What is the definition of the moment of inertia tensor used in the proof?

The definition of the moment of inertia tensor IijI_{ij} used in the proof, is: Iij=Vρ(xkxkδijxixj)dVI_{ij} = \int_V \rho (x_k x_k \delta_{ij} - x_i x_j) dV

This integral formula defines the components of the moment of inertia tensor for a rigid body rotating about an axis passing through the origin, where:

  • IijI_{ij} are the components of the moment of inertia tensor (a 3×33 \times 3 matrix).

  • V\int_V denotes the integral over the volume of the body.

  • ρ\rho is the mass density of the body (mass per unit volume).

  • xix_i are the Cartesian coordinates of a mass element within the body.

  • xkxk=k=13xk2x_k x_k = \sum_{k=1}^3 x_k^2 is the square of the distance from the origin, r2r^2.

  • δij\delta_{ij} is the Kronecker delta, which is 1 if i=ji=j and 0 if iji \neq j.

  • dVdV is the volume element.

In component form:

  • The diagonal components (i=ji=j) represent the moment of inertia about the ii-axis:

    Iii=Vρ(xkxkδiixixi)dV=Vρ(r2xi2)dVI_{ii} = \int_V \rho (x_k x_k \delta_{ii} - x_i x_i) dV = \int_V \rho (r^2 - x_i^2) dV

  • The off-diagonal components (iji \neq j) are the products of inertia:

    Iij=Vρ(xkxk0xixj)dV=VρxixjdVI_{ij} = \int_V \rho (x_k x_k \cdot 0 - x_i x_j) dV = - \int_V \rho x_i x_j dV

Brief audio

Last updated

Was this helpful?