- #1
Kate R
- 2
- 0
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.
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.