In the school laboratory, drop a small steel ball bearing into a tall jar of thick oil. For the first few millimetres it speeds up. Then it sinks at a slow, perfectly steady pace all the way to the bottom. Dropped in air from the same height, the same ball would reach the floor in a blink, still speeding up.
The drag on a sphere
A sphere moving through a fluid drags the fluid layers next to it, and they drag back. The result is a force against the motion, the viscous drag. What can it depend on? The fluid's viscosity \(\eta\), surely. The size of the sphere, its radius \(r\). And its speed \(v\) through the fluid. For slow motion the fluid's density hardly matters, because the flow is steady and the fluid is not being speeded up. So guess \(F = k\,\eta^a r^b v^c\), where \(k\) is a pure number.
\[ [\text{MLT}^{-2}] = [\text{ML}^{-1}\text{T}^{-1}]^a\,[\text{L}]^b\,[\text{LT}^{-1}]^c \]
Why. write the dimensions of each quantity: force, viscosity (from Lesson 1), radius and speed.
\[ [\text{MLT}^{-2}] = \text{M}^{a}\,\text{L}^{-a+b+c}\,\text{T}^{-a-c} \]
Why. collect the powers of M, L and T on the right.
\[ a = 1, \quad -a - c = -2, \quad -a + b + c = 1 \]
Why. the powers of M, of T and of L must be the same on both sides.
\[ a = 1, \quad c = 1, \quad b = 1 \]
Why. the first gives \(a\); the second then gives \(c = 1\); the third gives \(b = 1 + a - c = 1\).
\[ F = k\,\eta r v \]
Read it. every power is one. Dimensions cannot give the pure number \(k\).
The full calculation of the flow round the sphere was done by George Stokes in 1851. It gives \(k = 6\pi\). This is Stokes' law.
A sphere of radius \(r\) moving slowly at speed \(v\) through a fluid of viscosity \(\eta\) feels a viscous drag \(F = 6\pi\eta r v\). The drag acts opposite to its velocity, and the law holds for streamline flow.
A sphere of radius \(1\,\text{mm}\) moves at \(0.1\,\text{m/s}\) through a liquid of viscosity \(0.1\,\text{Pa s}\). Find the viscous drag on it.
\(F = 6\pi\eta r v = 6 \times 3.14 \times 0.1 \times 10^{-3} \times 0.1 = 1.9 \times 10^{-4}\,\text{N}\). The drag is tiny, about the weight of \(0.02\,\text{g}\).
A sphere moves through oil in streamline flow. Its radius is doubled and its speed is doubled. The viscous drag on it becomes
Three forces on a falling ball
Let go a sphere of radius \(r\) and density \(\rho\) in a liquid of density \(\sigma\) (sigma) and viscosity \(\eta\). Three forces act on it. Its weight \(W = \tfrac43\pi r^3\rho g\) pulls it down. The buoyant force of the chapter on Archimedes' Principle, \(F_B = \tfrac43\pi r^3\sigma g\), pushes it up. And the viscous drag \(F_v = 6\pi\eta r v\) acts against its motion, upward while it falls.
The weight and the buoyant force never change. The drag does: it grows with the speed. At the moment of release the ball is at rest, so the drag is zero. The net force is \(W - F_B = \tfrac43\pi r^3(\rho - \sigma)g\), and dividing by the mass \(\tfrac43\pi r^3\rho\) gives the starting acceleration. \[ a_0 = g\left(1 - \frac{\sigma}{\rho}\right) \] For a steel ball of \(7500\,\text{kg/m}^3\) in oil of \(1500\,\text{kg/m}^3\), \(a_0 = 10 \times (1 - 0.2) = 8\,\text{m/s}^2\).
Follow the arrows in Fig 3.1. The weight is \(8.5\,\text{mN}\) and the buoyant force \(1.7\,\text{mN}\), so \(6.8\,\text{mN}\) is left over to speed the ball up. As the speed grows, the drag arrow grows with it, and less force is left over. The acceleration falls, and the v-t graph bends. When the drag reaches \(6.8\,\text{mN}\), the three forces balance. The acceleration is zero, and the speed stays at \(0.1\,\text{m/s}\).
The approach is quick but gradual. The dashed line of Fig 3.1 shows that at the starting rate of \(8\,\text{m/s}^2\), the ball would reach \(0.1\,\text{m/s}\) in \(12.5\,\text{ms}\). In fact it has only \(63\,\%\) of that speed by then, because the drag is already growing. After about \(50\,\text{ms}\) it is within \(2\,\%\) of \(0.1\,\text{m/s}\), and it has fallen less than \(4\,\text{mm}\).
A glass marble of density \(2500\,\text{kg/m}^3\) is let go at rest in water of density \(1000\,\text{kg/m}^3\). Take \(g = 10\,\text{m/s}^2\). What is its acceleration at the instant it is let go, in \(\text{m/s}^2\)?
The steel ball of Fig 3.1 has reached \(0.05\,\text{m/s}\), half its final speed. What is its acceleration now?
