Maximum velocity of buoyant object

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SUMMARY

The discussion focuses on calculating the maximum velocity of a buoyant object, specifically a plastic ball with a diameter of 2.75 inches and a weight of 16 grams, submerged at a depth of 6 feet. The key takeaway is that terminal velocity is achieved when the buoyant force equals the viscous drag force, which may not occur within the 6-foot depth. The dynamic equation to consider is ma = f_b + f_drag, where f_b represents buoyancy and f_drag represents drag. Additionally, tethering 30 balls together with fishing line at 1/2 inch spacing requires the assumption of Stokes flow for each sphere to simplify the calculations.

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  • Understanding of buoyancy and drag forces
  • Familiarity with Stokes flow and its assumptions
  • Basic knowledge of dynamic equations in physics
  • Ability to interpret terminal velocity concepts
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  • Research terminal velocity calculations for multiple tethered objects
  • Explore the effects of depth on buoyancy and drag forces
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ralphamale
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I have a plastic ball 2.75" in diameter weighing 16 grams submerged at 6 feet depth. I am trying to determine the maximum velocity achieved when I release the ball? Also, if I tether (30) balls together with fishing line at 1/2" spacing how will that affect velocity? I really appreciate any help and input on this. Thank you!
 
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Terminal velocity is reached when the buoyant force equals the viscous drag force, which may not occur over the 6 feet. I would start with the full dynamic equation: ma = f_b + f_drag, where f_b is buoyancy and f_d the drag term. Assume Stokes flow for the drag term, and solve for the velocity as a function of depth. You will also have to assume that Stokes flow hold for each sphere in the chain in order to keep this problem easy.
 

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