Physics
Physics
5th Edition
ISBN: 9781260487008
Author: GIAMBATTISTA, Alan
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 9, Problem 76P
To determine

The terminal speed of an air bubble rising through water.

Expert Solution & Answer
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Answer to Problem 76P

The terminal speed of an air bubble rising through water is 2.9cm/s_.

Explanation of Solution

Given that the specific gravity of aluminium sphere is 2.7, and its terminal speed is 5.0cm/s.

The free-body diagram of the aluminium sphere and the air bubble is given in Figure 1.

Physics, Chapter 9, Problem 76P

Write the equilibrium condition on forces in y direction.

Fy=0 (I)

Apply the condition in equation (I) to the aluminium sphere.

F1D+F1Bm1g=0F1D=m1gF1B (II)

Here, F1D is the drag force on the aluminium sphere, F1B is the buoyant force on the aluminium sphere, m1 is the mass of aluminium sphere, and g is the acceleration due to gravity.

Apply the condition in equation (I) to the air bubble.

F2BF2Dm2g=0F2D=F2Bm2g (III)

Here, F2D is the drag force on the air bubble, F2B is the buoyant force on the air bubble, and m2 is the mass of air bubble.

Divide equation (III) by (II).

F2DF1D=F2Bm2gm1gF1B (IV)

Write the expression for the drag force on the aluminium sphere.

F1D=6πηrv1 (V)

Here, η is the viscosity of water, r is the radius of the sphere, and v1 is the speed of the sphere.

Write the expression for the drag force on the air bubble.

F2D=6πηrv2 (VI)

Here, v2 is the speed of the air bubble.

Divide equation (VI) by (V).

F2DF1D=6πηrv26πηrv1=v2v1 (VII)

Equate the right hand sides of equations (IV) and (VII).

F2Bm2gm1gF1B=v2v1 (VIII)

The buoyant force on aluminium sphere and the air bubble is equal to mwg, where mw is the mass of displaced water. Modify equation (VIII) using the expressions for buoyant force.

mwgm2gm1gmwg=v2v1v2=v1(mwm2m1mw)=v1(1m2mwm1mw1) (IX)

Replace ratios of mass in equation (IX) with densities.

v2=v1(1ρaρwρAlρw1) (X)

Here, ρa is the density of air, ρw is the density of water, and ρAl is the density of aluminium.

Since the specific gravity of aluminium is 2.7, the density of aluminium can be replaced as 2.7ρw. Thus, equation (X) becomes,

v2=v1(1ρaρw2.7ρwρw1)=v1(1ρaρw1.7) (XI)

Conclusion:

Substitute 1.20kg/m3 for ρa, 1001.8kg/m3 for ρw and 5.0cm/s for v1 in equation (XI) to find the terminal speed the air bubble.

v2=(5.0cm/s)(11.20kg/m31001.8kg/m31.7)=(5.0cm/s×1m100cm)(11.20kg/m31001.8kg/m31.7)=0.029m/s=0.029m/s×100cm1m=2.9cm/s

Therefore, the terminal speed of an air bubble rising through water is 2.9cm/s_.

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