You notice a small air bubble rising in a bottle of shampoo. The bubble has a ra

Question

You notice a small air bubble rising in a bottle of shampoo. The bubble has a radius of 3.00 mm and the density of air in the bubble is 1.8 kg/m^3. You determine that the bubble is rising at a constant speed of 4.00 cm/s. Take the density of the shampoo to be 900 kg/m^3. Use g=10 N/kg = 10 m/s^2. Also, note that the drag force on a ball moving through a fluid is: F = (6)(pi)(viscosity)(radius)(velocity). You can assume the air bubble stays a constant size as it rises, even though this is not quite true.

a) Let's say the force of gravity acting on the ball is F, directed down. We can then express all the forces in terms of F. In terms of F, what is the magnitude of the buoyant forceand the magnitude of the drag force?

b) Calculate the viscosity of the shampoo. Note that the units are Pa s.

c) There is another air bubble that is only 1.50mm in radius-assume the air density is the same as above. Calculate the value of this bubble's constant speed.

Explanation / Answer

part A:

Fb = mg = rho s * Vb * g

Fb = 900 * 1.13 e-2 * 10   

Fb = 1.02 e -3 N

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F = rh0 b Vb * g

F = 1.8 * 1.13 e -3 * 10

F = 2.03 e -6 N

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part B:

Fnet + Fdrag = Fb

Fdrag = F = Fb

Fdrag = 1.02 e-3 *-2.03 e-6

F = 1.015 e -3 N

---------------------------------
C.
Fdrag = 6 pi eta r V

eta = Fdrag/(6 pi * 3e -3 * 4 e-2)

eta = 0.449 pa s

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d. dv/dt = a = Fnet/m

a = ( rhod Vb g - rhob Vb g - 6 pi eta r V)/m

or dV/dt = 1/m *(1Fb - F - (6pi eta rV)

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