angular velocity of a box about a corner 1

[atayne]

I am attempting to find the angular velocity of a 3D box that is balanced on one of its corners and allowed to fall under the force of gravity alone at the instant the box impacts. Take the simplest case for example where the box is square on all sides and falls in a way that results in one of its faces hitting the ground flush (I believe this would result in the maximum instantaneous angular velocity as it has the furthest to fall in this direction). Once I figure out the simple case, I'll need to extrapolate it to other scenarios where the box is rectangular and the center of gravity is not located dead center.

I don't really know where to start on this because the moment is changing as the box falls and finding the mass moment of inertia for a complex shape such as this appears to be very difficult to do by hand. I know I need to calculate the angular velocity about an axis drawn through the CG and then use the Parallel Axis Theorem to project it down to the floor.

If you guys could get me started thinking about this the correct way, I would feel much better!

See sketch

 

[drawoh]

atayne,

If your box is balanced perfectly about its centre of gravity, it will not fall at all. The angular velocity will be zero. The falling behaviour of your box will be affected very much by its initial position.

This sounds like a calculus problem.

JHG

 

[atayne]

I realize that if perfectly balanced it would never fall. But say it's off center by some infinitesimally small amount, this should approach some kind of asymptotic limit I would think.

 

[rb1957]

initially it'll be zero (balanced), then it'll build up, and then become zero again (when it hits the ground).

so you've got a dynamic model ... box rotated about a corner to angle theta, forces ... weight balanced by inertia acceleration ... i suspect Adams could solve it quicker than you could, but where's the fun in that !?

 

[tygerdawg]

Model as perfect inverted pendulum with point mass and zero friction. This will give you velocity of the point from standard motion formulas (same as simple falling mass). Resolve that into angular velocity about the pivot.

TygerDawg
Blue Technik LLC
Virtuoso Robotics Engineering

http://www.bluetechnik.com

 

[atayne]

Tygerdawg: Thinking about it like that, it reminds me of the problem I had once back in dynamics class where I had to find ang. vel. of a rod which is fixed on one end and allowed to drop from horizontal. Found a thread discussing that here:

http://www.physicsforums.com/showthread.php?t=144774

Is that the method you are suggesting I use? What I'm not understanding is how this applies if the pendulum is perfectly vertical (It won't ever start falling).

 

[IRstuff]

I don't think that matters. You should be looking at the change in potential energy as the source of the angular momentum at impact. How it got started isn't relevant to your specific problem. However, if someone actually pushes it over, then you would need to add the initial momentum into the picture.

TTFN
faq731-376

 

[atayne]

I was wondering if it was as simple as finding the change in potential energy. It just seemed too simple to be true, but I won't complain about that! Thanks for the help guys

 

[zekeman]

I think that it is not so simple.

Unless you can prove that the moment of inertia in the rotational position is the minimum, then you cannot conclude that you get a maximum at impact.

E=1/2 I w^2=mgh

where h the height depends on the initial orientation.

I fail to see where this equation results in a maximum of w.

 

[electricpete]

I trust your opinion but fwiw ke=pe makes sense to me.

No work is done by the ground on the corner of the box. The only work done is by gravity which can be calculated from Change in m*g*delta-h where delta-h is change in height of center of mass.

Finding I requires a little work. Probably not in tables for box pivoting on corne

There needs to be assumption whet her the mass density is constant thru volume of box or the mass is concentrated on outside of box....same mass box would have different I under these two distributions


=====================================
(2B)+(2B)' ?

 

[electricpete]

Unless you can prove that the moment of inertia in the rotational position is the minimum

I would say moment of inertia of box about assumed pivot point at its lower corner is constant.

=====================================
(2B)+(2B)' ?

 

[zekeman]

That's not the point. You really don't known h since the box does not necessarily fall flush on its face; that happens in only one direction of an infinity of possible directions of fall.
Most of those impact the round on an edge.

 

[hydtools]

Say KE = PE, or energy = work done
then, 0.5*m*v^2 = m*g*h
where v is the vertical velocity of the cg after falling through vertical height h or delta-h as electricpete says. At perfectly flat-faced impact v = tangential velocity = r*angular velocity.

Ted

 

[rb1957]

don't forget that the box is rotating, so there are two KE terms ... translation and rotation ... no?

 

[electricpete]

I would say for kinetic energy we only need rotation...selecting the corner as the pivot point and angular velocity and moment of inertia about that point. Since the pivot point (center of rotation) is not moving, there is no need for a linear kinetic energy term.

Now I understand zekeman's point, it will be tough to predict which direction this will fall.

Maybe we can make an assumption that it falls in direction to land on a flat face (rather than an edge). Then I'd think this would be a conservative high estimate of velocity at time of first impact compared to any other orientation (center of mass falls farther when the flat face hits than if a corner edge hits).

=====================================
(2B)+(2B)' ?

 

[IRstuff]

"Maybe we can make an assumption that it falls in direction to land on a flat face (rather than an edge). "

That was aready stipulated by the OP in his first post. Since it results in the lowest position for the CG, it must have the highest KE.

TTFN
faq731-376

 

[electricpete]

Agreed.

=====================================
(2B)+(2B)' ?

 

[electricpete]

So, is this a uniform-volume-density box, or a hollow box with mass on the surfaces ?

I think anyone needs that info to figure out the mass moment of inertia I....(I'm not volunteering to calculate I.....might do it later if I get bored).

=====================================
(2B)+(2B)' ?

 

[electricpete]

It's easier to figure out I using uniform volume density. Set the lower left hand corner of the box at the origin 0,0,0 and the edges aligned along cooridnate axes.

I = int(rho * r(x,y,z)^2)dx dy dz
where r(x,y,z) = sqrt(x^2+y^2+z^2)

=====================================
(2B)+(2B)' ?

 

[zekeman]

Elsectricpete

I = int(rho * r(x,y,z)^2)dx dy dz
where r(x,y,z) = sqrt(x^2+y^2+z^2"

Incorrect, your r^2 term should be the crossproduct of the unit vector at the rotational axis and the vector r dotted against itself, or
(RxL).(RxL)
caps are vectors and L is the unit vector aligned with the axis of rotation.

I = int(rho * r(x,y,z)^2)dx dy dz
where r(x,y,z) = sqrt(x^2+y^2+z^2I = int(rho * r(x,y,z)^2)dx dy dz
where r(x,y,z) = sqrt(x^2+y^2+z^2


"Maybe we can make an assumption that it falls in direction to land on a flat face (rather than an edge). "

That was aready stipulated by the OP in his first post. Since it results in the lowest position for the CG, it must have the highest KE. "'

Not necessarily since we are after w, not KE and the I's vary with direction; could be a small I and impact on an edge.