- #1
bluekuma
- 7
- 0
The system is made of a disc the center of which is pinned to the origin (so the disc cannot translate), and some weights that can be stuck on the disc to make it tilt (weights do not translate on the disc) (see images attached).
There is no friction whatsoever. The only force is gravitational force, with direction opposite to the z-axis'.
Let's start with the disc at rest with its axis parallel to axis z. Now, if you put a weight on it, the disc starts oscillating just as if it was a pendulum. Then, at time t=t0, you put another weight on it.
If [itex]\vec{ω}[/itex] is the rotational speed vector and [itex]\vec{θ}[/itex] is the rotation vector of the disc (meaning the direction of the disc's axis is always the z-versor rotated by θ radians around [itex]\vec{θ}[/itex] ) what's the expression of [itex]\vec{f}[/itex](t,ω,θ) in:
d[itex]\vec{ω}[/itex]/dt = f(t,ω,θ),
d[itex]\vec{θ}[/itex]/dt = [itex]\vec{ω}[/itex]
Given the initial values [itex]\vec{ω}[/itex](t0) =[itex]\vec{ω}[/itex]0≠0 and [itex]\vec{θ}[/itex](t0=[itex]\vec{θ}[/itex]0≠0, d[itex]\vec{ω}[/itex](t0)/dt≠0,
that would give me a way to simulate the system's motion through a standard Runge-Kutta integration method.
MIGHT HELP TO KNOW:
- I'm pretty sure there is a way to divide the two vectorial equations in six (three systems of two) linear equations
- z-component of momentum [itex]\vec{M}[/itex] (where d[itex]\vec{ω}[/itex]/dt = [itex]\vec{M}[/itex]/I ) is always 0 (zero) as M is the result of a cross product between a vector r (x, y, x) and the gravitational force (0, 0, -mg), therefore z-component of [itex]\vec{ω}[/itex] and [itex]\vec{θ}[/itex] are also 0.
If my teacher is reading this: no, ultimately I could not do it alone, sorry. (This is not homework but I bet he'd get pissed if he knew I asked for help).
There is no friction whatsoever. The only force is gravitational force, with direction opposite to the z-axis'.
Let's start with the disc at rest with its axis parallel to axis z. Now, if you put a weight on it, the disc starts oscillating just as if it was a pendulum. Then, at time t=t0, you put another weight on it.
If [itex]\vec{ω}[/itex] is the rotational speed vector and [itex]\vec{θ}[/itex] is the rotation vector of the disc (meaning the direction of the disc's axis is always the z-versor rotated by θ radians around [itex]\vec{θ}[/itex] ) what's the expression of [itex]\vec{f}[/itex](t,ω,θ) in:
d[itex]\vec{ω}[/itex]/dt = f(t,ω,θ),
d[itex]\vec{θ}[/itex]/dt = [itex]\vec{ω}[/itex]
Given the initial values [itex]\vec{ω}[/itex](t0) =[itex]\vec{ω}[/itex]0≠0 and [itex]\vec{θ}[/itex](t0=[itex]\vec{θ}[/itex]0≠0, d[itex]\vec{ω}[/itex](t0)/dt≠0,
that would give me a way to simulate the system's motion through a standard Runge-Kutta integration method.
MIGHT HELP TO KNOW:
- I'm pretty sure there is a way to divide the two vectorial equations in six (three systems of two) linear equations
- z-component of momentum [itex]\vec{M}[/itex] (where d[itex]\vec{ω}[/itex]/dt = [itex]\vec{M}[/itex]/I ) is always 0 (zero) as M is the result of a cross product between a vector r (x, y, x) and the gravitational force (0, 0, -mg), therefore z-component of [itex]\vec{ω}[/itex] and [itex]\vec{θ}[/itex] are also 0.
If my teacher is reading this: no, ultimately I could not do it alone, sorry. (This is not homework but I bet he'd get pissed if he knew I asked for help).
