Sliding bar constraint equations

[v0id19]

Homework Statement

A uniform rod of length l rests on a horizontal floor and leans against a vertical wall, making an angle $$\theta$$ with the floor. It is initially held at rest. At t = 0, the rod is released and falls, sliding on the floor and the wall with no friction. The only forces acting on the rod are gravity and the forces of constraint.
1. Write the kinetic energy of the rod using the Cartesian coordinates of its center of mass and the angle $$\theta$$. Note that this calculation will require an integral.
2. Write the Lagrangian and the equations of constraint, again using $$\theta$$ and the Cartesian coordinates of the center of mass of the rod as your generalized coordinates.

Homework Equations

$$L = T - U$$
$$\lambda \frac{dg}{dq_i}=\frac{dL}{dq_i} - \frac{d}{dt} \frac{dL}{dq^{.}_i}$$
Where g is the constraint equation and q_i is the generalized coordinate.

The Attempt at a Solution

$$L = 1/2 m [x^{.}^2 + y^{.}^2 - 1/12 l^2 \theta ^{.}^2] - mgy$$
Where (x,y) is defined as the center of mass of the rod and m is the mass of the rod.
$$g_1 (y, \theta ) = l/2 sin( \theta ), g_2 (x, \theta ) = l cos( \theta ) - 2x = 0$$

If I do this, I substituted in g₁ for y to have the equation just in terms of x and $$\theta$$. This gives me a really ugly differential equation, so I feel like I'm doing something wrong. I eventually have to solve the equations to find when the end of the bar leaves the wall, which I would do by setting $$\lambda$$ equal to 1, but I haven't gotten that far yet because of the nastiness of the differential equations...
Any help (especially with getting the constraint equations) would be greatly appreciated.

 

[ehild]

You expressed the coordinates of the CM with the angle θ:
x=L/2 cosθ, y=L/2 sinθ.
Determine the components of the velocity, dx/dt and dy/dt in terms of dθ/dt (the angular velocity of rotation around the CM, shown in the picture, is d(π-θ)/dt=-dθ/dt. Substitute into the Lagrangian: it will be the function of the single variable θ. Show what you got.

ehild

 

[ashishsinghal]

You expressed the coordinates of the CM with the angle θ:
x=L/2 cosθ, y=L/2 sinθ.
Determine the components of the velocity, dx/dt and dy/dt in terms of dθ/dt (the angular velocity of rotation around the CM, shown in the picture, is d(π-θ)/dt=-dθ/dt. Substitute into the Lagrangian: it will be the function of the single variable θ. Show what you got.

ehild

Hey ehild,
You are in the same question once again.

Hey v0id19,
I have just finished on a thread very much similar to yours. Check that out for some help.

 

[ehild]

The previous thread is a bit messy. Let's start again, with a blank page.

ehild

 

[v0id19]

You expressed the coordinates of the CM with the angle θ:
x=L/2 cosθ, y=L/2 sinθ.
Determine the components of the velocity, dx/dt and dy/dt in terms of dθ/dt (the angular velocity of rotation around the CM, shown in the picture, is d(π-θ)/dt=-dθ/dt. Substitute into the Lagrangian: it will be the function of the single variable θ. Show what you got.

ehild

Thanks, this is what I wanted to do, but the fact that the question specifies cartesian coordinates is what's confusing me. I agree that it makes much more sense to do it this way.
I'll check the other thread and see if it helps.

 

[I like Serena]

Thanks, this is what I wanted to do, but the fact that the question specifies cartesian coordinates is what's confusing me. I agree that it makes much more sense to do it this way.
I'll check the other thread and see if it helps.

It's nice that we can calculate the kinetic energy with difficult lagrangian equations and differential equations, but isn't it easier to use conservation of energy?

That is:
E = T + V = constant ==> mgl/2 = T + mgy ==> T = mg(l/2 - y) = (1/2) mgl (1 - sin θ)

 

[ehild]

If it must be solved with Lagrangian and constrains, let it be.
In your Lagrangian function, the rotational kinetic energy has to be positive. (It is 1/2 I (-dθ/dt)²=1/2 I (dθ/dt)²)

The constrains are (I think you meant them in this way)

f1: y-L/2 sinθ = 0
f2: 2x-Lcosθ = 0

You get three Euler's differential equations: for y, for x and for θ, which contain the partial derivatives of both constrains, and have two λ-s, λ₁ for f1, λ₂ for f2.
Loosing contact with the wall means that λ₂=0.

The other thread uses the conservation of energy.

ehild