- #1

PhysicsRock

- 117

- 18

- Homework Statement
- We consider a coordinate transform where ##\vec{x}^\prime(t) = R(t) (\vec{a} + \vec{x}(t))## with a constant ##\vec{a}##.

Write the lagrangian in terms of ##\vec{x}## and ##\dot{\vec{x}}##.

- Relevant Equations
- Velocity in terms of ##\dot{\vec{x}}## and ##\vec{x}##: ##\dot{\vec{x}}^\prime = R \left[ \dot{\vec{x}} + \vec{\omega} \times (\vec{a} + \vec{x}) \right]##.

Lagrangian function ##L = T - V##.

I found a the answer in a script from a couple years ago. It says the kinetic energy is

$$

T = \frac{1}{2} m (\dot{\vec{x}}^\prime)^2 = \frac{1}{2} m \left[ \dot{\vec{x}} + \vec{\omega} \times (\vec{a} + \vec{x}) \right]^2

$$

However, it doesn't show the rotation matrix ##R##. This would imply that ##R^2 = R \cdot R = I##. ##R## is an orthogonal matrix, but I'm pretty sure that the square of such is not always equal to the identity.

So then how come the matrix doesn't show up in the expression for the kinetic energy?

$$

T = \frac{1}{2} m (\dot{\vec{x}}^\prime)^2 = \frac{1}{2} m \left[ \dot{\vec{x}} + \vec{\omega} \times (\vec{a} + \vec{x}) \right]^2

$$

However, it doesn't show the rotation matrix ##R##. This would imply that ##R^2 = R \cdot R = I##. ##R## is an orthogonal matrix, but I'm pretty sure that the square of such is not always equal to the identity.

So then how come the matrix doesn't show up in the expression for the kinetic energy?