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Physics / College Physics 1 / Chapter 12 / Problem 39

College Physics | 1st Edition | ISBN: 9781938168000 | Authors: Paul Peter Urone, Roger Hinrichs

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QUESTION:

Problem 39PE

Using the equation of the previous problem, find the viscosity of motor oil in which a steel ball of radius 0.8 mm falls with a terminal speed of 4.32 cm/s. The densities of the ball and the oil are 7.86 and 0.88 g/mL, respectively.

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QUESTION:

Problem 39PE

Using the equation of the previous problem, find the viscosity of motor oil in which a steel ball of radius 0.8 mm falls with a terminal speed of 4.32 cm/s. The densities of the ball and the oil are 7.86 and 0.88 g/mL, respectively.

ANSWER:

Solution 39PE

According to Stoke’s law, the viscous force experienced by the steel ball is $F{\ }_{v}=6\pi{}\eta{}rv$ , where $\eta{}=$ coefficient of viscosity, $r=$ radius of the steel ball, $v=$ velocity of the ball. When the ball attains terminal velocity, its weight is balanced by the viscous force and the buoyant force exerted by the oil. If $\rho{}$ is the density of the ball and $\sigma{}$ is the density of the oil, then the viscous force can be expressed as,

$viscous\ force=weight\ of\ the\ ball-\ buoyant\ force\ of\ the\ oil$

$6\pi{}\eta{}rv=\frac{4}{3}\pi{}r{\ }^{3}\rho{}g-\frac{4}{3}\pi{}r{\ }^{3}\sigma{}g$

$6\pi{}\eta{}rv=\frac{4}{3}\pi{}r{\ }^{3}(\rho{}-\sigma{})g$

$\eta{}=\frac{2}{9}\frac{r{\ }^{2}}{v}(\rho{}-\sigma{})g$ …..(1)

From this expression, we can calculate the coefficient of viscosity of the oil.

Step 1 of 2

Given data:

Radius $r=0.8$ mm

$=0.8\times{}1{0}^{-3}$ m

Terminal speed $v=4.32\$ cm/s

$=0.0432$ m/s

Density of the ball $\rho{}=7.86$ g/mL

$=7860$ kg/m3

Density of the oil $\sigma{}=0.88$ g/mL

$=880$ kg/m3

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