Density, upthrust and viscous drag flashcards

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Density
Mass per unit volume of a substance; ρ = m/V. Unit kg m⁻³.
Upthrust
The upward force a fluid exerts on a body wholly or partly immersed in it; equal to the weight of the fluid displaced.
Viscosity
A measure of a fluid's resistance to flow. Coefficient of viscosity η, unit Pa s (= N s m⁻² = kg m⁻¹ s⁻¹).
Laminar (streamline) flow
Flow in smooth layers that slide past each other without mixing; the velocity at each point is constant and streamlines never cross.
Turbulent flow
Flow with swirling eddies in which the layers mix and the velocity at a point changes irregularly in size and direction.
Terminal velocity of a sphere in a liquid
The constant velocity reached when weight = upthrust + viscous drag, so the resultant force is zero.
Hydrometer
A weighted float with a narrow graduated stem; the depth at which it floats gives the density of the liquid.
Density from mass and volume
ρ = m/V. (Yes)
Upthrust as a calculation
Upthrust = weight of fluid displaced, i.e. U = ρ_fluid × V_displaced × g. (No — learn it)
Stokes' law
F = 6πηrv (viscous drag on a small sphere moving slowly in laminar flow). (Yes)
Terminal velocity of a sphere (Stokes' law applies)
v = 2r²g(ρ_s − ρ_f) / 9η, from W = U + F. (No — derive it)
Fraction of a floating body below the surface
V_sub / V = ρ_body / ρ_fluid. (No — derive it)
Volume of a sphere
V = (4/3)πr³ — use the radius. (No — learn it)
Forces on a sphere falling at terminal velocity in a liquid
Weight down; upthrust and viscous drag both up; W = U + F.
v–t graph for a ball released from rest in a viscous liquid
Starts with gradient g(1 − ρ_f/ρ_s), less than g; the gradient decreases to zero as v approaches the terminal velocity.
Core Practical 2: graph of v against r²
Straight line through the origin; gradient = 2g(ρ_s − ρ_f)/9η, so η = 2g(ρ_s − ρ_f)/(9 × gradient).
Streamlines around a small sphere in laminar flow
Smooth curves that part round the sphere and close up behind it, never crossing; turbulent flow shows eddies behind it.
Hydrometer scale
Higher densities lower down the stem; marks crowd closer together towards the bulb (non-uniform scale).
Why is there an upthrust on a submerged body?
Fluid pressure increases with depth, so the push up on the bottom surface is greater than the push down on the top; the sideways pushes cancel.
When does a body float?
When its average density is less than the fluid's; it sinks until the weight of fluid displaced equals its own weight.
How does a hot-air balloon rise?
Heated air inside is less dense than the cool air outside. Upthrust (weight of cool air displaced) exceeds the total weight of hot air + envelope + load.
Conditions for Stokes' law
Small sphere, low speed, laminar flow (and far from container walls).
Effect of temperature on the viscosity of a liquid
Viscosity decreases (often steeply) as temperature rises, so terminal velocities in warm liquids are higher.
Explain how a ball bearing reaches terminal velocity in oil
Weight > upthrust, so it accelerates → drag rises with speed → resultant force and acceleration fall → when W = U + F, resultant = 0, velocity constant.
Core Practical 2: how do you know the ball is at terminal velocity?
Times for successive equal distances (e.g. between equally spaced bands) are the same.
Why does a steel ship float?
The hull encloses air, so the ship's average density is less than water's; it displaces its own weight of water.
A 54.0 g stone displaces 20.0 cm³ of water. Density in kg m⁻³?
ρ = 54.0/20.0 = 2.70 g cm⁻³ = 2.70 × 10³ kg m⁻³. (Yes)
Upthrust on a 2.5 × 10⁻⁴ m³ object fully under water (ρ = 1000 kg m⁻³)?
U = 1000 × 2.5 × 10⁻⁴ × 9.81 = 2.5 N (2.45 N). (No — learn upthrust = weight of fluid displaced)
Fraction of an ice block (917 kg m⁻³) submerged in sea water (1030 kg m⁻³)?
917/1030 = 0.89 — about 89% below the surface.
Drag on a sphere of radius 1.0 mm moving at 0.020 m s⁻¹ through a liquid of viscosity 1.2 Pa s?
F = 6π × 1.2 × 1.0 × 10⁻³ × 0.020 = 4.5 × 10⁻⁴ N. (Yes)
Terminal velocity of a steel ball (r = 0.50 mm, 7800 kg m⁻³) in oil (900 kg m⁻³, η = 0.30 Pa s)?
v = 2r²g(ρ_s − ρ_f)/9η = 2 × (0.50 × 10⁻³)² × 9.81 × 6900 / 2.7 = 0.013 m s⁻¹.
Balloon of 2000 m³; outside air 1.20 kg m⁻³, hot air 0.95 kg m⁻³. Maximum mass of envelope + load?
M = (1.20 − 0.95) × 2000 = 500 kg.
Graph of v against r² has gradient 1.20 × 10⁴ m⁻¹ s⁻¹; ρ_s = 7800, ρ_f = 1260 kg m⁻³. Viscosity?
η = 2g(ρ_s − ρ_f)/(9 × gradient) = 2 × 9.81 × 6540 / (1.08 × 10⁵) = 1.2 Pa s.
Same material, same liquid: ball B has twice the radius of ball A. Ratio of terminal velocities v_B/v_A?
v ∝ r², so 4.
Converting g cm⁻³ to kg m⁻³
Multiply by 1000 (not 10⁶, not 100): 1 g cm⁻³ = 1000 kg m⁻³.
Diameter given, Stokes' law needed
Halve it first: Stokes' law and (4/3)πr³ use the radius.
Using the whole volume for the upthrust on a floating body
Use only the volume below the surface.
Using Stokes' law for a skydiver or a car
Not valid: large, fast bodies cause turbulent flow and much larger drag.
Uncertainty in falling-ball times
Repeat readings spread more than the stopwatch resolution: use half the range, then ÷ mean × 100 for a percentage.

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