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I'm in the process of conducting research concerning free fall with air resistence. I'm dropping a beach ball filled with air from certain heigts, while measuring its speed and acceleration during the fall.

Theoretically, the initial (first few milliseconds) acceleration of the beach ball should be "g" (9,82 m/s^2). However, I am measuring an initial acceleration to be 6,5 m/s^2.

How come the empirical initial acceleration, does not equal the theoretical initial acceleration? My first guess was that the buoyancy has something to do with it, but there has to be more, since the buoyancy alone does not make up for it.

What can it be, that stops the beach ball from having "g" as its initial acceleration?

EDIT: I now realize, that I haven't done a very good job explaining my experiment, nor have I given any information concerning it. So here are the things that I haven't mentioned yet, maybe that can solve the problem.

First of all, I am actually NOT using a volleyball, but I am using a beach ball.

How I collect data: I install a motion sensor on the ground and connect it to a computer. I then lift my beach ball to a certain height, and bring it above the motion sensor. I then drop the ball, and let it land on the sensor. The sensor then spits out a lot of data, including velocity, position and acceleration during the entire fall of the ball.

Information concerning setup.

  1. Mass of ball with no air inside: 0,004856 kg
  2. Mass of ball full of air: 0,004924 kg
  3. Height ball was dropped from: 1,5 m
  4. Initial acceleration: 6,5 m/s^2
  5. Volume of beach ball full of air: 0,009655 m^3
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    $\begingroup$ You need to include the details of your experiment: how are you releasing the ball, how are you measuring its motion. Otherwise no one can help you. $\endgroup$ – Ben51 Nov 21 '18 at 14:21
  • $\begingroup$ BTW, theoretically, the ball will only have an acceleration of $g$ the very instant that you drop it, and not for any number of miliseconds after. But as Ben51 said, the specifics would depend on how you are determining all of this and what your setup is. $\endgroup$ – JMac Nov 21 '18 at 14:47
  • $\begingroup$ @JMac Are you saying that air resistance will make the acceleration less than g? Drag is proportional to velocity, so when the ball has near-zero velocity for the first few milliseconds, it experiences near-zero drag, meaning the acceleration should pretty much be g within experimental error. Theoretically, you are correct, but in practice there will be very little effect from drag at the start of the drop. $\endgroup$ – Nuclear Wang Nov 21 '18 at 16:35
  • $\begingroup$ @NuclearWang It was mostly a pedantic statement in response to his "in theory" statement. It's essentially true in practice; but not theory. $\endgroup$ – JMac Nov 21 '18 at 16:53
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    $\begingroup$ If you are just dropping the ball out of your hands, I am dubious you can be sure enough when it is released to make statements about what happens in the "first few milliseconds". In 5 ms, accelerating at g, it would fall only about a tenth of a millimeter; presumably that's smaller than the accuracy of your sensor. $\endgroup$ – Ben51 Nov 21 '18 at 20:35
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(a) I assume that you've checked your means of measuring acceleration, say by using it to measure the initial acceleration of a dropped stone.

(b) I'd have thought that buoyancy (Archimedian upthrust from the surrounding air) might decrease the acceleration significantly. As you know, the upthrust is equal and opposite to the pull of gravity on the air that the ball displaces, which, for a thin-skinned ball, will be almost the same as the pull of gravity on the air inside the ball. So the initial acceleration will be,$$a=\frac{\text{(mass of skin + mass of air inside)}\times g - \text{mass of air inside}\times g}{\text{mass of air + mass of skin}}$$So: $$a=\frac{\text{mass of skin}}{\text{mass of air + mass of skin}}\ \times g$$Is the mass of air negligible compared with the mass of the skin? To be honest, I don't think I've ever handled a beachball in my life, but I imagine that the skins are thin and light…

You've now provided some interesting data, but there seem to be inconsistencies in them. The volume looks about right, as it corresponds to a diameter of about 26 cm. But, at a density of $1.3\ \text{kg m}^{-3},$ the mass of air in the ball ought to be about 0.013 kg, which does not accord with your figures for masses of the beach ball empty and full. [It's difficult to measure the mass of the beach ball full, of course, on account of the upthrust on it!]

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  • $\begingroup$ The OP is using a volleyball, not a beach ball, which should weigh around 270 grams, about 50 times the mass of an equivalent volume of air. $\endgroup$ – Nuclear Wang Nov 21 '18 at 16:29
  • $\begingroup$ Yes, I believe you. Don't know why I though it was a beach-ball, but I've modified my answer accordingly, though you could still question its relevance! $\endgroup$ – Philip Wood Nov 21 '18 at 16:32
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    $\begingroup$ You should probably reverse that edit. OP actually provided more information, and surprisingly (or more likely, this is all a conspiracy), OP actually was using a beach ball for his experiment. So that's a pretty weird coincidence. $\endgroup$ – JMac Nov 21 '18 at 21:17
  • $\begingroup$ @JMac Thanks! I suspect that the question did originally say 'beach ball, but was soon modified to 'volleyball', before returning to the original! I've added to my post some concerns about the extra information. $\endgroup$ – Philip Wood Nov 21 '18 at 22:43

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