AP EAPCET 07-Jul-2022 Shift 1 Paper

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. According to conservation of angular momentum, total initial angular momentum $=$ total final angular momentum $\Rightarrow I_1 \omega_1 + I_2 \times 0 = (I_1+I_2) \omega$ $\Rightarrow \omega = \frac{I_1 \omega_1}{I_1+I_2} = \frac{I_1 \omega_1}{I_1+3 I_1} \Rightarrow \omega = \frac{\omega_1}{4}$ Now, initial rotational energy of system is $K_1 = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 (0)^2$ $\Rightarrow K_1 = \frac{1}{2} I_1 \omega_1^2$ Final rotational kinetic energy of system is $K_2 = \frac{1}{2} (I_1+I_2) \omega^2$ $= \frac{1}{2} (I_1+3 I_1) \cdot \frac{\omega_1^2}{16}$ or $K_2 = \frac{1}{4} K_1$ $= \frac{1}{2} \times 4 I_1 \times \frac{\omega_1^2}{16}$ $= \frac{1}{4} \left( \frac{1}{2} I_1 \omega_1^2 \right)$ Hence, fraction of lost energy $= \frac{\Delta K}{K_1} = \frac{K_1 - K_2}{K_1} = \frac{\frac{3}{4} K_1}{K_1}$ $= \frac{3}{4} = 0.75$