IPhO 2022 · Theory problem 3

Scaling laws

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Scaling laws describe the functional relationship between two physical quantities that scale with each other over a significant interval. This functional relationship can be a power law, but there are other possibilities, too. Oftentimes, exact expressions are beyond reach, but scaling laws can still be derived.

Part A. Spaghetto (2.0 points)

A.1 2.0pt

A spaghetto straw of diameter \(d\) is being balanced horizontally from its middle. If \(d = 1\,\mathrm{mm}\), the straw breaks under its own weight once its length reaches \(l = 50\,\mathrm{cm}\). What is the maximum length \(l'\) of the straw of diameter \(d' = 1\,\mathrm{cm}\) before it breaks under its own weight?

Figure 1

Part B. Sand castle (2.0 points)

B.1 2.0pt

The average grain volume of coarse-grained sand is 10 times as large as that of fine-grained sand. Wet fine-grained sand and wet coarse-grained sand have both optimal water content (i.e. assuming the maximal strength of the constructions from it) and are used to build two cylinders of exactly the same shape and size. The strength of each cylinder is tested by pressing it between two parallel plates. The cylinder made of coarse-grained sand gets destroyed once the force applied to press the plates reaches \(F_{\mathrm c} = 10\,\mathrm{N}\). How large is the force \(F_{\mathrm f}\) needed to destroy the cylinder made of fine-grained sand? You may ignore the effects of gravity.

Part C. Interstellar travel (2.0 points)

C.1 2.0pt

The spaceship of an interstellar expedition travels at a constant magnitude of the proper acceleration \(g = 10\,\mathrm{m/s^2}\), i.e., this is the acceleration of the spaceship in the inertial frame of reference where it is instantaneously at rest. The passengers must be able to return to Earth within their remaining expected lifetime of 50 years. The maximum distance from Earth reached by the spaceship is \(d\). If the acceleration is increased to \(g' = 15\,\mathrm{m/s^2}\), the spaceship can reach a farther distance \(d'\). What is the ratio \(d'/d\)?

Hint 1. You may wish to use the relativistic velocity addition formula, however, there are also other approaches.

Hint 2. You may need to deal with hyperbolic functions defined as follows: \[ \cosh x=\frac12(e^x+e^{-x}),\qquad \sinh x=\frac12(e^x-e^{-x}),\qquad \tanh x=\frac{e^x-e^{-x}}{e^x+e^{-x}}. \]

Hint 3. Depending on your approach, you may need one or more of these integrals: \[ \int\frac{\mathrm dx}{1-x^2}=\operatorname{atanh}x+C,\qquad \int\frac{\mathrm dx}{\sqrt{1+x^2}}=\operatorname{asinh}x+C,\qquad \int\sinh x\,\mathrm dx=\cosh x+C, \] where \(\operatorname{asinh}x\) and \(\operatorname{atanh}x\) are the inverse functions of the respective hyperbolic functions.

Part D. That sinking feeling (2.0 points)

D.1 2.0pt

A solid wooden ball of radius \(r_0\) is floating in the water. Ignoring frictional effects, the frequency of small oscillations would be \(\omega_0\), but because of viscous friction, after being displaced vertically, the frequency of decaying oscillations is actually \(0.99\,\omega_0\). What is the minimum radius \(r_{\min}\) of a wooden ball floating in water that undergos small oscillations when displaced? Hint: the viscous drag force acting on a given body is proportional to its speed relative to the bulk of the fluid, and to the viscosity \(\eta\) of the fluid it is moving in. The unit of the viscosity is \(\mathrm{kg/(m\cdot s)}\).

Official English text of the problem.

Answers

  • A.1: \(\displaystyle l'=l\sqrt{\frac{d'}d}=0.50\sqrt{10}\,\mathrm m\approx1.6\,\mathrm m.\)
  • B.1: \(\displaystyle F_{\mathrm f}=10^{1/3}F_{\mathrm c}=10^{4/3}\,\mathrm N\approx22\,\mathrm N.\)
  • C.1: With \(T=50\,\mathrm{years}\) measured by the passengers, \[ \boxed{\frac{d'}d= \frac{g}{g'}\, \frac{\cosh\!\left(g'T/(4c)\right)-1} {\cosh\!\left(gT/(4c)\right)-1} \approx4.8\times10^2.} \]
  • D.1: \(\displaystyle r_{\min}=(1-0.99^2)^{1/3}r_0\approx0.27r_0.\) This is the critical radius; oscillations require \(r>r_{\min}\).

