4 ms·
> Without the distribution of the weight of the can, I don't think we can get an actual number. Because the distribution of weight in the can isn't changing, y
by Strilanc 12y ago
> Without the distribution of the weight of the can, I don't think we can get an actual number.
Because the distribution of weight in the can isn't changing, you can treat it as a point mass as far as the center of mass is concerned. The fluid's distribution is changing, but we know how.
- thedufer 12y agoHrmmm. I did a full write-up assuming that worked, and then managed to convince myself it didn't. Looks like you're right, though. Here it is: Let's call C the weight of the can, L the weight of the liquid, and x the proportion of the can that's full. Then the center of mass is (C/2 + xL/2)/(C+L). We want where this is equal to x. Solving for x gives us x=C/(2C+L). This passes the sniff test: for C>>L, x->1/2, and for L>>C, x->0, as we'd expect. For the given weights, we get x=0.1, so when the can is 10% full.
- Strilanc 12y agoThe liquid's center of mass isn't xL/2, it's x L^2 /2. One factor of L for getting lighter, and one for getting lower. I made the same mistake like five times.
- thedufer 12y agoIn my formulation, x is the proportion and L the total weight, so I believe it would be x^2 L/2. I also just realized that the denominator ought to contain xL, not L. My solution is a mess, clearly. The original idea is right, though.