Density, upthrust and viscous dragEdexcel International A Level Physics: Revision notes

The density of a material is its mass per unit volume:

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

The SI unit is kg m⁻³. Note that $1\ \text{g cm}^{-3} = 1000\ \text{kg m}^{-3}$ and $1\ \text{cm}^3 = 10^{-6}\ \text{m}^3$. For a cube of side $2.0\ \text{cm}$ and mass $21.6\ \text{g}$: $V = (0.020)^3 = 8.0 \times 10^{-6}\ \text{m}^3$ and $\rho = \frac{0.0216}{8.0 \times 10^{-6}} = 2.7 \times 10^3\ \text{kg m}^{-3}$.

To find the density of an irregular solid, measure its mass on a balance and its volume by displacement of water in a measuring cylinder.

A body in a fluid experiences an upward force called upthrust, because the pressure on the lower surface is greater than on the upper surface. The upthrust is equal to the weight of the fluid displaced by the body:

$U = \rho_f V g$

where $\rho_f$ is the density of the fluid and $V$ is the volume of fluid displaced. For a fully submerged body, $V$ is the volume of the body. A body floats when its upthrust equals its weight, so it displaces fluid with a weight equal to its own. A fully submerged cube of volume $1.0 \times 10^{-3}\ \text{m}^3$ in water has $U = 1000 \times 1.0 \times 10^{-3} \times 9.81 = 9.8\ \text{N}$.

A body moving through a fluid experiences a viscous drag force opposing the motion. Viscosity $\eta$ measures how much a fluid resists flow, with unit Pa s (kg m⁻¹ s⁻¹). For a small sphere of radius $r$ moving at speed $v$, Stokes' law gives:

$F = 6\pi\eta r v$

Stokes' law applies only to small spherical objects moving at low speeds with laminar flow (no turbulence). The drag increases with speed. Viscosity depends on temperature: for a liquid it decreases as temperature rises (oil flows more easily when warm).

When a ball is released in a viscous liquid, it accelerates until the drag and upthrust together balance the weight. The resultant force is then zero and the ball moves at constant terminal velocity:

$W = U + F$

$\frac{4}{3}\pi r^3 \rho_s g = \frac{4}{3}\pi r^3 \rho_l g + 6\pi\eta r v$

So $v = \frac{2r^2(\rho_s - \rho_l)g}{9\eta}$ and $\eta = \frac{2r^2(\rho_s - \rho_l)g}{9v}$. For a steel ball of radius $2.0\ \text{mm}$ in glycerol ($\eta = 0.95\ \text{Pa s}$), this gives $v = 0.060\ \text{m s}^{-1}$.

To measure the viscosity of a liquid:

• Measure the ball's diameter with a micrometer in several places and average, then find $r$.
• Find the ball's density and the liquid's density from masses and volumes.
• Drop the ball down the axis of a tall transparent tube, with marks placed well below the surface so that the ball has reached terminal velocity.
• Time the fall between two marks with a stopwatch (or light gates). $v =$ distance / time. Check the speed between successive marks is constant.
• Repeat with several balls and average, then calculate $\eta = \frac{2r^2(\rho_s - \rho_l)g}{9v}$.

Record the temperature of the liquid, since viscosity depends on it. Use small balls so that the speed is low and the flow is laminar.