1.20 In addition to the downward force of gravity (weight) and drag, an object f

Question

1.20 In addition to the downward force of gravity (weight) and drag, an object falling through a fluid is also subject to a buoyancy force which is proportional to the displaced volume (Archimedes' principle). For example, for a sphere with diameter d (m), the sphere's volume is V d76, and its projected area is A -nd14. The buoyancy force can then be computed as Fb--pVg. We neglected buoyancy in our derivation of Eq. (1.8) because it is relatively small for an object like a bungee jumper moving through air. However, for a more dense fluid like water, it becomes more prominent. (a) Derive a differential equation in the same fashion as Eq. (1.8), but include the buoyancy force and represent the drag force as described in Sec. 1.4. (b) Rewrite the differential equation from (a) for the special case of a sphere (c) Use the equation developed in (b) to compute the terminal velocity (i.e., for the steady-state case). Use the follow- ing parameter values for a sphere falling through water: sphere diameter = 1 cm, sphere density = 2700 kg/m3 water density = 1000 kg/m3, and C-0.47. (d) Use Euler's method with a step size of t 0.03125 s to numerically solve for the velocity from t = 0 to 0.25 s with an initial velocity of zero

Explanation / Answer

mdvdt=mgkvVmdvdt=mgkvV

mdvdt=mg12v10mdvdt=mg12v10

Since gravity is not considered, wouldn't it be:

mdvdt=12v10mdvdt=12v10

v=40et20v=40et20

Setting to 0 and solving for t, we get t=ln(2) sec.

Integrate to get the position function and sub in the time just found.

d=40et20td=40et20t

d=33.86 md=33.86 m

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