Recent content by Nutcracker

Now I don't understand, how this can be true. Equation \dot{\vec{L}} = \vec{M} where \vec{M} is torque, holds true. So when torque is exerted on the body, the angular momentum vector will certainly move. It cannot be frozen in one direction. The image shows, that the force is rotating...

Now everyone is referring to this gimbal lock thing, but that's not what I have problem with. \varphi = \Omega_{\varphi}t + \varphi_0, \theta = const., \psi = 0 is not an example of loss of the degree of freedom, when \theta \neq 0 . Okay, so if there's no apparent problem, try this: you...

D H: The Eulerian angles I used are defined: take the principal axes aligned with laboratory axes, which are x y z. Now rotate principal axes (with whole body) along z axis couterclockwise by angle \varphi . You've just created new coordinate system, x', y', z'. It's clear, that z = z'. Now...

Thanks for your answer, but now I think I absolutely don't understand the Eulerian angles. It's said that Eulerian angles gives position in space relative to some inertial coordinate system (usually laboratory z-axis). But result for spherically symmetric torque-free body gives \varphi =...

Hello everyone, as I know the regular precession of torque-free symmetric top is such a cliche, I'll try to keep derivations short. The goal is, as I pointed out, to inspect a behaviour of the torque-free symmetric top in terms of precession rate, rotation rate and nutation angle. One way is to...