A wheel is rotating freely at some angular speed. A second wheel initially at rest and with thrice the rotational inertia of the first, is suddenly coupled to the first wheel. The fraction of the original rotational kinetic energy lost is

Let, $I_1=$ Inertia of first (rotating) wheel $\omega_1=$ Angular speed of first wheel $I_2=3 I_1=$ Inertia of second wheel $\omega_2=0=$ Angular speed of second wheel let $\omega=$ Angular speed of coupled wheels. let $\omega=$ Angular speed of coupled wheels. According to conservation of angular momentum, total initial angular momentum = total final angular momentum $\begin{aligned} & \Rightarrow I_1 \omega_1+I_2 \times 0=\left(I_1+I_2\right) \omega \\ & \Rightarrow \omega=\frac{I_1 \omega_1}{I_1+I_2}=\frac{I_1 \omega_1}{I_1+3 I_1} \Rightarrow \omega=\frac{\omega_1}{4}\end{aligned}$ Now, initial rotational energy of system is $K_1=\frac{1}{2} I_1 \omega_1^2+\frac{1}{2} I_2(0)^2$ $\Rightarrow \quad K_1=\frac{1}{2} I_1 \omega_1^2$ Final rotational kinetic energy of system is $K_2=\frac{1}{2}\left(I_1+I_2\right) \omega^2$ $=\frac{1}{2}\left(I_1+3 I_1\right) \cdot \frac{\omega_1^2}{16}$ $\begin{aligned} & =\frac{1}{2} \times 4 I_1 \times \frac{\omega_1^2}{16} \\ & =\frac{1}{4}\left(\frac{1}{2} I_1 \omega_1^2\right)\end{aligned}$ or $\quad K_2=\frac{1}{4} K_1$ Hence, fraction of lost energy $\begin{aligned} & =\frac{\Delta K}{K_1}=\frac{K_1-K_2}{K_1}=\frac{\frac{3}{4} K_1}{K_1} \\ & =\frac{3}{4}=0.75\end{aligned}$