Kinetic Energy Calculation for Rigid Body Lagrangians with Two Particles

[castusalbuscor]

Homework Statement

two particles, $$m_{1}$$, and $$m_{2}$$ are connected together by a thin masses rod of length d. The system moves under a uniform potential function $$U(r_{1}^{\rightarrow}, r_{2}^{\rightarrow})$$.

What is the Kinetic energy of the system in cartesian coordinates?

Homework Equations

The constraint equation we were given:

$$f(r_{1}^{\rightarrow}, r_{2}^{\rightarrow}) = |r_{1}^{\rightarrow}, r_{2}^{\rightarrow}| = d$$​

The Attempt at a Solution

I am not sure how to approach the problem, any hints on how I can start this would be great!

 

[Jhero]

Hello,

To solve this problem, you will need to use the constraint equation given to you. This equation represents the constraint on the motion of the particles, which is that they are connected by a thin rod of length d. This means that the particles can only move in a way that maintains the distance between them at d.

To find the kinetic energy of the system, you will need to use the formula for kinetic energy in cartesian coordinates, which is KE = 1/2 * m * (v_{x}^2 + v_{y}^2 + v_{z}^2). Here, m represents the mass of the particles and v_{x}, v_{y}, and v_{z} represent the velocities in the x, y, and z directions, respectively.

Since the particles are connected by a rod, their velocities will be related to each other. This means that you will need to use the constraint equation to express the velocities of the particles in terms of each other. Once you have done this, you can substitute these expressions into the formula for kinetic energy to get the final answer.

I hope this helps. Good luck with your problem!