Physics question: 12 oz can and some wind

[dnooman]

At times, I've had a twelve ounce can on a ledge, and the wind was enough to knock it over, other times it was not. The weight/level of liquid in the can varied, as did the wind speed.

Many variables are in play clearly. My question is about how one goes about making determinations about the forces needed to topple a can in the wind.

The weight of the full can, the height of the liquid in it, the center of gravity of the can, should all factor in no? Wind speed, area of the can surface, leverage over the can's COG?

Take a 12oz can that's half full of water. What wind speed at a 90 degree angle would it take to topple the can? Any other similar situation works too. This is not a homework question.

[The Niply Elder]

Yes to all.

There are a handful of textbook problems that fit this. The can is simplified to a cylinder. Air speed is used to calculate an idealized distributed force over the height of the can. The fundamental equation to balance is moment. The moment produced by air is equal to the total air force times the force centroid (near the middle of the can height) distance from the bottom. The moment on the other side of the equation is equal to the mass of the object times the distance from the center of mass minus the overhang distance along the horizontal axis (to a maximum equal to the radius of the cylinder if ther is no overhang).

Eta: probably no on the height of the liquid. The height of the center of gravity is not important as long as it is in the axis of rotation f the cylinder.

[Mangetout]

Originally Posted by The Niply Elder

Eta: probably no on the height of the liquid. The height of the center of gravity is not important as long as it is in the axis of rotation f the cylinder.

I think it must be a factor - because the tipping point is when the centre of mass moves outside the footprint of the can - if it's high, then the tipping angle at which this occurs is smaller than when it's low (although when it's low, the whole system is lighter anyway - and the centre of mass will be able to shift as the liquid moves during tipping)

[YamatoTwinkie]

Originally Posted by Mangetout

I think it must be a factor - because the tipping point is when the centre of mass moves outside the footprint of the can - if it's high, then the tipping angle at which this occurs is smaller than when it's low (although when it's low, the whole system is lighter anyway - and the centre of mass will be able to shift as the liquid moves during tipping)

True, but assuming constant wind speeds, once the moment due to the wind is greater than the restoring moment from the weight and cg location of can in the radial direction, the can *will* tip over. The height of the cg just affects how fast the "tip" will be once it starts tipping.

[Chessic Sense]

Since we have a constant wind, if the can moves at all, it'll tip, as Yamato said. The CG only has to pass the outside support when the force is delivered all at once, like in a bowling ball/pin scenario.

So imagine a can sitting there with no wind. Imagine the pivot point on the leeward side of the can at the bottom. With no wind, the CG of the can causes the can to rotate "windward" until it comes into contact with the surface it's sitting on. Then normal force keeps it from rotating.

Now we add wind. If we had a small scale under the windward edge of the can, it'd measure some weight. As we added wind, it would get less and less as the wind got stronger and stronger. Once it measures zero, the wind is stronger than the weight of the can and it'll tip.

So all you have to do is measure the torque from the CG of the can on the leeward pivot point, then calculate a wind equal to that.

[Mangetout]

Originally Posted by YamatoTwinkie

True, but assuming constant wind speeds, once the moment due to the wind is greater than the restoring moment from the weight and cg location of can in the radial direction, the can *will* tip over. The height of the cg just affects how fast the "tip" will be once it starts tipping.

Less so with a lower CG, I think, because the path the CG must take is an arc, (radius of the back corner of the can, to the CG) - so the wind isn't just pushing the can over, it's lifting it a bit - and as the can tips back, it has a smaller wind profile - which, for the precisely correct value of wind speed (and also assuming unrealistically low degrees of turbulence etc) should self-correct so as to tip the can only partially.

[MikeS]

The problem with using moment of inertia for this situation is that it assume that we have a rigid body. The liquid in the can isn't rigid, of course, and so we'll get some kind of free surface effect due to the motion of the liquid.

Originally Posted by Chessic Sense

Now we add wind. If we had a small scale under the windward edge of the can, it'd measure some weight. As we added wind, it would get less and less as the wind got stronger and stronger. Once it measures zero, the wind is stronger than the weight of the can and it'll tip.

It'll tip, but it might not tip over. There'll be a range of wind speeds (perhaps small) where the center of mass will still be over the original footprint of the can, and so if the wind suddenly let up the can would right itself. If the wind was in this range of speeds, the equilibrium situation would have the can tilted slightly, with only one point on its edge touching the table and the torques due to gravity and the wind about that point balanced out. This probably wouldn't be a very stable equilibrium, but it would be an equilibrium.

[Chronos]

The fact that most of the weight is liquid should probably make the can easier to tip, since when it's tipped, the center of gravity will move in that same direction, relative to the rigid can.

[Chessic Sense]

Originally Posted by Mangetout

Less so with a lower CG, I think, because the path the CG must take is an arc, (radius of the back corner of the can, to the CG) - so the wind isn't just pushing the can over, it's lifting it a bit - and as the can tips back, it has a smaller wind profile - which, for the precisely correct value of wind speed (and also assuming unrealistically low degrees of turbulence etc) should self-correct so as to tip the can only partially.

I don't know about that. If you lift the can a little bit, then the air will hit the underside of the can which causes a lifting action. This is assuming the back pivot point is braced against something. Otherwise the can slides instead of tips and I call shenanigans.

Originally Posted by MikeS

It'll tip, but it might not tip over. There'll be a range of wind speeds (perhaps small) where the center of mass will still be over the original footprint of the can, and so if the wind suddenly let up the can would right itself. If the wind was in this range of speeds, the equilibrium situation would have the can tilted slightly, with only one point on its edge touching the table and the torques due to gravity and the wind about that point balanced out. This probably wouldn't be a very stable equilibrium, but it would be an equilibrium.

Why would the wind let up? And why would it matter if the CG is over the footprint? If the CG moves backward, it'll keep moving backward. There's no point where the can would start to rotate but then somehow increase its windward torque. There's a reason people lean into strong winds.

[Mangetout]

Originally Posted by Chessic Sense

I don't know about that. If you lift the can a little bit, then the air will hit the underside of the can which causes a lifting action. This is assuming the back pivot point is braced against something. Otherwise the can slides instead of tips and I call shenanigans.

True - I didn't think about the wind profile of the newly-exposed bottom of the can. I was going to say that this additional area might not compensate for the reduction in area caused by the tilting of the can, but on reflection, I think it probably does - because by the time the can is sufficiently tilted that the wind profile is on the decrease, it's past the tipping point.