A disc of moment of inertia $4 \mathrm{~kg}-\mathrm{m}^2$ revolving with $16 \mathrm{rad} / \mathrm{s}$ is placed on another disc of moment of inertia $8 \mathrm{Kg}-\mathrm{m}^2$ revolving $4 \mathrm{rad} / \mathrm{s}$. The angular frequency of composite disc

The principle of conservation of angular momentum states that the total angular momentum of the system remains constant when no external torque acts on it.

Let $I_1 = 4 \text{ kg-m}^2$ and $\omega_1 = 16 \text{ rad/s}$ be the moment of inertia and angular velocity of the first disc.

Let $I_2 = 8 \text{ kg-m}^2$ and $\omega_2 = 4 \text{ rad/s}$ be the moment of inertia and angular velocity of the second disc.

The initial total angular momentum $L_i$ is given by $L_i = I_1 \omega_1 + I_2 \omega_2$.

$L_i = (4 \times 16) + (8 \times 4) = 64 + 32 = 96 \text{ kg-m}^2/\text{s}$.

When the two discs are placed together, they form a composite disc with a total moment of inertia $I = I_1 + I_2 = 4 + 8 = 12 \text{ kg-m}^2$.

Let $\omega$ be the final angular velocity of the composite disc. The final angular momentum $L_f$ is $L_f = (I_1 + I_2) \omega = 12 \omega$.

Equating $L_i = L_f$, we get $96 = 12 \omega$.

Solving for $\omega$, we have $\omega = \dfrac{96}{12} = 8 \text{ rad/s}$.

Answer: $8 \mathrm{rad} / \mathrm{s}$