# Lagrangian question - rough ball on moving wedge

1. Jun 22, 2013

### venerium

1. The problem statement, all variables and given/known data
(context: I'm studying for a test, and this is a question from a past exam paper.)

"A wedge of mass M with angle $\phi$ is free to slide on a frictionless horizontal table. A solid ball of radius a and mass m is placed on the slope of the wedge. The contact between the ball and the wedge is perfectly rough.
(a) Derive an expression for the total kinetic energy of this system."

3. The attempt at a solution

I've chosen X to represent the horizontal displacement of the wedge, and x as the displacement of the ball along the surface of the wedge.

T for the wedge is easy enough $\frac{1}{2}M\dot{X}^2$.

Total T for the ball must be: T due to rotational motion of the ball + T due to linear velocity (down the wedge + due to the wedge's own velocity).

We're given I = $\frac{2}{5}ma^2$ for the ball, so I get:
$T_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}\frac{2}{5}ma^2\omega^2 = \frac{1}{5}ma^2\frac{\dot{x}^2}{a^2} = \frac{m\dot{x}^2}{5}$
(Am I correct in only using the $\dot{x}$ velocity component for $\omega = \frac{v}{r}$?)

Using my coordinates, the total velocity of the ball is:
$\vec{v} = \dot{X}\widehat{I} + \dot{x}\widehat{i}$ (i being the unit vector in the direction of x)

From worked solutions to similar problems with blocks I've found online, I think the linear T for the ball must be:

$T = \frac{1}{2}m\vec{v}\cdot\vec{v} = \frac{1}{2}m\dot{X}^2 + \frac{1}{2}m\dot{x}^2 + m\dot{X}\dot{x}\cos{\phi}$

But I'm not sure where that third $2\dot{X}\dot{x}\cos{\phi}$ term is coming from in the evaluation of the dot product. I understand that it couldn't just be $\dot{x}^2 + \dot{X}^2$, but how do you go about evaluating the dot product in this case?

Any hints would be really appreciated, I'm still trying to develop an intuition for these Lagrangian problems. Cheers!

2. Jun 22, 2013

### TSny

Welcome to PF, venerium!

For any two vectors A and B,

(A + B)$\cdot$(A + B) = A$\:\cdot$A + 2A$\:\cdot$B + B$\:\cdot$B.

You can show this using the more basic identity:

A$\:\cdot$(B + C) = A$\:\cdot$B + A$\:\cdot$C

3. Jun 22, 2013

### venerium

Ah, thank you! Thought it would be something simple - I'm a bit rusty on my math. Does the rest of my solution/thinking look fine?

And thanks for the welcome, I've been a lurker for years now so it's about time I joined. :P

4. Jun 22, 2013

### TSny

It all looks good to me.

However, $X$ and $x$ are not independent variables. So, you would need to go on and write T in terms of just one of the variables $\dot{X}$ or $\dot{x}$ before using it in the Lagrangian.