Moments of Inertia and Bernoulli's Lemniscate

Key Concepts

Moment of Inertia (MI)

• MI about x-axis: ( I_x = \rho \int y^2 , dx , dy )
• MI about y-axis: ( I_y = \rho \int x^2 , dx , dy )
• Choosing the axis defines which variable gets squared in the integration.

Lemniscate of Bernoulli

• Equation: [ r^2 = a^2 \cos 2\theta ]
• This curve resembles the infinity symbol ((\infty)).

Problem: Find MI of One Loop of the Lemniscate About the Initial Line

1. Given Lemniscate Equation: ( r^2 = a^2 \cos 2\theta )
2. Coordinate Transformation:
   • ( x = r \cos \theta )
   • ( y = r \sin \theta )
   • ( dx , dy = r , dr , d\theta )
3. MI (about x-axis): ( I_x = \int \rho , y^2 , dx , dy )
4. Transform Polar to Cartesian Coordinates in MI:
   • ( \rho r^2 \sin^2 \theta , r , dr , d\theta )
   • Limits for ( r ): ( 0 ) to ( a \sqrt{\cos 2 \theta} )
   • Limits for ( \theta ): ( 0 ) to ( \pi/4 )
   • Multiply by 2 because of symmetry.

Integration Steps

• Integrand: ( 2 \rho \int_{0}^{\pi/4} \int_{0}^{a \sqrt{\cos 2 \theta}} r^3 \sin^2 \theta , dr , d\theta )
• Integration w.r.t ( r ):
  • ( \int_{0}^{a \sqrt{\cos 2 \theta}} r^3 , dr = \left. \frac{r^4}{4} \right|_{0}^{a \sqrt{\cos 2 \theta}} )
  • Result: ( \frac{\rho}{2} \int_{0}^{\pi/4} a^4 \cos^2 2 \theta \sin^2 \theta , d\theta )
• Simplification:
  • Use identity ( \sin^2 \theta = 1 - \cos^2 \theta )
  • Substitution: ( u = 2\theta ), leading to adjusted limits.
• Final Integral Calculation:
  • ( \frac{\rho a^4}{8} \int_{0}^{\pi/2} (1 - \cos u) \cos^2 u , du )
  • Apply reduction formula/evaluation techniques.

Convert MI in Terms of Mass (M)

• Given ( M = \int \rho , dA = \int \rho r , dr , d\theta )

• Evaluate ( M ) and find ( \rho ) in terms of ( M ).

• Substitute ( \rho ) back in MI expression.

• Final Result:

  • ( I = \frac{M a^2 (3\pi - 8)}{48} )