Solution

A.1

Assume the spaghetti has the same density and breaking stress at both diameters, and use the slender-beam approximation. Its cross-sectional area scales as \(A\propto d^2\), so its weight per unit length is \[ q=\rho g_{\mathrm E}A\propto d^2, \] where \(g_{\mathrm E}\) is the gravitational acceleration.

Consider one half of the straw. Its weight is \(q l/2\), acting a distance \(l/4\) from the central support. Thus the largest bending moment, occurring at the middle, is \[ M_{\max}=\frac{ql}{2}\frac l4=\frac{ql^2}{8} \propto d^2l^2. \]

In beam bending, the maximum longitudinal stress is \[ \sigma_{\max}=\frac{M_{\max}(d/2)}{I}, \] where the second moment of area satisfies \(I\propto d^4\). Consequently, \[ \sigma_{\max}\propto\frac{l^2}{d}. \]

Breaking occurs at the same material stress, so \[ \frac{l'^2}{d'}=\frac{l^2}{d}, \qquad l'=l\sqrt{\frac{d'}d}. \] Since \(d'/d=10\), \[ \boxed{l'=0.50\sqrt{10}\,\mathrm m =1.581\ldots\,\mathrm m\approx1.6\,\mathrm m.} \]

B.1

Wet sand is held together by capillary forces from water bridges between grains. Let \(a\) be a characteristic grain size and \(\gamma\) the water surface tension. Assume similar grain shapes, packing and wetting properties. At optimal water content, the bridge geometry scales with \(a\).

The capillary force at a contact scales as \[ f_{\mathrm{cap}}\propto\gamma a. \] This follows either from surface tension acting along a contact line of length proportional to \(a\), or from a capillary pressure of order \(\gamma/a\) acting over an area of order \(a^2\).

The number of contacts transmitting force across a fixed macroscopic area \(S\) scales as \(S/a^2\). Hence the characteristic cohesive stress scales as \[ \sigma_{\mathrm{coh}} \propto \frac{f_{\mathrm{cap}}}{a^2} \propto\frac{\gamma}{a}. \] For cylinders of the same shape and size, the compressive failure load has the same scaling: \[ F_{\mathrm{break}}\propto\frac1a. \]

Because grain volume is proportional to \(a^3\), \[ \frac{a_{\mathrm c}}{a_{\mathrm f}} =\left(\frac{V_{\mathrm c}}{V_{\mathrm f}}\right)^{1/3} =10^{1/3}. \] Therefore, \[ \frac{F_{\mathrm f}}{F_{\mathrm c}} =\frac{a_{\mathrm c}}{a_{\mathrm f}} =10^{1/3}, \] giving \[ \boxed{F_{\mathrm f}=10\,10^{1/3}\,\mathrm N =21.54\ldots\,\mathrm N\approx22\,\mathrm N.} \]

C.1

Treat Earth as an inertial reference frame, neglect gravitational fields, and consider a straight-line journey starting and ending at rest relative to Earth. The available \(50\) years is the passengers’ proper time, denoted by \(T\). Acceleration reversals are idealized as instantaneous.

First derive the displacement under constant proper acceleration. Define the rapidity \[ \theta=\operatorname{atanh}(v/c). \] During a proper-time interval \(\mathrm d\tau\), the velocity change in the instantaneous rest frame is \(g\,\mathrm d\tau\). Relativistic velocity addition gives, to first order, \[ \mathrm dv =g\left(1-\frac{v^2}{c^2}\right)\mathrm d\tau, \] and therefore \[ \frac{\mathrm d\theta}{\mathrm d\tau}=\frac gc. \]

Starting from rest, \[ \theta=\frac{g\tau}{c}, \qquad v=c\tanh\theta, \qquad \frac{\mathrm dt}{\mathrm d\tau}=\cosh\theta. \] It follows that \[ \frac{\mathrm dx}{\mathrm d\tau} =v\frac{\mathrm dt}{\mathrm d\tau} =c\sinh\!\left(\frac{g\tau}{c}\right). \] Integrating, \[ x(\tau)=\frac{c^2}{g} \left[\cosh\!\left(\frac{g\tau}{c}\right)-1\right]. \]

To maximize the outward distance in a given proper time while starting and finishing the outward leg at rest, the ship accelerates outward for half that time and decelerates for the other half. Indeed, for an outward leg of proper duration \(L\), its rapidity cannot exceed \[ \theta(\tau)\leq\frac gc\min(\tau,L-\tau). \] The accelerate-then-decelerate trajectory attains this bound, maximizing the displacement because \(\mathrm dx/\mathrm d\tau=c\sinh\theta\) increases with \(\theta\).

