Piston falling in a cylinder scenario?

Question

Say you have a fitted but freely moving piston falling in a large cylinder. The piston weighs 100kg but only weighs 1 kg under water. The cylinder is filled with water, but there is plumbing going from the bottom of the cylinder to the top. The plumbing is also filled with water. As the piston moves down in the cylinder the water is free to move from beneath it, through the plumbing, and back to the top of the cylinder in a continuous loop.Lets assume the plumbing is of the appropriate shape and design to allow the water to travel as freely as possible. What would the velocity of the piston be after a 20m freefall?

Answer 1

Work this problem like this: calculate the net downward force on the piston using that 1kg, but then keep in mind that the force is applied to the whole 100 kg mass. This will give you an acceleration and then you can use the ol' x=1/2 a * t^2 to get the time of the fall, and then t*a to get the velocity.

Now that is the way you are SUPPOSED to solve it.


There are other ways too using Kinetic Energy, (which you probably have not been introduced to this early in the semester.)

The total KE change will be that force multiplied over the distance over which it is applied (20m). Then you can just use the definition of KE=1/2mv^2 (where m is the total mass again) to find the velocity a little more directly.

Another tricky-dicky way, certain to annoy your TA: The mass/force ratio is 1/100th what it is if there was no fluid. Since v goes as the square root of the total KE, the velocity will by the SQRT of 1/100 (1/10th) of what it would be if there was no fluid.

WARNING: solving these problems is the best practice for a test which might have different kinds of problems. I recommend you solve it the first way for your own benefit.

EDIT: George contacted me and assured me this is some project he is working on.

I calculate that the thing will impact at 1.98 m/s in the idealized case. In reality, the water is getting denser as the object falls. If the object is not compressible then eventually it will reach a point of neutral bouancy, probably before 20m since it appears to be just a little bit heavier than the water it is displacing (bouyant weight 1% of its real weight).

This still neglects friction with the water and such. I think the piston will sink a little ways and then just stop.

Answer 2

If the plumbing is ideal and lossless, it's the same as releasing the piston in a swimming pool. Then, the velocity depends almost entirely on the shape of the cylinder. Any real plumbing will offer resistance to the flow of the water, slowing the descent even more.

Answer 3

Hi. Ignoring water resistance and inertia, the velocity would be the same as any object in free fall - about 20 meters/second (2 second fall time). Remember that the way you describe the conditions makes the resistance of the water cancel because it also applies a force to the top of the piston.

Answer 4

Mr. Quark's approach seemed right at first. (Except I would have used V^2 = Vo^2 + 2as, fewer steps.)

But something Cirric said made me think differently. When the force of the 1Kg weight accelerates the 100Kg of the piston it ALSO has to accelerate some water. We don't know how much water.

I think this is a problem we don't have info to handle. Anybody agree with me? I'll watch.

Answer 5

Uh..don't know! Sry. But, I do know a little something about the Detroit Pistons!!!

Answer 6

take water out piston will fall faster

Answer 7

There are too many variables in your question to permit an answer.