What is the optimal amount to fill a cylindrical cup so that it requires the most force to tip it over?

Answer 1

I guess you are asking about the stability - translates to the lowest CG position. The C.G of the system should be as low as possible.

Assume the Cup is empty at first. The cup is made of ceramic say. Then the C.G is just below the middle in the vertical direction. Why ? beacuse there is material at the bottom. But If the bottom thickness is small, then we can assume the C.G is at the middle height.

Then pour some liquid say hot coffee. Imagien pouring it gently. As you pour, the C.G of the WHOLE system is affected. How ?. In this case it will further lower from the middle towards the bottom as we pour hot coffee. Why ? because as the liquid is more denser than air (note that we have neglected bottom thickness, we assume it is not there).
So, as we keep pouring, the C.G goes does down and down. At the same the liquid surface comes up and up. The C.G will keep going down until it meets the liquid surface. After that point, the C.G will increase. So the point where the C.G meets the coffee surface, is the point of lowest C.G. At this point, the cup wont topple over.

If we take into account the bottom thickness, the lowest C.G height will differ. Why ? beacuse coffee and ceramic have different densities. Assume an extreme case, a ceramic cup with neglible bottom thickness and pouring supe dense liquid. What happens. ? In this case, the lowest C.G height will be so low !. The rate of lowering of the C.G with respect to the increase in liquid level is very high. So this means the C.G meets the liquid surface quickly.

Hope this answers your question

Answer 2

I agree with BekkiB's analysis. I can't think of an easy way of answering without scribbling out a lot of calculus equations; but here is how I'd approach it:

1. Assume the cylinder is weightless (to make the math easier).
2. Write a function that gives the x and y coordinates of the water's center of mass, as a function of the water's (initial) height h, the cup's radius r, and the tip angle θ. This'll probably be tricky because the shape of the water's mass is unusual (no longer a simple cylinder) once you start tipping.
3. Determine, as a function of h and r, the critical angle θc at which the water's cm is directly over the point where the cup is touching the table.
4. Using (2) and (3), figure out how high the cm is lifted (compared to its original position) when the cup is tilted to the critical angle. (This will still be a function of h and r). The amount of work done is the weight of the water times the vertical rise of the cm.
5. Take the derivitive of the function with respect to h (or more likely, h/r), to figure out how to maximize the work done.

Answer 3

The answer, assuming you are filling it with a typical heavier-than-air material such as water and assuming that the cup's opening is on the circular end and not in the "side" somewhere, is always to fill it as full as possible. The more material inside, the more gravity will pull the cup back toward the normal resting position when it is tipped slightly. Maximum force occurs when the cup is first tipped. Beyond that point the required force decreases to zero and then becomes negative as the cup reaches and passes the point where it will fall on its own and a restoring force would be needed to return the cup to an upright position.

Answer 4

I don't think your question is precisely stated. You don't really want to maximize force, but the amount of work to tip it.

If you calculate the moment of force to start a cup tipping, that quantity will increase the higher you fill the cup. So the first answerer is actually correct for the exact question you asked. But as you fill the cup, the range of stability goes down. You don't have to tip it far before it falls. So you can optimize the fill level to maximize the work (torque * angle) needed to tip the cup.

Edit: If you agree with me that what you really want to maximize is work, Rick's answer below gives you the first steps. I was too lazy to write all that (and actually doing what he says will be an even bigger pain).

Answer 5

Filled to just under the center of gravity. Any liquid above that line will start the curve to being 'top heavy'. All weight below the center of gravity is considered ballast.

Answer 6

Fill it. Any empty space in the cup will assist you in tipping it over.