# Time-dependent Lagrangian problem

1. Nov 27, 2011

### saul goodman

1. The problem statement, all variables and given/known data
Q) A child, Alice, on a playground merry-go-round can be modelled as a point mass m on a homogeneous horizontal disc of mass M and radius a. The disc rotates without friction about a vertical axis through its center. Alice clings to a straight railing that extends from the center of the disc to its perimeter. Alice's distance R(t) from the centre is a function of time determined by her muscles, while the angle θ between the railing and (say) the East is a dynamical variable

Find the Lagrangian for the system. Deduce from Lagrangian that pθ (momentum) is conserved

2. Relevant equations
The disc's (merry-go-round) momentum of inertia is 0.5ma^2

3. The attempt at a solution
In all honesty, I haven't been able to give a serious attempt at this. In lectures we have done no time-dependent examples. Obviously I have to use the formula L=T-V (kinetic - potential energy) however I don't know how I would begin to work out the kinetic energy. Should I start with working out the center of mass?

2. Nov 27, 2011

### physicsvalk

You can figure out the T based on the center of mass of the system, but it's much easier to break the T into two (Alice and the merry go round), and then sum them up.

There wouldn't be any gravitational V since (we hope) she doesn't fall. Because of this the Lagrangian should give you a conservation of the generalized momentum.

3. Nov 27, 2011

### saul goodman

Okay thanks a lot. Well if we do it like that I get:

Kinetic energy for Alice:

T=0.5 m R2 + 0.25 m R2 θ2

Kinetic energy for the merry go round:

T = 0.5 m a2 θ2 + 0.25 m a2 θ2 = 0.75 m a2 θ

Although I'm not confident with these answers. In my notes kinetic energy in a system is defined as T= 0.5 M R2 + 0.5 I θ`2 where I is the moment of inertia, but I'm not sure if the moment of inertia for the girl is the same as that for a disc...

Oh and I'm guessing since there isnt any gravity, V=0

4. Nov 27, 2011

### physicsvalk

What's the moment of inertia of one point particle, rotating about an axis? (It might help to know that Alice's I is the same as the I for a ring of negligible thickness, rotating about an axis perpendicular to its center.)

5. Nov 27, 2011

### saul goodman

I=mr2 for a particle rotating about an axis, so unless I'm missing something the moment of inertia is simply I=mR^2 for Alice? (which is what I wrote in my previous post)

6. Nov 27, 2011

### physicsvalk

For that, yes. Now you can simply apply the Lagrangian and get your answer :D

7. Nov 27, 2011

### saul goodman

Thank you for the help!