Equations of Motion for a Cylinder Riding on a Cart (2DOF) Using Lagrange's Equations
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- Опубликовано: 26 авг 2024
- Deriving the equations of motion (EOMs) using the method of LaGrange's equations for the a cylinder riding on a cart. Two degree-of-freedom system.
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Hello sir,
Why didn't you do the following?
1) in the kinetic energy, instead of replacing theta_dot by (X2_dot - X1_dot) / r we replace X2_dot by X1_dot + R * theta_dot
2) in the potential energy we write the energy of spring K2 as 0.5 * K2 * R^2 * theta^2
therefore the EOM will be in terms of X1 and theta
Yes, you can use x and theta as coordinates instead. This is acceptable since it fully describes the system. The equations of motion (the matrices) will look slightly different, however, certain characteristics will be preserved (like the natural frequencies). The choice of coordinate is typically dependent on what one is trying to investigate since some coordinates may be more convenient depending on the problem.
why is kinetic energy of the cylinder not 0.5*m*(x2_dot+x1_dot)^2?
This is because x2 is the absolute displacement of the cylinder and NOT the displacement of the cylinder relative to the cart.
In the start of equation 6, it should be (M+m/2) not 3m/2.
Both M and m are different....right?
It is correct as written. Yes, M and m are different, however, M does not appear at all in the equation for the x2 coordinate (since it only multiplies x1_dot and not x2_dot).
If you go back one step and look at the expression for the kinetic energy, T, you will find that there are two terms that yield a "m" after differentiating with respect to x2_dot. These come from the second and third terms in the K.E. expression. The first term of the K.E. which contains the M is eliminated when differentiating with respect to x2_dot. I think if you write out this step on paper, you'll see it instantly.
@@Freeball99 oh yes. I get it now.
Thank You very much "GURU DEV".
How would this change if there was an external force acting only on the cylinder?
Assuming the force, F, was applied at the center of the cylinder in the horizontal direction, the second equation of motion (eqn. 6) would have F on the right-hand side instead of zero.
Hi may I know how is R Theta = x2-x1, I don't get the x2 -x1, x2 represents the mass cylinder and x1 is the trailer
x1 = absolute displacement of cart
x2 = absolute displacement of cylinder
r·θ = displacement of cylinder relative to cart
= x2 - x1
Sir, what app are you using in creating these lecture videos?
App is called "Paper" by WeTransfer. Running on an iPad Pro 13 inch and using an Apple Pencil.
The friction between the cart and the cylinder causes the no slip condition then why the force along generalized coardinate x2 is neglected ............
The force caused in the x2 direction by the no-slip condition is a kinematical constraint - it's an internal force. It is included in the sense that it couples the coordinates between two rigid bodies, but it is not an external force in the x2 direction.
Why you are using dev for both x1 and x2 while deving only x1
you can see in 5:50 min
According to Lagrange's equations, I needs to take 2 derivatives. First with respect to x1_dot and then with respect to time. In this case I did both in one step.
Taking the derivative of m/4 * (x2_dot - x1_dot)^2 with respect to x1_dot (using the chain rule) results in:
-m/2 * (x2_dot - x1_dot)
Taking the derivative of -m/2 * (x2_dot - x1_dot) with respect to time results in:
-m/2 * (x2_ddot - x1_ddot)
@@Freeball99 What about the potential energy when you differentiate isn't it xdot instead of x.
@@tanmanpeng5675 Potential energy is never dependent on velocity. Lagrange’s equations does not require us to differentiate the potential with respect to time. Only with respect to the coordinates.