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Is it possible to disregard drag force of water with no viscosity that affects lightweight pop-up ball (its weight is assumed to be zero)? There is a discussion, on which I consider that although weight is small, the influence of drag force does not depend on ball but on water, therefore this force must be taken into account. What can you say?

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what is the motion of the ball? Does it somehow create lift force or moves with acceleration? In either of this two cases there will be drag. – Yrogirg Jan 18 '13 at 12:04
Original task says that a ball is put into the water on some deep and is released. So it is necessary to find out an acceleration, considering the weight of ball to be zero and that the water has no viscosity. I think, the motion of ball must be found by solving motion equation... I have found solution (maybe erroneous) according to which the acceleration will have a logariphmic-like look. – ASten Jan 18 '13 at 12:10
up vote 1 down vote accepted

In the situation you have described in the comment there is a drag force acting on a ball even if water is assumed to be inviscid. However, one should carefully calculate it in order to decide whether it matters at all.

The effect responsible for the drag is called added mass. Actually I refer you to that wikipedia article to find out the explanation. What matters is whether this drag is strong enough for your case.

The force should be

$$\boldsymbol F = -\frac{1}{2} \rho_w V \boldsymbol a$$

where $\boldsymbol a$ is the ball's acceleration, $\rho_w$ is water density and $V$ is the ball's volume. So if the acceleration is comparable to $\boldsymbol g$ then the drag is comparable to the buoyant force and could not be neglected. Since you assume ball's density to be zero then you have to take the drag into account because otherwise it acceleration would be equal to g.

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Thank you for the answer! I found the task on the similar forum (in russian) and answer shown there said that an acceletation must be $2g$ (!). Furthermore the topicstarter of that task tells me that taking this force into account is erroneous due to weight of ball is equal (or tends) to zero. – ASten Jan 18 '13 at 13:16

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