- #1
etotheipi
- Homework Statement
- Show that the moment of inertia tensor satisfies $$\frac{dI_{ij}}{dt} \omega_j = \varepsilon_{ijk} \omega_j I_{kl} \omega_l$$for a body undergoing arbitrary rotation about the origin.
- Relevant Equations
- N/A
We start from the definition$$I_{ij} = \int_V \rho(x_k x_k \delta_{ij} - x_i x_j) dV \iff \dot{I}_{ij} = \int_V \rho (2 x_k \dot{x}_k \delta_{ij} - \dot{x}_i x_j - x_i \dot{x}_j ) dV$$Now since the rigid body rotation satisfies ##\dot{\vec{x}} = \vec{\omega} \times \vec{x} \iff \dot{x}_i = \varepsilon_{ilm} w_l x_m##, and also since ##\vec{\omega}## is the same at all points within ##V## at a given time ##t##, we have$$
\begin{align*}
\dot{I}_{ij} &= \int_V \rho(2\delta_{ij} \varepsilon_{klm} x_k x_m w_l - \varepsilon_{ilm} x_j x_m w_l - \varepsilon_{jlm} x_i x_m w_l) dV \\
\dot{I}_{ij} w_j &= w_l w_j \int_V \rho (2 \delta_{ij} \varepsilon_{klm} x_k x_m - \varepsilon_{ilm} x_j x_m - \varepsilon_{jlm} x_i x_m) dV\end{align*}$$I don't know what to do next! We need to show that the integrand reduces to$$\rho (2 \delta_{ij} \varepsilon_{klm} x_k x_m - \varepsilon_{ilm} x_j x_m - \varepsilon_{jlm} x_i x_m) = \rho \varepsilon_{ijk}(x_m x_m \delta_{kl} - x_k x_l)$$Any help is appreciated, thanks.
\begin{align*}
\dot{I}_{ij} &= \int_V \rho(2\delta_{ij} \varepsilon_{klm} x_k x_m w_l - \varepsilon_{ilm} x_j x_m w_l - \varepsilon_{jlm} x_i x_m w_l) dV \\
\dot{I}_{ij} w_j &= w_l w_j \int_V \rho (2 \delta_{ij} \varepsilon_{klm} x_k x_m - \varepsilon_{ilm} x_j x_m - \varepsilon_{jlm} x_i x_m) dV\end{align*}$$I don't know what to do next! We need to show that the integrand reduces to$$\rho (2 \delta_{ij} \varepsilon_{klm} x_k x_m - \varepsilon_{ilm} x_j x_m - \varepsilon_{jlm} x_i x_m) = \rho \varepsilon_{ijk}(x_m x_m \delta_{kl} - x_k x_l)$$Any help is appreciated, thanks.