential
$$
V=-mg(q_1+q_2)\tan(\theta)
$$
where I have replaced $x_i$ in the diagram with $q_i$.
This yields the Lagrangian
$$
L=K-V=\frac12M\dot q_1^2 + \frac12 m ((\dot q_1 + \dot q_2)^2\tan^2(\theta) + \dot q_2^2)+mg(q_1+q_2)\tan(\theta)
$$
which, when processed with the Euler-Lagrange equations gives a solution ('after a little simplification')
$$
\ddot q_1=\frac{mg\sin(\theta)\cos(\theta)}{M+m\sin^2(\theta)}
$$
(I did my best to process it with Sympy in python and got that solution as
$$
\ddot q_1=\frac{mg\tan(\theta)}{M \tan^2(\theta) + M + m\tan^2(\theta)}
$$
I think I see why it is equivalent after dividing top and bottom by $\cos^2(\theta)$, so I feel good that I have things set up right.)
To crosscheck things I also wanted to apply the Hamiltonian procedure, and so after replacing the $\dot q_1=\frac{p_1}{M}$ and $\dot q_2=\frac{p_2}{m}$
I had the Hamiltonian
$$
H=K+V=-mg(q_1 + q_2)\tan(\theta) + \frac12 m ((p_2/m + p_1/M)^2\tan^2(\theta) + p_2^2/m^2) + \frac12 p_1^2/M
$$
But processing this with Hamilton's equation $\dot p=-\frac{\partial H}{\partial q}$
I think that should mean that $\frac{\ddot q_1}{M}=\dot p_1=mg\tan(\theta)$ which (I think?) is not the same as the one suggested in the solution in the book. This seems to be what I get both by hand and symbolically with Sympy.
So at the moment (no pun intended) I'm stumped as to what I'm doing wrong. Can I not use the time derivative of the momentum to check the acceleration like this?
Best Answer
The misconception OP is having is the approach to the Hamiltonian method.
The Hamilton equation $\dot{p_i} = - \frac{\partial H}{\partial q_i}$ does not apply to the "ordinary momentum" $p_i$. It instead applies to the canonically conjugate momentum, which is defined as $p_i = \frac{\partial L}{\partial \dot{q_i}}$.
In simple cases, the part of the Lagrangian that depends on $\dot{q}_i$ is $\frac{1}{2} m_i \dot{q_i}^2$, so the canonically conjugate momentum is indeed $m q_i$. In this case, however, the canonically conjugate momenta are
$p_1 = M \dot{q_1} + m \tan^2\theta (\dot{q_1} + \dot{q_2})$
$p_2 = m (\dot{q_2} + \tan^2 \theta (\dot{q_1} + \dot{q_2}))$
As you can see, there is some annoying entanglement going on here between the two variables. This is reflected by the fact that energy depends on $(q_1 + q_2)^2$, in part.