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[da_nang]

Nope, it's a vector. In proper notation, rxv = ωr² where the scalar r = ||r|| and r is the position vector.

[Skyrunner]

So in the above case, what would the angular velocity vector look like? I can't quite picture it. Assume that the yellow dot is the position of the pivot.

[da_nang]

By the right hand rule, it will point into the paper at that specific time.

EDIT: Assuming it's going down at the moment, of course.

[Skyrunner]

Then what sort of vector computations would you need to do to move the rod to the moment immediately after this one? Assume that I have positional vectors for both the pivot and the end of the rod.

[MagmaMcFry]

Rodrigues' rotation formula is what you need here.

[frostshotgg]

Also, just on a generic side note, since it came up: Right hand rule is dumb. It makes absolutely no sense why everything in the universe is right handed, except for like 3 things. Damn right hand rule.

[Skyrunner]

Rodrigues' rotation formula is what you need here.

I don't get it. Where is my angle theta in the equation

Nope, it's a vector. In proper notation, rxv = ωr² where the scalar r = ||r|| and r is the position vector.

Also, linking a math page is pretty useless if the reader doesn't know about the maths in question. Just saying
Somehow, Wikipedia manages to perfectly describe a mathematical concept, but not actually explain it. >__>

[frostshotgg]

That's the thing. There is no explanation. The universe is right handed. It points like your right hand would.

[da_nang]

Rodrigues' rotation formula is what you need here.

I don't get it. Where is my angle theta in the equation

Nope, it's a vector. In proper notation, rxv = ωr² where the scalar r = ||r|| and r is the position vector.

ω = dθ/dt

But if this is a physics problem as in the picture, you're better off using moment of inertia and torque since that is what's going to define ω and in turn θ and r.

[MagmaMcFry]

Rodrigues' rotation formula is what you need here.

I don't get it. Where is my angle theta in the equation

Nope, it's a vector. In proper notation, rxv = ωr² where the scalar r = ||r|| and r is the position vector.

Also, linking a math page is pretty useless if the reader doesn't know about the maths in question. Just saying
Somehow, Wikipedia manages to perfectly describe a mathematical concept, but not actually explain it. >__>

Usually you'd describe a rotation of an object using a direction of the axis the object rotates about (as an unit vector), and a scalar angle. But you can combine those two values into a single 3D vector ω by multiplying the axis direction vector by the angle of rotation. To convert such a vector back into the axis-angle form, the angle is the length of ω, and the axis direction is the direction of ω.

Both descriptions are useful in different situations, for example you can add two angular momentums by adding their ωs, and you can apply a rotation with Rodrigues' formula, which needs the axis-angle description. You'll just need to convert between those descriptions occasionally.

That's the thing. There is no explanation. The universe is right handed. It points like your right hand would.

Actually, it's just because someone once needed to assign signs to the two possible orientations of three-dimensional space. He said that the right-hand orientation gets a + sign and the left-hand orientation gets a - sigh, and everyone went and adopted that practice. Really it's an arbitrary choice, and you can actually swap the signs of all orientation-dependent values, and all the equations will still hold true.

[Skyrunner]

Cool, that makes a lot more sense. O.o I guess I'll implement it tomorrow, though I might have to upgrade my vector2s to vector3.

Also, that × means cross product, right?

[da_nang]

Note that for the particular scenario in the picture, consider the coordinate system relative to the pivot. Assuming that the rod of mass m has a center of mass at p_CM (relative to the pivot) then by Steiner's theorem, the moment of inertia when rotating the rod around the pivot is

where I_CM is the moment of inertia when the pivot is at the center of mass.

The net torque then is

However, the force acting on it is gravity and the torque it creates (under classical uniform gravity conditions) is

Since this is the only torque, then this is also the net torque.

Solving for the angular acceleration gives us

However, if you dig deeper into it, it's not as straightforward as integrating it. In fact, this is a nonlinear ODE of the second order:


Note the minus sign. The angle theta is the angle from the equilibrium with positive being counter-clockwise. The vector form is also lost or one can imagine it as limiting only to movement along the axis of rotation i.e. this is the ODE for the magnitude of the component parallel to the axis of rotation.

[Skyrunner]

This is all pretty complex I thought I would have just needed to convert linear acceleration to angular acceleration, instead of considering moment of inertia... 'll have to edit my code a bunch. It does all make sense though, especially the part where you find the inertia of the rod. How is the m.o.i found when the pivot is in the center, though?

I actually have a more complicated question, which is "how do the two rods in a double pendulum interact" >_> I would appreciate an explanation that doesn't represent the system relative to the pivot angles, because (in case it wasn't obvious) I'm actually doing a sort of element simulation. O.o

The part I can't conceptualize is how the pivot between the two rods would move. How do the rods influence it? First question is more important!

[da_nang]

Depends on the shape and mass distribution. A few of them are here.

And for double pendula, they behave rather chaotically.

[frostshotgg]

Double pendulums are suffering. Never ask about them, never be told about them, save everybody some time.

Ninjaedit: Meh. Now you can see why they're terrible.