The return journey over the same distance requires the same minimum proper time. Thus the optimal round trip consists of four equal intervals:

  1. Accelerate away from Earth.
  2. Decelerate to rest at the greatest distance.
  3. Accelerate toward Earth.
  4. Decelerate to rest at Earth.

Each interval lasts \(T/4=12.5\) years. The two outward intervals cover equal distances, so \[ d(g)=\frac{2c^2}{g} \left[\cosh\!\left(\frac{gT}{4c}\right)-1\right]. \] Hence \[ \boxed{\frac{d'}d= \frac{g}{g'}\, \frac{\cosh\!\left(g'T/(4c)\right)-1} {\cosh\!\left(gT/(4c)\right)-1}.} \]

Using \(c\simeq3.00\times10^8\,\mathrm{m/s}\) and \(1\,\mathrm{year}\simeq3.156\times10^7\,\mathrm s\), \[ \frac{gT}{4c}\simeq13.15, \qquad \frac{g'T}{4c}\simeq19.72. \] Both arguments are large, so \(\cosh x-1\simeq e^x/2\). Thus \[ \frac{d'}d \simeq\frac{g}{g'} \exp\!\left(\frac{(g'-g)T}{4c}\right) \simeq\frac23e^{6.575}. \] The result is \[ \boxed{\frac{d'}d\approx4.8\times10^2.} \]

D.1

Assume all balls are made of the same wood, buoyancy determines their equilibrium immersion, and the stated linear viscous-drag model applies.

For a small downward displacement \(z\), the additional submerged volume is \(A_{\mathrm w}z\), where \(A_{\mathrm w}\) is the area of the sphere’s cross-section at the water surface. The additional buoyancy provides a restoring force \[ F_{\mathrm{rest}}=-kz, \qquad k=\rho_{\mathrm w}g_{\mathrm E}A_{\mathrm w}. \]

The equilibrium submerged fraction is fixed by the wood-to-water density ratio. The floating configurations are therefore geometrically similar, giving \[ A_{\mathrm w}\propto r^2,\qquad k\propto r^2,\qquad m\propto r^3. \] The undamped angular frequency consequently satisfies \[ \Omega^2=\frac{k}{m}\propto r^{-1}. \]

Write the viscous force as \[ F_{\mathrm{visc}}=-b\dot z. \] Since it is proportional to viscosity and velocity, dimensional analysis for geometrically similar bodies gives \[ b=C\eta r\propto r, \] where \(C\) is a dimensionless geometric factor.

The equation of motion is \[ m\ddot z+b\dot z+kz=0. \] Define the damping rate \[ \beta=\frac{b}{2m}\propto r^{-2}. \] In the underdamped regime, the oscillation frequency is \[ \omega_{\mathrm d}=\sqrt{\Omega^2-\beta^2}. \]

At radius \(r_0\), the given frequencies imply \[ (0.99\,\omega_0)^2=\omega_0^2-\beta_0^2, \] so \[ \frac{\beta_0^2}{\omega_0^2} =1-0.99^2=0.0199. \]

Combining the radius scalings, \[ \frac{\beta^2(r)}{\Omega^2(r)} =0.0199\left(\frac{r_0}{r}\right)^3. \] Oscillations occur when this ratio is less than one. The critical radius therefore satisfies \[ 0.0199\left(\frac{r_0}{r_{\min}}\right)^3=1, \] which yields \[ \boxed{r_{\min}=(0.0199)^{1/3}r_0 =0.27099\ldots\,r_0\approx0.27r_0.} \]

At equality the motion is critically damped and does not oscillate. Thus \(r_{\min}\) is the limiting minimum radius, with oscillatory motion occurring for \(r>r_{\min}\).

The solution was checked against the official solution.