Density upthrust and viscosity

Density of a Material

Key Equations

Density equation:

$$\rho =\frac{m}{V}$$

Variables: - $\rho $ = density - $m$ = mass - $V$ = volume

SI unit: kg m⁻³

Rearrangements:

$$m=\rho V$$

$$V=\frac{m}{\rho }$$

ProportionalityFor a fixed volume, mass is directly proportional to density. For a fixed mass, volume is inversely proportional to density.

Density is the mass per unit volume of a substance and describes how tightly matter is packed within a given space. A denser material has more mass packed into the same volume than a less dense one. Density depends on the material and on temperature, because thermal expansion changes the volume.

Solids and liquids have densities of order 10³ kg m⁻³, while gases at room conditions are typically around 1 kg m⁻³. Water has a density of 1000 kg m⁻³, which makes it a useful reference value in calculations. To find density experimentally, mass is measured with a balance and volume is found from regular dimensions or by displacement of water.

MisconceptionDensity is not the same as weight or “heaviness.” A small piece of lead is denser than a large block of foam, even though the foam may weigh more overall. Density compares mass per unit volume, not total mass.

Exam TipAlways check whether the question asks for mass, volume or density before substituting.

Upthrust on a Submerged Object

The upthrust on an object immersed in a fluid equals the weight of the fluid that the object displaces. This statement is known as Archimedes’ principle and applies in both liquids and gases.

The upthrust arises because the fluid pressure on the lower surface of an object is greater than the pressure on the upper surface. The pressure difference produces a net upward force on the object. This force does not depend on the object’s own weight — only on the volume of fluid pushed aside and the density of that fluid.

Key Equations

Upthrust equation:

$$U={\rho }_{\text{fluid}} {V}_{\text{fluid}} g$$

Variables: - $U$ = upthrust - ${\rho }_{\text{fluid}}$ = density of the fluid - ${V}_{\text{fluid}}$ = volume of fluid displaced - $g$ = gravitational field strength

SI unit: N

ProportionalityUpthrust is directly proportional to both the density of the fluid and the volume displaced. Doubling the volume submerged doubles the upthrust.

An object floats when its weight equals the upthrust, which happens once enough of the object has sunk to displace its own weight in fluid. An object sinks when its weight exceeds the maximum possible upthrust, meaning the fluid cannot displace enough mass even when the object is fully submerged.

Viscous Drag and Stokes’ Law

When a small sphere moves slowly through a fluid, it experiences a viscous drag force that opposes its motion. The size of this drag depends on the radius of the sphere, its speed and a property of the fluid called viscosity.

Viscosity [symbol η, “eta”] measures the resistance of a fluid to flow. A fluid with high viscosity, such as honey, flows slowly because internal layers slide past each other with difficulty. A fluid with low viscosity, such as water, flows easily. Viscosity is strongly temperature dependent: for liquids, viscosity decreases as temperature rises, because faster molecules can move past each other more easily.

Key Equations

Stokes’ Law:

$$F=6\pi \eta rv$$

Variables: - $F$ = viscous drag force - $\eta $ = viscosity (coefficient) of the fluid - $r$ = radius of the sphere - $v$ = speed of the sphere through the fluid

SI unit of F: N SI unit of η: Pa s (pascal second), equivalent to N s m⁻²

Rearrangements:

$$\eta =\frac{F}{6\pi rv}$$

$$v=\frac{F}{6\pi \eta r}$$

ProportionalityThe drag force is directly proportional to radius, speed and viscosity. Doubling the speed doubles the drag; doubling the radius also doubles the drag.

Stokes’ Law applies only when three conditions are met simultaneously: - The object is small and spherical. - The object moves at a low speed through the fluid. - The flow around it is laminar [smooth layers of fluid sliding past each other], with no turbulence.

When speed becomes too high, the flow becomes turbulent and Stokes’ Law breaks down. Drag then increases more rapidly than predicted by the equation.

A sphere falling through a viscous fluid eventually reaches terminal velocity. At this point the drag force increases until it equals the weight minus the upthrust, so the resultant force is zero and therefore the acceleration is zero. The sphere continues to fall at a constant speed.

