Helix Moment

Classical Mechanics Level 3

x = cos ⁡ θ , y = sin ⁡ θ , z = θ , 0 ≤ θ ≤ 2 π x = \cos \theta , \qquad y = \sin \theta, \qquad z = \theta, \qquad 0 \leq \theta \leq 2 \pi

Consider a helical section, with parameters given above. The mass of the helical section is M M . The moment of inertia about the z z -axis can be expressed as α M \alpha M . Determine the value of α \alpha .

Details and Assumptions:

  • The mass is distributed uniformly as a function of arc length.
  • Neglect units.


The answer is 1.

Steven Chase by Steven Chase , 37, USA

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1 solution

Chew-Seong Cheong
Aug 23, 2016

As far as z z -axis is concerned, the mass of the helix is uniformly distributed at radius R = 1 R=1 away. The moment of inertia is therefore that of a thin ring with radius R = 1 R=1 , which is given by I = M R 2 = M I = MR^2 = M . Therefore, α = 1 \alpha = \boxed{1} .

Similar result if we used calculus.

I = ∫ 0 2 π σ R ⋅ R 2 d θ where σ is mass per unit length of the helix. = σ R 3 θ ∣ 0 2 π = 2 π σ R ⋅ R 2 Note that 2 π σ R = M = M R 2 Note that R = 1 = M \begin{aligned} I & = \int_0^{2 \pi} \color{#3D99F6}{\sigma} R\cdot R^2 \ d\theta & \small \color{#3D99F6}{\text{where }\sigma \text{ is mass per unit length of the helix.}} \\ & = \sigma R^3\theta \bigg|_0^{2 \pi} \\ & = \color{#3D99F6}{2 \pi \sigma R}\cdot R^2 & \small \color{#3D99F6}{\text{Note that }2\pi \sigma R = M} \\ & = \color{#3D99F6}{M}\color{#D61F06}{R^2} & \small \color{#D61F06}{\text{Note that }R = 1} \\ & = M \end{aligned}

⟹ α = 1 \implies \alpha = \boxed{1} .

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