How to Find the Lagrangian for a Child on a Merry-Go-Round?

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To find the Lagrangian for a child on a merry-go-round, the kinetic energy must be calculated for both the child and the disc, with the potential energy being zero due to the absence of gravity. The kinetic energy for Alice is expressed as T = 0.5 m R'² + 0.25 m R² θ'², while the kinetic energy for the merry-go-round is T = 0.75 m a² θ'². The moment of inertia for Alice, modeled as a point mass, is I = mR². The Lagrangian, defined as L = T - V, leads to the conclusion that the generalized momentum pθ is conserved. This analysis highlights the dynamics of the system and the conservation laws applicable to it.
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Homework Statement


Q) A child, Alice, on a playground merry-go-round can be modeled 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

Homework Equations


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

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?
 
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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.
 
physicsvalk said:
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.

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

Kinetic energy for Alice:

T=0.5 m R`2 + 0.25 m R2 θ`2

Kinetic energy for the merry go round:

T = 0.5 m a2 θ`2 + 0.25 m a2 θ`2 = 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 R`2 + 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 isn't any gravity, V=0
 
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.)
 
physicsvalk said:
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.)

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)
 
For that, yes. Now you can simply apply the Lagrangian and get your answer :D
 
Thank you for the help!
 
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