Rewrite Lagrangian's equation of motion as a 1st order difeq

[noon0788]

I'm most of the way through this problem. I've already found the lagrangian equations of motion as a second order differential equation. I'm just stuck at the end...

Homework Statement

Derive a first order differential equation for the motion of a hoop in a bigger hoop.

Homework Equations

The Attempt at a Solution

I found the equation of motion to be:

0=g Sin(θ) + 3/2*R*Doubledot[θ]

I think a conserved quantity is angular momentum = 3/2*m*k²*Dot[θ]

How do I write θ as a first order difeq?

 

[mathfeel]

You don't state all your variables clearly and it is difficult for me to figure out what system you are solving for.

If you are correct and$\frac{3}{2} m k^2 \dot{\theta}$ is indeed a COM (I don't know what $k$ is since you did not state in your original question and I am going to assume that it does not vary with time), then: $\ddot{\theta} = 0$, which would reduce the EOM to just $0 = g \sin\theta$.

 

[noon0788]

Sorry, here's some clarification:

Basically, it's a hoop that's allowed to move in a larger fixed hoop without slipping.

k is the radius between the center of the free hoop and the bigger hoop that it's in. It does not vary with time.

If the EOM is 0=gsinθ, isn't that just trivial?

 

[mathfeel]

Sorry, here's some clarification:

Basically, it's a hoop that's allowed to move in a larger fixed hoop without slipping.

k is the radius between the center of the free hoop and the bigger hoop that it's in. It does not vary with time.

There is no such thing as the radius between two things? You mean the distance between the center of the two loops? And I suppose your $\theta$ is the angle of the line that connects the two centers makes with the vertical? How did you eliminate the angular variable for the small loop? How did you apply the equation of constrain (which I think is what this problem is all about)? Why don't you start from the beginning and show us what you have done.

If the EOM is 0=gsinθ, isn't that just trivial?

That is why I think you are wrong.

 

[paranormal]

The first step in rewriting the Lagrangian's equation of motion as a first order differential equation is to define a new variable, let's call it x, which represents the position of the hoop. Then, we can express the second derivative of θ as the first derivative of x, since the second derivative of x represents the acceleration of the hoop.

So, we can rewrite the equation of motion as:

0 = g Sin(θ) + 3/2*R*(d²x/dt²)

Next, we can use the chain rule to express the first derivative of x in terms of θ and its derivative:

dx/dt = R*Dot[θ]

Substituting this into the equation of motion, we get:

0 = g Sin(θ) + 3/2*R²*(d²θ/dt²)

Now, we can use the conserved quantity, angular momentum, to eliminate the second derivative of θ. Rearranging the equation, we get:

d²θ/dt² = -2*g Sin(θ)/(3*R)

This is now a first order differential equation for the motion of the hoop in a bigger hoop, with the variable θ as the dependent variable. We can also express this in terms of x, the position of the hoop, by using the chain rule again:

d²x/dt² = R*d²θ/dt² = -2*g*R*Sin(θ)/3

This is the final form of the first order differential equation for the motion of the hoop in a bigger hoop.