Irodov 1.339: The Relaxation Time to Terminal Speed – Amit & Deepali Sinha Foundation

Problem Irodov 1.339

A steel ball of diameter $d = 3.0$ mm is released from rest and sinks in olive oil of viscosity $\eta = 0.90$ P. The ball’s speed rises from zero toward a steady (terminal) sinking speed. Find the time $t$ after release at which the ball’s instantaneous speed falls short of that steady-state speed by a fraction $n = 1.0\%$ of it.

The steel ball of diameter d, released from rest in the olive oil and left to sink. — The steel ball of diameter $d$ , released from rest in the olive oil and left to sink.

Given

$d = 3.0$ mm
$\eta = 0.90$ P
$n = 1.0\%$
steel density $\rho = 7.8\times10^3$ kg/m $^3$ (tables)

Find

Find

t

, the time for the ball’s speed to come within

n

of its steady-state value

Solution

Step 1The equation of motion, with Stokes drag

The free body this step writes down, at a general instant before steady sinking: the net weight (weight minus buoyancy) exceeds the viscous drag, so the ball is still speeding up. The instantaneous velocity v(t) is drawn alongside, off the body. — The free body this step writes down, at a general instant before steady sinking: the net weight (weight minus buoyancy) exceeds the viscous drag, so the ball is still speeding up. The instantaneous velocity $\vect v(t)$ is drawn alongside, off the body.

Let $\rho$ be the ball’s own density and $\rho_0$ the oil’s, with radius $r = d/2$ ; write $m$ for the ball’s own mass and $m_0$ for the mass of oil it displaces, so the net weight driving it down is $(m-m_0)g$ , the quantity the figure marks. Newton’s second law, with the net weight (weight minus buoyancy) driving the ball down and Stokes drag $6\pi\eta r v$ opposing its motion, reads

\frac{4}{3}\pi r^3\rho\,\frac{dv}{dt} = \frac{4}{3}\pi r^3(\rho-\rho_0)g - 6\pi\eta r v.

Dividing through by $(4/3)\pi r^3\rho$ isolates the acceleration,

\frac{dv}{dt} = (\rho-\rho_0)\dfrac{g}{\rho} - \frac{9\eta}{2\rho r^2}\,v

(1)

Step 2Steady-state speed and the relaxation rate

At steady state the ball no longer accelerates, so setting $dv/dt=0$ in (1) leaves the two terms on the right in balance,

0 = (\rho-\rho_0)\frac{g}{\rho} - \frac{9\eta}{2\rho r^2}\,v_\infty ,

and solving that for the steady-state speed gives

v_\infty = \frac{2r^2(\rho-\rho_0)g}{9\eta}

(2)

The coefficient multiplying $v$ in (1) is a constant with the dimensions of a reciprocal time, and it is worth a name of its own:

k \equiv \frac{9\eta}{2\rho r^2}

(3)

Its reciprocal $1/k$ is the relaxation time of the motion. Since $kv_\infty = (\rho-\rho_0)g/\rho$ by (2), the constant term in (1) is exactly $kv_\infty$ , and the equation of motion collapses into a statement about the shortfall $v_\infty - v$ alone:

\frac{dv}{dt} = k\,(v_\infty - v)

(4)

Step 3Integrate from rest, then apply the

n

condition

Separating the variables in (4) and integrating from $v=0$ at $t=0$ ,

\int_0^v \frac{dv'}{v_\infty-v'} = \int_0^t k\,dt' .

Carrying out both integrals, the left one as a logarithm of the shortfall,

\begin{align*}\Bigl[-\ln\left(v_\infty-v'\right)\Bigr]_0^v &= kt \\ -\ln\left(v_\infty-v\right)+\ln v_\infty &= kt \\ -\ln\!\left(\frac{v_\infty-v}{v_\infty}\right) &= kt .\end{align*}

The ball’s speed differs from $v_\infty$ by the fraction $n$ exactly when $v_\infty-v = nv_\infty$ , so the logarithm’s argument is simply $n$ :

\begin{align*}-\ln n &= kt \\ t &= -\frac{\ln n}{k} .\end{align*}

Substituting the relaxation rate (3) for $k$ ,

t = -\frac{2\rho r^2}{9\eta}\,\ln n .

The ball is given by its diameter rather than its radius, so put $r = d/2$ in that prefactor:

\frac{2\rho r^2}{9\eta} = \frac{2\rho}{9\eta}\cdot\frac{d^2}{4} = \frac{\rho d^2}{18\eta},

which puts the time in terms of the given data alone.

t = -\frac{\rho d^2}{18\eta}\,\ln n

(5)

Step 4Evaluate

With the steel ball’s density $\rho = 7.8\times10^3$ kg/m $^3$ from the tables, $d = 3.0\times10^{-3}$ m, $\eta = 0.090$ Pa $\cdot$ s and $n=0.010$ , (5) gives

t = -\frac{(7.8\times10^3)(3.0\times10^{-3})^2}{18(0.090)}\ln(0.010)\ \text{s},

t = -\dfrac{\rho d^2}{18\eta}\ln n \approx 0.20\ \text{s}

(6)

Check

Dimensionally $[\rho][d^2]/[\eta] = (\text{kg/m}^3)(\text{m}^2)/(\text{Pa}\cdot\text{s}) = (\text{kg/m})/(\text{kg}/(\text{m}\cdot\text{s})) = \text{s}$ , a time, as required. As $n\to1$ (no time at all needed to be within $100\%$ , since the ball starts at $v=0$ ) $\ln 1=0$ gives $t=0$ , matching the release instant; as $n\to0$ (demanding the ball reach $v_\infty$ exactly) $t\to\infty$ , consistent with the exponential approach never actually closing the gap.

Remark

Because the relaxation rate (3) carries the ball’s own density $\rho$ but not the oil’s $\rho_0$ , two identical steel balls released into two different liquids of the same viscosity but different density would reach their own, different, terminal speeds along the same relaxation timescale $t\propto -\ln n$ ; only how far each ball ultimately falls per second depends on the fluid it displaces, not how quickly it settles into doing so. One honesty note about the model, since Stokes’ law was assumed at the outset: with the table values for steel and olive oil the terminal speed comes out near $0.38$ m/s, which puts the Reynolds number of the flow past the ball at roughly $11$ , an order of magnitude above the creeping-flow range in which Stokes’ law is actually derived. A real ball this size would settle somewhat differently; the problem is posed inside the Stokes idealization, and the time above is the answer that idealization gives.