Multi-Dimensional Body Mass Scaling (Revised)

Classical Mechanics Level 3

An exotic creature has a body plan consisting of three parts:

  1. A 3-dimensional cylindrical torso of radius R R and height H H
  2. A 1-dimensional (line segment) neck of length L L
  3. A 2-dimensional head consisting of a circular disk of radius r . r.

Each individual part has a uniform mass density as a function of length (1D), area (2D), or volume (3D). Originally, each body part has mass M M , so the total mass is 3 M 3M .

The creature then grows in such a way that it becomes a proportionally scaled-up version of its old self. During this growth process, the mass densities of its body parts remain constant: R ′ = α R H ′ = α H L ′ = α L r ′ = α r . \large{R' = \alpha \, R \\ H' = \alpha \, H \\ L' = \alpha \, L \\ r' = \alpha \, r}. For what value of α \alpha will the creature's total mass be twice its starting value?

Give your answer as ⌊ 1000 α ⌋ \lfloor 1000 \alpha \rfloor .

Note: ⌊ ⋅ ⌋ \lfloor \cdot \rfloor denotes the floor function .


The answer is 1389.

Steven Chase by Steven Chase , 37, USA

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1 solution

Brian Moehring
Jul 20, 2018

Since the mass densities are constant, the new mass of the 3D piece will be α 3 M \alpha^3 M , the new mass of the 2D piece will be α 2 M \alpha^2 M and the new mass of the 1D piece will be α M \alpha M .

For the creature's total mass to be twice its starting value means α 3 M + α 2 M + α M = 2 ( 3 M ) ⟹ α 3 + α 2 + α = 6 \alpha^3 M + \alpha^2 M + \alpha M = 2(3M) \implies \alpha^3 + \alpha^2 + \alpha = 6 Solving this numerically gives α ≈ 1.38919 ⟹ ⌊ 1000 α ⌋ = 1389 \alpha \approx 1.38919 \implies \lfloor 1000\alpha\rfloor = \boxed{1389}

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