# A.5 Mathematical analysis for a rotating rigid body such as rim type flywheels

In contradiction to the analysis given in Fig 17B & 19 and Appendix 2, this analysis simply considers the Earth as an eccentrically rotating solid body such as an unbalanced flywheel. As such it does not describe the circumferential forces which are thought to be linked to the tectonic forces resulting in plate movements, but does describe the situation that would occur if the lithosphere were treated as a thin shell sphere subjected to an internal pressure with a developed ‘vertical force P’ (Fig C within Fig 19).

**Notations and Value****s**

R = Radius (m); Value: 6.4x10^{6} metres

Δr = Eccentricity (m); Value: 1.0 metres

T = Thickness (m); Value: 1.0 metres

ρ = Density (kgm^{-3}); Value: 2.8 x10^{3} kgm^{-3}

ω = Angular velocity (radsec^{-1}); Value: 7.27x10^{-5} radsec^{-1}

σ = Hoop Stress (Nm^{-2})

Consider a cylinder of mean radius r and thickness t rotating at an angular velocity ω about its axis (Fig.17A):

The mass of the portion Rδθ = ρRδθ.t

The radial force on the element = mass x acceleration = (ρRδθ.t) Rω^{2}

This will produce the Hoop Stress σ

Resolving radially

2σt.sin½δθ= ρR^{2}ω^{2}.tδθ (as sin½ θ → ½θ)

Therefore σ = ρR^{2}ω^{2}

If the centre of rotation is displaced δr from the centre of mass (Fig 9) then the tensile force on the ‘heavier side’ will be increased by the following amount:

Thus, the increase in tensile stress = ρω^{2} ((R+δr)^{2}-R^{2})

= ρω^{2} ((R^{2}+2δr.R+δr^{2})-R^{2})

= ρω^{2} (2δr.R+δr^{2})

Substituting the values stated above:

The additional Tensile Stress = 2.5 x 10^{3} x (7.27x10^{-5 })^{2} x (2x10^{3 }x 6.4 x 10^{6} +10^{6})

= 1.89 x 10^{5} Nm^{-2}

On the opposite side the decrease in the Tensile Stress will be as follows:

Thus the ‘decrease’ in tensile stress = ρω^{2} ((R – δr)^{2}-R^{2})

= ρω^{2} ((R^{2}-2δr.R+δr^{2})-R^{2})

= ρω^{2} (δr^{2 }-2δr.R)

Substituting the numerical values, Tensile Stress will have a negative value

The Tensile Stress is thus = -1.89x10^{5} Nm^{-2}

This Negative Tensile Stress is the Compression Stress = 1.89 x10^{5} Nm^{-2}

As stated above, the rigid body approach while clearly demonstrating the differential stress due to eccentricity is not considered as the model for tectonic movement. The model for tectonic movement as defined in Section 7 is based on having relative movement between the outer rim or crust and the main body or mantle.