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

 

[electricpete]

I was coming back to mention my expression r was incorrect. zekeman beat me to it.

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(2B)+(2B)' ?

 

[hydtools]

I must correct my post stating v is the vertical velocity of the cg. It is just the velocity v of the cg. And angular velocity w = v/r.
Solving the geometry of the cube. Length of a side is L. The length of the diagonal of the cube is L*sqrt(3). The diagonal of the cube is vertical at the start. The cg falls a distance of r-L/2 where L/2 is the distance of the cg above the horizontal just at face impact. r = one half the diagonal = L*sqrt(3)/2 The cg is located at the geometric center of the box. r is the distance from the box corner to the cg and is the radius of the arc of motion.

w = 2*sqrt[g(sqrt3 - 1)/3L]

Like a falling body where velocity v is only dependent on the distance of the fall and not the weight, the rotational velocity is only dependent on the size of the box and not the weight. Interestingly, the rotational velocity varies inversely as the sqrt of the box size. A larger box rotates slower than a smaller box. An empty box will rotate at the same angular velocity as a completely full box.

Ted

 

[Walterke]

Assuming the cube is situated on it's edge, under 45°:
see added image
m.g.(h2-h1)=m.w²/2
g.(h2-h1)=w²/2
w=sqrt(2.g.(h2-h1))

(h2-h1)=sqrt2.L-L/2

w=sqrt(2.g.(sqrt(2).L-L/2))

NX 7.5
Teamcenter 8

 

[electricpete]

m.g.(h2-h1)=m.w²/2

I presume m*w^2/2 is supposed to represnet kinetic energy.
However unit analysis shows m*w^2 will have units of kg/sec^2... not energy.

If we wanted to simplify the problem such that the mass acted like it was concentrated as some radius of gyration k (for purposes of rotating inertia about the corner), then we could write kinetic energy as:
m*k^2*w^2/2
and now the units are kg * m^@ / sec^2... energy.

Determining k may take some work... it is equivalent to determining moment of inertia as discussed above.

As a first approximation, you might assume k is equal to distance to the center of mass at center of cube. That will give a low estimate of k, since the mass further away needs to have a higher weighting than the mass close to the pivot point.

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[electricpete]

....the falsely-low estimate of k obtained as above (by assuming m placed at center of mass) leads to falsely high estimate of w. Since that's conservative...may be acceptable, depending on your purpose, and how much time you want to spend on the problem.

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(2B)+(2B)' ?