Calculating the Rotational Inertia — Question 9

Question 9

Two allowed radii turn inertia into a mass-allocation constraint

A rotor has total adjustable mass $4\,\mathrm{kg}$ carried only on two thin concentric rings of radii $0.20$ and $0.50\,\mathrm m$. The supporting structure has negligible mass. Required inertia is $0.58\,\mathrm{kg\,m^2}$.

1. Find the mass assigned to each ring.

2. Give the full attainable inertia interval and test a requested value of $1.1\,\mathrm{kg\,m^2}$.

Question 9 - Solution

Two allowed radii turn inertia into a mass-allocation constraint

(a) Derivation and calculation

Let $m_o$ be the outer-ring mass and $4-m_o$ the inner-ring mass: $$I=(4-m_o)(0.20)^2+m_o(0.50)^2=0.16+0.21m_o.$$ $$0.58=0.16+0.21m_o\Rightarrow \boxed{m_o=2.00\ \mathrm{kg},\qquad m_i=2.00\ \mathrm{kg}}.$$

(b) Result and interpretation

Nonnegative allocations require $0\le m_o\le4$, giving $$\boxed{0.16\le I\le1.00\ \mathrm{kg\,m^2}}.$$ A target of 1.1 would require $m_o=(1.10-0.16)/0.21=4.476$ kg and negative inner mass. It is unattainable under the stated constraints.

Physical check

The inertia depends linearly on outer mass, so the extrema occur when all mass occupies one ring.