A Copper Ball in Oil

Q. The terminal velocity of a copper ball of radius 2.0 mm falling through a tank of oil at 20 °C is 6.5 cm/s. Density of oil is 1.5 × 10³ kg/m³, density of copper is 8.9 × 10³ kg/m³. Taking g as 9.8 m/s², what is the viscosity of the oil at 20 °C, in kg/(m s)? expects 0.99 kg m⁻¹ s⁻¹

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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)?

The playground

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

The solution

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 Ruby

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

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

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

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)