Examiner InsightWhen asked to justify Stokes’ Law in a question, both the size and speed conditions must be stated, plus the requirement of laminar flow. Mentioning only “small sphere” loses marks.

Exam TipState all three conditions — small, spherical, slow, laminar flow.

Worked Example: Drag on a Steel Ball Bearing

Scenario

A steel ball bearing of radius 1.5 mm falls through glycerol of viscosity 1.49 Pa s at a speed of 0.040 m s⁻¹. Calculate the viscous drag force acting on the ball.

Step 1 — Convert given values to SI units.

Radius conversion:

$$r=1.5 \text{mm}$$

$$r=1.5\times {10}^{-3} \text{m}$$

Step 2 — Apply Stokes’ Law.

$$F=6\pi \eta rv$$

$$F=6\pi \times 1.49\times (1.5\times {10}^{-3})\times 0.040$$

$$F=1.685\times {10}^{-3} \text{N}$$

Final answer

$F\approx 1.7\times {10}^{-3} \text{N}$

Interpretation

The drag force is small but significant compared to the ball’s weight, which is why ball bearings reach a low terminal velocity in viscous liquids.

CORE PRACTICAL 2 — Falling-Ball Viscosity

A falling-ball method uses the terminal velocity of a small sphere descending through a liquid to determine the liquid’s viscosity. At terminal velocity, the resultant force on the ball is zero, so the equations for weight, upthrust and Stokes’ drag combine to give viscosity directly.

Aim

To determine the viscosity of a viscous liquid (such as glycerol or wallpaper paste) by measuring the terminal velocity of a small ball bearing falling through it.

Method

1. Use a micrometer to measure the diameter of the ball bearing in three places and average to find the radius r.
2. Find the mass of the ball with a digital balance, and calculate its density using ρ = m/V.
3. Fix two rubber bands around the cylinder, separated by a measured distance d. Position the upper band well below the liquid surface so the ball reaches terminal velocity before passing it.
4. Record the temperature of the liquid with a thermometer.
5. Release the ball gently from forceps just at the liquid surface and start a stopwatch as it crosses the upper marker.
6. Stop the stopwatch as the ball crosses the lower marker. Record the time t.
7. Calculate terminal velocity using v = d/t. Repeat for several balls and take a mean.
8. Use the equilibrium condition at terminal velocity to calculate viscosity.

Variables

• Independent variable (IV): Identity (and therefore radius) of the ball bearing, varied between trials by using balls of different sizes; measured in m.
• Dependent variable (DV): Time t for the ball to fall the distance d between the markers, measured with a digital stopwatch in seconds; converted to terminal velocity v = d/t.
• Control variables (CV): Temperature of the liquid (kept constant by allowing the liquid to reach room temperature and monitoring with a thermometer); the same liquid throughout; vertical orientation of the cylinder (checked with a set square).

Expected results

At terminal velocity, weight equals upthrust plus drag, so:

$$4/3\pi {r}^{3}{\rho }_{\text{ball}} g=4/3\pi {r}^{3}{\rho }_{\text{fluid}} g+6\pi \eta rv$$

Rearranging for viscosity:

$$\eta =\frac{2{r}^{2}g({\rho }_{\text{ball}}-{\rho }_{\text{fluid}})}{9v}$$

A larger ball reaches a higher terminal velocity, but the calculated η should be the same within experimental uncertainty.

Precaution

The ball must reach terminal velocity before crossing the upper marker, otherwise the calculated v is too low and η too high. Position the upper band several centimetres below the liquid surface and check that the ball spacing between successive markers is constant. Avoid using balls so large that flow becomes turbulent, since Stokes’ Law would no longer apply.

Measurement technique

Use a micrometer (resolution 0.01 mm) for ball radius, taking the mean of three readings around the sphere. Use a digital stopwatch (resolution 0.01 s) for time, but note that human reaction time gives an uncertainty of about ±0.2 s. Increase the distance d between markers to reduce the percentage uncertainty in time. Repeat the drop at least three times for each ball and average the results.

SafetyGlycerol and similar liquids can stain skin and clothes. Wear safety goggles. Clean spills immediately, since the floor becomes very slippery.