The time derivative of the moment of inertia tensor, $\dot{I}{ij}$, is calculated by taking the time derivative and applying the product rule for differentiation. The initial definition is: $I{ij} = \int_V \rho (x_k x_k \delta_{ij} - x_i x_j) dV$
Since the integral is over a rigid body, the volume of integration ($V$) and the density ($\rho$) are constant with respect to time. Therefore, the time derivative $\dot{I}_{ij}$ acts only on the coordinates $x_i$ and $x_j$ (and $x_k$):
$$ \dot{I}{ij} = \frac{d}{dt} \int_V \rho (x_k x_k \delta{ij} - x_i x_j) dV = \int_V \rho \frac{d}{dt} (x_k x_k \delta_{ij} - x_i x_j) dV $$
The derivative of $x_k x_k$ (which is $r^2$, the square of the distance from the origin) is found using the product rule, noting that $\dot{x}_k = \frac{dx_k}{dt}$:
$$ \frac{d}{dt} (x_k x_k) = \left(\frac{d}{dt} x_k\right) x_k + x_k \left(\frac{d}{dt} x_k\right) = \dot{x}_k x_k + x_k \dot{x}_k = 2 x_k \dot{x}_k $$
Since $\delta_{ij}$ is a constant tensor, the derivative of the first term is:
$$ \frac{d}{dt} (x_k x_k \delta_{ij}) = 2 x_k \dot{x}k \delta{ij} $$
The derivative of the product $x_i x_j$ is found using the product rule:
$$ \frac{d}{dt} (x_i x_j) = \left(\frac{d}{dt} x_i\right) x_j + x_i \left(\frac{d}{dt} x_j\right) = \dot{x}_i x_j + x_i \dot{x}_j $$
Substituting these results back into the expression for $\dot{I}_{ij}$ gives the final formula:
$$ \dot{I}_{ij} = \int_V \rho \left(2 x_k \dot{x}k \delta{ij} - (\dot{x}_i x_j + x_i \dot{x}_j)\right) dV $$
This result expresses the time rate of change of the inertia tensor in terms of the position ($x$) and velocity ($\dot{x}$) of every mass element in the rigid body.
How is the time derivative of the moment of inertia tensor calculated-L.mp4