A Copper Ball in Oil

solved StokesLaw

Q. The terminal velocity of a copper ball of radius 2.0 mm $a = 2.0$ mm falling through a tank of oil at 20 °C is 6.5 cm/s $v_{t} = 6.5$ cm/s. Density of oil is 1.5 × 10³ kg/m³ $σ = 1.5 \times 10^{3}$ kg/m³, density of copper is 8.9 × 10³ kg/m³ $ρ = 8.9 \times 10^{3}$ kg/m³. Taking g as 9.8 m/s² $g = 9.8$ m/s², what is the viscosity of the oil at 20 °C, in kg/(m s)?

Inputs

radius

a

2 mm

terminal

v_{t}

6.5 cm/s

oil

σ

1500 kg/m³

copper

ρ

8900 kg/m³

gravity

g

9.8 m/s²

Outputs

viscosity

η

How it's solved

Solved with StokesLaw

Given

$a = 2.0$ mm
$v_{t} = 6.5$ cm/s
$σ = 1.5 \times 10^{3}$ kg/m³
$ρ = 8.9 \times 10^{3}$ kg/m³
$g = 9.8$ m/s²

Find $η$ .

Working

$a = 2.0$ mm $= 0.002$ m
$v_{t} = 6.5$ cm/s $= 0.065$ m/s
terminal velocity of a sphere(ball) $v_{t} = \frac{2 \cdot a^{2} \cdot (ρ - σ) \cdot g}{9 \cdot η}$
$0.065 = \frac{2 \cdot {0.002}^{2} \cdot (8900 - 1500) \cdot 9.8}{9 \cdot η}$
$η = 0.9917$ kg m⁻¹ s⁻¹
$η = 0.991726496$ Pa s $= 0.991726496$ kg m⁻¹ s⁻¹
$η \approx 0.99$ kg m⁻¹ s⁻¹to 2 significant figures

Conclusion

At its terminal velocity the ball's weight, less the oil's buoyancy, is matched by Stokes' drag, which grows with the viscosity; so the speed it settles to measures the viscosity.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :a_copper_ball_in_oil do4  kind :numerical5  describes "The terminal velocity of a copper ball of radius %{radius} falling through a tank of oil at 20 °C is " \6            "%{terminal}. Density of oil is %{oil}, density of copper is %{copper}. Taking g as %{gravity}, " \7            "%{viscosity}, in kg/(m s)?"8  given radius: "2.0 mm"9  given terminal: "6.5 cm/s"10  given oil: "1.5e3 kg/m3"11  given copper: "8.9e3 kg/m3"12  given gravity: "9.8 m/s2"13  asks viscosity: "what is the viscosity of the oil at 20 °C"14end

solution.rb

1require_relative "../questions/a_copper_ball_in_oil"2require_relative "../science/mechanical_properties_of_fluids/stokes_law"34solve :a_copper_ball_in_oil, using: StokesLaw do5  objects :ball67  take radius: ball.radius8  take terminal: ball.terminal_velocity9  take oil: sigma10  take copper: ball.density11  take gravity: g1213  viscosity = eta14  answer viscosity:, in: "kg m-1 s-1", figures: 21516  conclusion "At its terminal velocity the ball's weight, less the oil's buoyancy, is matched by Stokes' " \17             "drag, which grows with the viscosity; so the speed it settles to measures the viscosity."18end

playground.rb

1require_relative "../solutions/a_copper_ball_in_oil"23play_with :a_copper_ball_in_oil do4  title "A Copper Ball in Oil"5end

The laws it drew on

quantity	written	units	within	from
gravity	$g$	m/s²	—	StokesLaw
radius	$a$	m	0 … ∞	StokesLaw
density	$ρ$	kg/m³	0 … ∞	StokesLaw
density of the fluid	$σ$	kg/m³	—	StokesLaw
terminal velocity	$v_{t}$	m/s	—	StokesLaw
coefficient of viscosity	$η$	Pa s	—	StokesLaw
terminal velocity of ground	$v_{t,g}$	m/s	—
radius of ground	$a_{g}$	m	0 … ∞
density of ground	$ρ_{g}$	kg/m³	0 … ∞
terminal velocity of ball	$v_{t}$	m/s	—
radius of ball	$a$	m	0 … ∞
density of ball	$ρ$	kg/m³	0 … ∞

v_{t,b} = \frac{2 \cdot {a_{b}}^{2} \cdot (ρ_{b} - σ) \cdot g}{9 \cdot η}

terminal velocity of a sphere

v_{t,g} = \frac{2 \cdot {a_{g}}^{2} \cdot (ρ_{g} - σ) \cdot g}{9 \cdot η}

terminal velocity of a sphere(ground)

v_{t} = \frac{2 \cdot a^{2} \cdot (ρ - σ) \cdot g}{9 \cdot η}

terminal velocity of a sphere(ball)