49-52 - Density, Upthrust and Viscosity (CP4)

49-52 - Density, Upthrust and Viscosity (CP4)

Modelling what happens when an object is in a fluid: how much matter is packed into a volume, how the displaced fluid pushes back, and how a viscous liquid resists motion. The core practical uses a falling sphere to turn those ideas into a measurement of viscosity. The lesson covers density, upthrust, Stokes' law and the falling-ball method (Core Practical 4).

Density as mass per volume

Imagine two spheres with the same radius: one is glass, one is steel. They displace the same volume of liquid, but the steel one has more mass packed into that volume. Density is the quantity that captures that difference:

Density

ρ=mV \rho = \frac{m}{V}

ρ\rho: density (kg m⁻³); mm: mass (kg); VV: volume (m³). For a sphere V=43πr3V = \frac{4}{3}\pi r^{3}.

The most common trap is mixing units. If a diameter is measured in mm, convert to metres before finding the volume. If a mass is measured in g, convert to kilograms before using SI density.

For a sphere,

V=43πr3V = \frac{4}{3}\pi r^3

Worked example: a small sphere has diameter 10.0 mm10.0\,\mathrm{mm} and mass 4.10 g4.10\,\mathrm{g}.

The radius is half the diameter:

r=5.00 mm=5.00×10−3 mr = 5.00 \text{ mm} = 5.00 \times 10^{-3}\text{ m}

The volume is:

V=43π(5.00×10−3)3=5.24×10−7 m3V = \frac{4}{3}\pi(5.00 \times 10^{-3})^3 = 5.24 \times 10^{-7}\,\mathrm{m^{3}}

The mass is:

m=4.10 g=4.10×10−3 kgm = 4.10 \text{ g} = 4.10 \times 10^{-3}\text{ kg}

So:

ρ=4.10×10−35.24×10−7=7.83×103 kg m−3\rho = \frac{4.10 \times 10^{-3}}{5.24 \times 10^{-7}} = 7.83 \times 10^{3}\text{ kg m}^{-3}

The decision point was not the algebra; it was noticing that density compares mass with the actual volume occupied, so the sphere's radius had to be cubed after converting to metres.

Upthrust as displaced fluid

When an object is submerged, the pressure on its lower surface is greater than the pressure on its upper surface because pressure increases with depth. The sideways forces cancel, but the vertical forces do not. The result is an upward force called upthrust.

Upthrust

The upward force a fluid exerts on an object in it, equal to the weight of fluid displaced by the object.

For a fully submerged object, the displaced volume is the object's volume. If the fluid has density ρf\rho_f, then:

Upthrust

U=ρfVg U = \rho_{\mathrm{f}} V g

UU: upthrust (N); ρf\rho_{\mathrm{f}}: density of the fluid, not the object (kg m⁻³); VV: volume of fluid displaced (m³); gg: 9.81 N kg⁻¹.

A useful misconception to catch early: upthrust is not a reward for floating. It acts on a sinking object too. A ball bearing sinks because its weight is larger than the upthrust, not because upthrust is absent.

Worked example: a ball of mass 0.0400 kg0.0400\,\mathrm{kg} and volume 5.20×10−6 m35.20 \times 10^{-6}\,\mathrm{m}^{3} is fully submerged in washing-up liquid of density 1020 kg m−31020\,\mathrm{kg}\,\mathrm{m}^{-3}.

The ball's weight is:

W=mg=0.0400×9.81=0.392 NW = mg = 0.0400 \times 9.81 = 0.392\text{ N}

The upthrust uses the density of the liquid, not the density of the ball:

U=ρfVg=1020×5.20×10−6×9.81=5.20×10−2 NU = \rho_f Vg = 1020 \times 5.20 \times 10^{-6} \times 9.81 = 5.20 \times 10^{-2}\text{ N}

Before viscous drag is considered, the downward force left over is:

W−U=0.392−0.0520=0.340 NW - U = 0.392 - 0.0520 = 0.340\text{ N}

That result tells the physical story: the liquid does push upward, but not enough to balance the ball's weight.

Viscous drag and terminal speed

A liquid such as glycerol or washing-up liquid does not simply let a ball pass through it. Layers of liquid have to move around the ball, and the liquid's internal resistance produces viscous drag.

Viscosity

Dynamic viscosity η\eta measures how strongly a fluid resists flow (internal friction between its layers); unit Pa s, equivalent to N s m⁻².

For a small sphere moving slowly through a fluid with laminar flow, Edexcel require Stokes' law:

Stokes' law

F=6πηrv F = 6\pi\eta r v

FF: viscous drag (N); η\eta: viscosity (Pa s); rr: sphere radius (m); vv: speed (m s⁻¹). Only for a small sphere moving slowly with laminar (non-turbulent) flow.

The direction matters even though the equation gives a magnitude. If the sphere is falling downward, viscous drag acts upward, opposite to the motion.

Worked example: a sphere of radius 2.50 mm2.50\,\mathrm{mm} falls through a liquid of viscosity 1.20 Pa s1.20\,\mathrm{Pa}\,\mathrm{s} at 0.0400 m s−10.0400\,\mathrm{m}\,\mathrm{s}^{-1}.

Convert the radius:

r=2.50×10−3 mr = 2.50 \times 10^{-3}\text{ m}

Then:

F=6π(1.20)(2.50×10−3)(0.0400)=2.26×10−3 NF = 6\pi(1.20)(2.50 \times 10^{-3})(0.0400) = 2.26 \times 10^{-3}\text{ N}

The choice of equation is conditional. Stokes' law is not a general drag law for every falling object. It applies here only because the object is a small sphere, the speed is low, and the flow is laminar rather than turbulent.

As the ball starts falling, its speed is small, so viscous drag is small. As speed increases, Stokes' drag increases. Eventually the upward forces, upthrust plus viscous drag, balance the downward weight:

W=U+FW = U + F

The resultant force is then zero, so the ball moves at constant terminal velocity.

At terminal velocity weight = upthrust + viscous drag: the forces are balanced, not absent.

Falling-ball viscosity method

Core Practical 4 uses the terminal velocity of a sphere to determine the viscosity of a liquid. The apparatus is deliberately simple: a tall transparent tube of liquid, small spheres, rubber bands or marks on the tube, a metre ruler, a timer, and a micrometer or vernier callipers for the sphere diameter.

[DIAGRAM: asset_name: falling ball apparatus; asset_slug: 9ph0_falling_ball; recommended_method: matplotlib; description: Wide liquid column, central sphere, timing marks below release, and terminal force diagram showing upward upthrust and drag balanced by weight.]
Diagram

The timing marks are not placed right at the surface. The top timing mark must be far enough below the liquid surface that the sphere has already reached terminal velocity by the time it gets there. Check this by comparing transit times over successive equal vertical distances. Then the measured speed between marks is a terminal speed, not an average speed from release.

A practical route:

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