The terminal velocity
The steady speed at which the forces balance is the terminal velocity, \(v_t\). At \(v_t\) the acceleration is zero, so the weight is balanced exactly by the buoyant force and the drag together. Set up that balance and solve for \(v_t\).
The terminal velocity of a body falling through a fluid is the constant speed it finally reaches. There the drag has grown to balance the weight less the buoyant force, so the net force on it is zero.
\[ W - F_B - F_v = m a \]
Why. Newton's second law for the falling sphere, with down as positive.
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Your turn. which line holds at the terminal velocity?
\[ 6\pi\eta r v_t = \tfrac43 \pi r^3 (\rho - \sigma) g \]
Why. move the buoyant force to the left side and take out the common factor.
\[ v_t = \dfrac{2r^2(\rho - \sigma)g}{9\eta} \]
Read it. divide both sides by \(6\pi\eta r\); \(\tfrac{4}{3 \times 6} = \tfrac29\).
Three things follow from the formula. The terminal velocity grows as the square of the radius: a ball twice as wide falls four times as fast. It grows with the difference of densities, \(\rho - \sigma\). And it is inversely proportional to the viscosity. If \(\rho = \sigma\), the sphere hangs at rest anywhere in the liquid.
Move the slider in the sim above from \(3\,\text{mm}\) to \(1.5\,\text{mm}\). The curve now flattens at \(0.025\,\text{m/s}\), a quarter of \(0.1\,\text{m/s}\). The weight fell eight times, but the drag at a given speed fell only two times. The smaller ball also settles four times sooner, in a few milliseconds. Its starting acceleration is the same \(8\,\text{m/s}^2\), so a quarter of the speed takes a quarter of the time.
The steel ball of Fig 3.1 has radius \(3\,\text{mm}\) and density \(7500\,\text{kg/m}^3\). The oil has density \(1500\,\text{kg/m}^3\) and viscosity \(1.2\,\text{Pa s}\). Take \(g = 10\,\text{m/s}^2\). Find the terminal velocity, and check that Stokes' law holds.
\(r = 3 \times 10^{-3}\,\text{m}\), \(\rho - \sigma = 6000\,\text{kg/m}^3\), \(\eta = 1.2\,\text{Pa s}\).
\(v_t = \dfrac{2r^2(\rho - \sigma)g}{9\eta}\), and Stokes' law needs \(R_e = \sigma v_t\,(2r)/\eta\) below about \(1\).
\(v_t = \dfrac{2 \times 9 \times 10^{-6} \times 6000 \times 10}{9 \times 1.2} = \dfrac{1.08}{10.8} = 0.1\,\text{m/s}\). Then \(R_e = 1500 \times 0.1 \times 0.006/1.2 = 0.75\).
The ball falls steadily at \(0.1\,\text{m/s}\), as in Fig 3.1. With \(R_e = 0.75\) the flow round it is streamline, so Stokes' law was fair to use.
A glass ball of radius \(3\,\text{mm}\) and density \(2500\,\text{kg/m}^3\) falls through glycerine of density \(1250\,\text{kg/m}^3\) and viscosity \(1.25\,\text{Pa s}\). Take \(g = 10\,\text{m/s}^2\). Find its terminal velocity, in m/s.
A steel ball of radius \(6\,\text{mm}\), twice the radius of the ball in the example, falls through the same oil. If Stokes' law still holds, its terminal velocity is
Stokes' law holds only for slow, streamline flow round the sphere, a Reynolds number below about \(1\). A cricket ball in air, a stone in water or a raindrop is far outside this. At high speed the drag grows nearly as \(v^2\), not as \(v\). The terminal velocity still exists, but the formula \(2r^2(\rho - \sigma)g/9\eta\) gives far too large a value.
How fast is the terminal velocity reached? The equation of motion is \(m\,dv/dt = (W - F_B) - 6\pi\eta r v\). Its solution is \(v = v_t\left(1 - e^{-t/\tau}\right)\), with the time constant \(\tau = m/(6\pi\eta r) = 2r^2\rho/9\eta\). For the steel ball, \(\tau = 12.5\,\text{ms}\). This is also \(v_t/a_0\), the time the dashed line of Fig 3.1 takes to reach \(v_t\).
At Cambridge, George Stokes wanted to know why a pendulum swinging in air slows down. The drag of the air, and even the air it carries with it, change its period by a small but measurable amount. His paper of 1851 on the internal friction of fluids worked out the flow of a viscous fluid round a moving sphere. The drag \(6\pi\eta r v\) was one result. Stokes then used it to explain why clouds float.
- Leaving out the buoyant force when finding the terminal velocity in a liquid.At the terminal velocity, weight = buoyant force + drag. Only for a drop in air is the buoyant force negligible.
- Thinking the ball's acceleration is g until it reaches its terminal velocity.At release the acceleration is \(g(1 - \sigma/\rho)\), and it falls steadily as the drag grows.
- Taking the terminal velocity to be proportional to the radius.The weight grows as \(r^3\) and the drag as \(r\), so \(v_t\) grows as \(r^2\).