Note that three of the four KE terms involving ##m_1## have factors that naturally combine to form a perfect square.ChiralSuperfields said:
Thanks for your reply @haruspex!haruspex said:
No, ignore your highlighting.ChiralSuperfields said:
I think that is part of my confusion. I am unsure how the wedge accelerates to the right and why it does, since as far as I know, there is no horizontal force acting on it since all contact in the system is frictionless. Maybe it is from the weight of m_1 that causes m_2 to accelerate?kuruman said:
Sorry, I am not sure how there are three ##m_1## terms for KE that can be combined into a expression of the form ##\frac 12m_1(…)^2##. I only get two, which are ##\frac{1}{2}m_1([\frac{dx_2}{dt}]^2 + [\frac{dy_1}{dt}\sinβ]^2)## since the ##m_1## moves at the same speed as the block in the positive ##x_2## direction which moves at ##\frac{dx_2}{dt}## and ##-\frac{dy_1}{dt}\sinβ## from block moving down the incline since only a component is parallel to the wedge.haruspex said:
What about the force from the block on the wedge?ChiralSuperfields said:
The weight on m1 is vertical and acts only on m1. The contact force between block and wedge though …ChiralSuperfields said:
Please write out your full derivation as suggested by @kuruman earlier. In particular, can you write the horizontal position ##x_1## as a function of ##x_2## and ##y_1##?ChiralSuperfields said:
##\frac 12m_1\dot x_2^2##, ##\frac 12m_1\dot y_1^2##, ##\frac 12m_1\frac 1{\tan^2(\beta)}\dot y_1^2##, ##m_1\frac 1{\tan(\beta)}\dot y_1\dot x_2##.ChiralSuperfields said:
Sorry, I don't understand the technique of introducing a third variable to describe the two generalized variables ##x_2## and ##y_1##. We have not been taught that. May you please explain why it is needed?Orodruin said:
Sorry do you please know how you got the later three terms?haruspex said:
It is not a third variable as it can be expressed in terms of the other two. It is a help function that you will differentiate with respect to time to find the horizontal velocity of the block (which is necessary to write down its kinetic energy).ChiralSuperfields said:
Sorry, I should have made it clear that I am working backwards from the expression you are supposed to arrive at. (You had not posted your own equation, so that is all I could have been referring to.)ChiralSuperfields said:
In addition, it is many times useful to consider an expression in some particular case. In this scenario, it could be instructive to consider the case where the wedge does not move, i.e., ##\dot x_2 = 0##. In that case only two of these terms will be non-zero and it can be instructive to try to figure these out and why they appear.haruspex said:
It doesn't. Why do you think it does? The displacement, velocity and acceleration may be measured as positive to the right, but that just means the values will be negative.ChiralSuperfields said:
It can of course in general, but not if released from rest as in the specific problem.haruspex said:
You don’t need to make any assumptions on how the block moves. All you need to do is to introduce a set of coordinates that describe the system fully. Then you need to figure out how the speed of the block and wedge can be expressed in those coordinates to find the kinetic energy and how the potential can be written in those same coordinates. That’s all there is to it. After that you apply the EL equations to find the equations of motion.ChiralSuperfields said:
haruspex said:
Thank you for your replies @haruspex and @Orodruin!Orodruin said:
The Lagrangian of a block on an inclined plane is given by the difference between its kinetic energy (T) and potential energy (V). For a block of mass m on an inclined plane with an angle θ, the Lagrangian L can be expressed as L = T - V. The kinetic energy can be calculated based on the velocity of the block along the incline, and the potential energy depends on the height of the block relative to a reference level.
The kinetic energy T of a block moving down an inclined plane can be derived from its velocity. If the block moves with a velocity v along the incline, the kinetic energy is given by T = (1/2)mv². If we relate the velocity to the angle of the incline and the position along the incline, we can express v in terms of the displacement along the incline.
The potential energy V of a block on an inclined plane is given by V = mgh, where h is the height of the block above a reference level. For a block on an incline at angle θ and distance d along the incline, the height can be expressed as h = d sin(θ), leading to the potential energy being V = mg(d sin(θ)).
To apply the Euler-Lagrange equation, you first need to express the Lagrangian L in terms of the generalized coordinates and velocities. The equation is given by d/dt(∂L/∂q̇) - ∂L/∂q = 0, where q is the generalized coordinate (displacement along the incline) and q̇ is its time derivative (velocity). By substituting the expressions for T and V into the Lagrangian and differentiating, you can derive the equations of motion for the block.
In the analysis of a block on an inclined plane using Lagrangian mechanics, several assumptions are typically made: the block is treated as a rigid body, the incline is frictionless (unless otherwise stated), the angle of inclination is constant, and the motion occurs in a vertical plane. Additionally, it is assumed that the effects of air resistance are negligible, and the gravitational field is uniform.