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Parallel-axis theorem checked on two equal point masses on the third axis

Example

Let $m > 0$ , $r > 0$ , and let

$ν : = m δ_{r e_{3}} + m δ_{- r e_{3}}$

be the weighted sum of the two Dirac measures at $\pm r e_{3}$ (The Dirac set function at a point, Nonnegative scalar multiples and countable weighted sums of measures are measures), a bounded finite Borel measure and hence a mass distribution of total mass $2 m$ (Mass distributions on $R^{3}$ : total mass, first moment and inertia matrix). Assume the Axiom of Countable Choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), used only to make the displayed integrals finite (Finiteness and a uniform bound for the entries of the inertia tensor about an arbitrary point, Finiteness of the moments of a bounded finite Borel measure on $R^{3}$ ). Then:

$ν$ is centred, $μ (ν) = m r e_{3} + m (- r e_{3}) = 0$ , and its inertia matrix about the origin is $I (ν) = d i a g (2 m r^{2}, 2 m r^{2}, 0)$ ;
the inertia tensor about $a = c e_{3}$ , computed directly from the definition, is $I_{ν} (c e_{3}) = d i a g (2 m (r^{2} + c^{2}), 2 m (r^{2} + c^{2}), 0)$ , and the parallel-axis identity of Parallel-axis theorem: the inertia tensor about a displaced point of a centred mass distribution, $I_{ν} (a) = I (ν) + 2 m (∥ a ∥_{2}^{2} I_{3} - a a^{T})$ , reproduces exactly this value.

All entries are exact polynomial expressions in $m, r, c$ ; no measurement is involved.

Facts & Assumptions

Given: The parameters $m > 0$ , $r > 0$ , a real $c$ , the measure $ν = m δ_{r e_{3}} + m δ_{- r e_{3}}$ , and the notation of Mass distributions on $R^{3}$ : total mass, first moment and inertia matrix and The inertia tensor of a finite bounded Borel measure about any point of $R^{3}$ .

[L1]

The measure and its integrals (The Dirac set function at a point, A Dirac set function is a probability measure, Nonnegative scalar multiples and countable weighted sums of measures are measures, Nonnegative scalar multiples and countable weighted sums of measures): $ν (E) = m δ_{r e_{3}} (E) + m δ_{- r e_{3}} (E)$ for every Borel set $E$ , so $ν$ is a measure with $ν (R^{3}) = 2 m$ and $ν ({r e_{3}}) = ν ({- r e_{3}}) = m$ ; for every real measurable $f$ bounded on the two-point support one has $\int f d ν = m f (r e_{3}) + m f (- r e_{3})$ (The nonnegative Lebesgue integral, The integral of a nonnegative simple function, Integral over a measurable subset, Integrable real and complex functions, and their integrals, Monotonicity and nonnegative homogeneity of the nonnegative integral, The Lebesgue integral is linear on $L^{1} (μ)$ ).

[L2]

The two inertia objects (Mass distributions on $R^{3}$ : total mass, first moment and inertia matrix, The inertia tensor of a finite bounded Borel measure about any point of $R^{3}$ ): $I (ν)_{j k} = \int (∥ ρ ∥_{2}^{2} δ_{j k} - ρ_{j} ρ_{k}) d ν$ and $I_{ν} (a)_{j k} = \int (∥ ρ - a ∥_{2}^{2} δ_{j k} - (ρ_{j} - a_{j}) (ρ_{k} - a_{k})) d ν$ ; all entries are finite real numbers under ${A C}_{ω}$ (Finiteness and a uniform bound for the entries of the inertia tensor about an arbitrary point, Finiteness of the moments of a bounded finite Borel measure on $R^{3}$ ).

[L3]

The parallel-axis theorem (Parallel-axis theorem: the inertia tensor about a displaced point of a centred mass distribution): for a centred mass distribution of total mass $m_{t o t}$ one has $I_{ν} (a) = I (ν) + m_{t o t} (∥ a ∥_{2}^{2} I_{3} - a a^{T})$ .

[L4]

Matrix data (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , The vector space $M_{m \times n} (F) : = F^{m \times n}$ of $m$ by $n$ matrices over a field, with entrywise operations, Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes, Laws of finite sums and finite products): for $ρ = \pm r e_{3}$ one has $ρ ρ^{T} = r^{2} e_{3} e_{3}^{T}$ and $∥ ρ ∥_{2}^{2} = r^{2}$ ; for $a = c e_{3}$ and $ρ = r e_{3}$ one has $ρ - a = (r - c) e_{3}$ while for $ρ = - r e_{3}$ one has $ρ - a = - (r + c) e_{3}$ ; and $I_{3} - e_{3} e_{3}^{T} = d i a g (1, 1, 0)$ .

Verification

Proof technique: direct; evaluate the two atomic integrals and compare the two sides of the parallel-axis identity.

1.1givenL1L2algebra

Centredness and total mass: by the atomic formula of [L1], $μ (ν) = m \cdot r e_{3} + m \cdot (- r e_{3}) = 0$ and $ν (R^{3}) = 2 m$ , so $ν$ is a centred mass distribution of total mass $2 m$ and [L3] applies.

2.1step 1.1L1L2L4algebra

Inertia about the origin: by [L1], [L2] and [L4], $I (ν) = m (r^{2} I_{3} - r^{2} e_{3} e_{3}^{T}) + m (r^{2} I_{3} - r^{2} e_{3} e_{3}^{T}) = 2 m r^{2} (I_{3} - e_{3} e_{3}^{T}) = d i a g (2 m r^{2}, 2 m r^{2}, 0) .$

2.2step 1.1L1L2L4algebra

Inertia about $c e_{3}$ , computed directly: for $ρ = r e_{3}$ the shifted integrand is $(r - c)^{2} (I_{3} - e_{3} e_{3}^{T})$ and for $ρ = - r e_{3}$ it is $(r + c)^{2} (I_{3} - e_{3} e_{3}^{T})$ , so by [L1], [L2] and [L4], $I_{ν} (c e_{3}) = m ((r - c)^{2} + (r + c)^{2}) (I_{3} - e_{3} e_{3}^{T}) = 2 m (r^{2} + c^{2}) (I_{3} - e_{3} e_{3}^{T}) = d i a g (2 m (r^{2} + c^{2}), 2 m (r^{2} + c^{2}), 0) .$

3.1step 1.1step 2.1step 2.2L3L4algebra∎

The theorem side: by [L3] with $m_{t o t} = 2 m$ and $a = c e_{3}$ , $I_{ν} (c e_{3}) = I (ν) + 2 m (c^{2} I_{3} - c^{2} e_{3} e_{3}^{T})$ , which by step 2.1 and [L4] equals $2 m (r^{2} + c^{2}) (I_{3} - e_{3} e_{3}^{T})$ , in exact agreement with the direct computation of step 2.2: the parallel-axis identity holds at this data.

Parallel-axis theorem checked on two equal point masses on the third axis

Example

Facts & Assumptions

Verification

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