Solution
= Solution
The initial relation implies $L^2=2I_2T$, and conservation preserves it. With $I_3=I_1+I_2$, this identity reduces to
$$
\omega_3^2=\frac{I_2-I_1}{I_1+I_2}\omega_1^2.
$$
The energy then gives
$$
\omega_2^2=\frac{2T}{I_2}-\omega_1^2.
$$
The first Euler equation is $\dot\omega_1=-\omega_2\omega_3$. Squaring and substituting the preceding identities yields
$$
\boxed{\dot\omega_1^2=\left(\frac{2T}{I_2}-\omega_1^2\right)
\frac{I_2-I_1}{I_2+I_1}\omega_1^2.}
$$