# Setting Up Lagrangian, David Morin 6.25

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1. Jan 24, 2016

### Timeia

1. The problem statement, all variables and given/known data

A rigid “T” consists of a long rod glued perpendicular to another rod of length l that is pivoted at the origin. The T rotates around in a horizontal plane with constant frequency ω. A mass m is free to slide along the long rod and is connected to the intersection of the rods by a spring with spring constant k and equilibrium length zero. Find r(t), where r is the position of the mass along the long rod. There is a special value of ω; what is it, and why is it special? Figure 6.27 in David Morin Classical Mechanics

2. Relevant equations
L = T - V

3. The attempt at a solution

I am having trouble setting up the Lagrangian for this system so that it is in appropriate coordinate system.

So far T = ½mv^2 + ½ω^2 + ½kx^2

V = -kx - mghsin(ωt)

2. Jan 25, 2016

### Fred Wright

Your expression for the Lagrangian is wrong. I think that the kinetic energy term should be:
(1/2)Iω2 + (1/2)mv2 and the potential energy should be
-(1/2)kr2. I is the moment of inertia of the rod and v is the velocity of the mass in the horizontal plane. I suggest writing the velocity in terms of the rotational angle and the x and y components in the plane, i.e.
x = rcos(ϑ), y = rsin(ϑ),
dx/dt = -rdϑ/dt sin(ϑ) +dr/dtcos(ϑ),
dy/dt =rdϑ/dt + dr/dtsin(ϑ),
and v2=((dx/dt)2 + (dy/dt)2)

3. Jan 31, 2016

### Timeia

okay thanks! Except l is changing with time as well based on where the mass is on the rod, so l = sqrt(l2 + r2)