Unformatted text preview: o great that it manages to balance the object’s
weight and any other forces. The weight of a sphere of radius R and density ρs is
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Fw = πR3 ρs g
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i (2) If you came up with a slightly diﬀerent area, that is not a problem  the geometry is not strictly deﬁned in the
problem, so you had a choice to either add or subtract 5 cm from the given dimensions. Either way is ﬁne, it does
not change the conclusion. If we are in a surrounding ﬂuid of density ρ, we must also account for the buoyant force, equal to
the weight of the displaced ﬂuid. This is the same as the expression above if we substitute ρs → ρ
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B = πR3 ρg
3 (3) The drag and buoyant forces will act in one direction, the weight of the object opposing them. A
force balance yields, at terminal velocity,
0 = Fd + B − Fw
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0 = 6πηRv + πR3 ρg − πR3 ρs g
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6πηRv = πR (ρs − ρ) g
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2g 2
v=
R (ρs − ρ)
9η (4)
(5)
(6)
(7) The fact that terminal velocity depends on particle size has many interesting technological applications. (As a quick forinstance: http://en...
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 Fall '08
 StanJones
 Physics, Force, Mass, Simple Harmonic Motion, Work, pivot point

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