I'm reviewing some classical mechanics and came across a problem from Colombia Universities physics Ph.D. qualification exam. The question is this:

A mass $m$ slides down a circularly curved surface on an object with mass $M$. Mass $M$ is free to slide on a frictionless surface. What are the final speeds of the two masses after $m$ separates from $M$?

enter image description here

Now this problem can be solved easily using conservation of momentum and energy. i.e., \begin{equation} mgR = \dfrac{1}{2} mv^2 + \dfrac{1}{2} MV^2 \end{equation} and since the horizontal momentum is conserved, \begin{equation} mv - MV = 0 \end{equation}

Combining these two gives the desired velocities. I won't post them here since my question isn't about the velocities, rather the Lagrangian of the system.

In the problem's solution, the kinetic energy term of the Lagrangian is \begin{equation} T = \dfrac{1}{2}M\dot{x}_M^2 + \dfrac{1}{2}m (\dot{x}_M^2 + 2R\dot{\theta} \dot{x}_M \sin{\theta} + R^2 \dot{\theta}^2 ). \end{equation}

For the small block, $m$, I'm specifically confused about the center cross term $2R\dot{\theta} \dot{x}_M \sin{\theta}$. Where does this come from? For a more thorough response, what are the generalized coordinates used here and how do they build the kinetic energy terms in the Lagrangian?

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    $\begingroup$ Answering your primary question requires understanding what you request as a "more thorough response"; only by understanding the generalized coordinates can you make sense of the Lagrangian. My approach would be understanding the coordinates first and then checking that understanding by getting to the KE term independently. $\endgroup$ Aug 5, 2019 at 23:14
  • $\begingroup$ I think the crux of your confusion is what $x_m$ is? It looks to be the displacement of the left edge of the larger block from some reference point. Then the $x$ coordinate of the smaller block can be written $x_s = x_m - R \cos(\theta) + const$. (The constant is irrelevant but for neatness sake you could work it out and then pick your reference position to make it zero.) $\endgroup$
    – jacob1729
    Aug 5, 2019 at 23:33

2 Answers 2


It is easier to start with Cartesian coordinates.

The first term is just the kinetic energy of $M$


The kinetic energy of $m$ is just

$$\frac {1} {2} m((\dot{x}-\dot{x}_M)^2+\dot{y}^2)$$

where $x$ and $y$ represent the position of $m$ with respect to the block.


$$x=R\cos(\theta),\quad y=R\sin(\theta)$$

Make the substitution for $x$ and $y$ and you will get the given Lagrangian.

I know that it would take someone much more intuitive than I to be able to write down the given Lagrangian by inspection. Writing the Cartesian coordinate Lagrangian and converting is fairly straightforward, however.


enter image description here

Step by Step:

I) choose inertial frame

II) write the position vectors:

$$\vec{R}_M=\begin{bmatrix} x_M \\ 0 \\ \end{bmatrix}$$

$$\vec{R}_m=\begin{bmatrix} x_M +R\,\cos(\theta)\\ R\,\sin(\theta) \\ \end{bmatrix}$$

III) calculate the velocities $\vec{v}=\vec{\dot{R}}$ thus : $$\vec{\dot{R}}_M=\begin{bmatrix} \dot{x}_M \\ 0 \\ \end{bmatrix}$$

$$\vec{\dot{R}}_m=\begin{bmatrix} \dot{x}_M -R\,\dot{\theta}\,\,\sin(\theta)\\ R\,\dot{\theta}\,\cos(\theta) \\ \end{bmatrix}$$

IV) Kinetic energy

$$T=\frac{1}{2}\,M\,\vec{\dot{R}}_M\cdot \vec{\dot{R}}_M+ \frac{1}{2}\,m\,\vec{\dot{R}}_m\cdot \vec{\dot{R}}_m$$

$$T=\frac{1}{2}\,M\, \dot{x}_M^2+ \frac{1}{2}\,m\,\left(\dot{x}_M^2+2\,\dot{x}_M\,\dot{\theta}\,r\,\sin(\theta)+R^2\,\dot{\theta}^2\right)$$

V) Potential energy:


we have two degree of freedom $x_m$ and $\theta$ but just one generalized coordinate, so we need one constraint equation:

VI) constraint equation


from this equation we can eliminate one degree of freedom:


$$x_M=-2\,R\,\cos(\theta)$$ $$\dot{x}_M=2\,R\,\dot{\theta}\,\sin(\theta)$$

now we can work with the generalized coordinate $\theta$ thus

$T=T(\theta\,,\dot{\theta})\quad$ and $\quad\,U=U(\theta)$

you can now obtain the equation of motion with EL method

VII) The equation of motion:

$$\boxed{\frac{d^2 \theta(\tau)}{d\tau^2}={\frac {\cos \left( \theta \left( \tau \right) \right) \left( 4\,R \left( {\frac {d}{d\tau}}\theta \left( \tau \right) \right) ^{2}M \sin \left( \theta \left( \tau \right) \right) +mg \right) }{R \left( -4\,M-m+4\, \left( \cos \left( \theta \left( \tau \right) \right) \right) ^{2}M \right) }}} $$


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