Two discs of moment of inertia I _ 1 = 4kg m ^ 2 and I _ 2 = 2kg m...

Applying conservation of angular momentum, we get

${I}_{1}{\omega}_{1} + {I}_{2}{\omega}_{2} = \left({I}_{1} + {I}_{2}\right){\omega}_{0} ⇒ 4 \times 10 + 2 \times 4 = \left(4 + 2\right){\omega}_{0} ⇒ {\omega}_{0} = 8rad{s}^{- 1}$

Now, initial rotational kinetic energy is ${E}_{1} = \dfrac{1}{2}{I}_{1}{\omega}_{1}^{2} + \dfrac{1}{2}{I}_{2}{\omega}_{2}^{2} = 216J$

and final rotational kinetic energy is

${E}_{2} = \dfrac{1}{2}\left({I}_{1} + {I}_{2}\right){\omega}_{0}^{2} = 192J$

Required, energy loss $\DeltaE = {E}_{1} - {E}_{2} = 24J$