Writeup: 2025 USAPhO A1

23 Apr, 2025

It's been a while since I've seriously thought about physics. I decided to look at this year's USAPhO for fun after being reminded by a junior friend of mine. I was surprised by the volume of mechanics questions on the test -- 4 mechanics questions on one test is very atypical. A1 was one of them.

Problem Statement:

Most hairy mammals shake after getting wet; shaking the water off is energetically advantageous to waiting for it to dry on its own. In this problem, we investigate the mechanics of shaking off water. For simplicity, we will model animals as solid cylinders with the sides covered in fur. You may ignore gravity throughout the problem.

(a) We can model the shaking by assuming that the angular position of the cylinder undergoes a sinusoidal oscillation: $θ = A \cos (ω t)$ . (The rotation is about the axis of rotational symmetry.) Letting the radius of the cylindrical animal be R, derive an expression for the magnitude of the acceleration of a point on the surface of the animal under such a motion.

Solution: In Cartesian coordinates, the position $p$ of a point is given by

p = (R \cos (A \cos (ω t)), R \sin (A \cos (ω t))) .

We differentiate twice w.r.t $t$ to find the acceleration.

\frac{d^{2}}{d t^{2}} p = R (A ω^{2} (\cos (ω t) \sin (A \cos (ω t)) - A \sin^{2} (ω t) \cos (A \cos (ω t))), - A ω^{2} (A \sin^{2} (ω t) \sin (A \cos (ω t)) + \cos (ω t) \cos (A \cos (ω t)))) .

Taking the magnitude of this vector yields the magnitude of the acceleration as

R A ω^{2} \sqrt{A^{2} s i n^{4} (ω t) + c o s^{2} (ω t)} .

(b) For the rest of this problem, rather than modeling sinusoidal motion, we will assume that the cylindrical animal is simply spinning at constant angular speed $ω$ . Wet fur tends to separate into cylindrical clumps of radius $r ≪ R$ . A droplet of water on the end of a clump of fur will be separated from it if the centripetal force overcomes the forces due to surface tension. Derive an approximate relationship (valid up to scalar numerical constants) between the surface tension $σ$ , the radius $r$ of the fur clump, the radius $R$ of the animal, the angular speed $ω$ , and the density $ρ$ of water. The diagram below show the cross-section of the animal, showing four clumps of fur with droplets on their end.

Solution: We know that $σ$ has units $\frac{N}{m}$ and $r$ has units $m$ , so the force due to surface tension must follow $σ r$ .

Similarly we know that the centripetal force must be $R ω^{2} (mass of droplet)$ , and using dimensional analysis, we can conclude that $mass of droplet ~ ρ r^{3}$ , so our relation is

σ r ~ R ω^{2} ρ r^{3} .

(c) Experimentally, it is observed that all mammalian fur forms clumps of similar radius. Under the assumption that different animals all have the same density and are simply scaled copies of each other (i.e. larger animals are both longer and fatter), the relationship between an animal’s mass $M$ and its angular velocity of shaking $ω$ is of the form $ω ~ M^{n}$ : find the value of $n$ .

Solution: Note that $M ~ R^{3}$ , so if we wish to keep $R ω^{2} ρ r^{3} ~ 1$ (i.e. constant) we must have $ω ~ M^{- \frac{1}{6}}$ so $n = - \frac{1}{6}$ .

(d) Shaking requires energy, which we can crudely model as the rotational energy of the corresponding cylinder. An alternative strategy for the animal is to simply air-dry their fur, which requires energy to evaporate the water. Assume a wet animal has approximately $5$ of their body weight in water, and has to supply all the energy for evaporating the water. Our model predicts that for some animal sizes, it will be energetically advantageous to air dry themselves: estimate the range of animal masses for which this is true. You may use the following facts: a mouse weighs $20$ $g$ , has a radius of $1 cm$ , and shakes itself with angular velocity $ω = 30 rad / s$ . The latent heat of vaporization of water at room temperature is $λ = 2430 J / g .$

Solution: Note: I used a calculator. Not sure if I'd be actually allowed to in an exam.

Using ©, we see that the energy of shaking scales at a factor of $M^{\frac{4}{3}}$ while the energy of air drying scales linearly with $M$ .

For a mouse, energy of shaking comes out to

\frac{1}{2} ω^{2} \frac{1}{2} M R^{2} = 0.00045 J

and the energy of evaporating is

0.05 M λ = 2430 J .

For an animal with mass $X$ , we must have

(\frac{X}{M})^{\frac{1}{3}} > \frac{2430}{0.00045} = 5400000

so $\frac{X}{M} > 1.57 \cdot 10^{20}$ which is a clearly a large and unfeasible requirement for any reasonable animal weighing mass $X$ . So no animals satisfy this requirement.