Solving Lagrangian Mechanics: Rod and Bean System Explained

[Kate R]

I'm stuck on a problem with lagrangian mechanics.

Here's the problem;

One end of a rod slides along a vertical pole while the other end
slides a long a horizontal pole. At the same time a bean slides a long
the rod. Find the lagrangian for the system.

And this is what I worked out so far;

The kinteic energy for the rod would be;

T = 1/2* M Vcm^2 + 1/2* Icm(theta dot)^2

Where M is the mass of the rod
Vcm is the velocity for the center of mass
Icm is the moment of intertia for center of mass; Icm = M/12*(L/2)^2
L is the length of the rod
theta is the angle between the vertical pole and the rod

The potential energy for the rod would be;

V = MgL*cos(theta)

So far I think it's ok because the velocity of the rod is relative to a fixed intertial frame, but I don't know what to do with the bean.

The beans velocity would be the veclocity of the rod plus the beans velocity relative to the rod, right?

I would be very greatful if someone could give me a little help with
this.

 

[Meir Achuz]

Yes, jusst add the two velocites vectorially.

 

[Kate R]

And that is easier said than done (at least to me).The bean is in an accelerating frame of reference that is rotating around it's own axis AND in a circle (with the redius L/2).
So is it not really three velocities? The beans velocity, the velocity of the rod around it's own axis and the velocity of the rod in the circle

And how do I find the beans velocity relative to the fixed frame of reference?

 

[siddharth]

And that is easier said than done (at least to me).The bean is in an accelerating frame of reference that is rotating around it's own axis AND in a circle (with the redius L/2).
So is it not really three velocities? The beans velocity, the velocity of the rod around it's own axis and the velocity of the rod in the circle

And how do I find the beans velocity relative to the fixed frame of reference?

Picture it this way. At any instant of time, let B denote the point on the rod right under the bean, A denote the rod.

(i) v_B = v_A + v_(B/A), where v_B is vel of the point on the rod under the bean, v_A is the vel of the bean, and v_(B/A) is due to the rotation of the rod.
(ii) Now, the relative velocity of the bead wrt to the rigid point on the rod under it is always going to be in a direction along the rod. So, the absolute velocity of the particle is going to be v_B + v_(P/B), where v_B as before, is the vel of the point on the rod under the bean, and v_(P/B) is the vel of the particle with respect to the point (which is constrained to move along the rod.

You need to use this absolute velocity in the Lagrangian of the system.