According to the work-energy theorem, the work done on the flywheel is equal to the change in its rotational kinetic energy. $W = \Delta K = \frac{1}{2} I \omega_f^2 - \frac{1}{2} I \omega_i^2 = \frac{1}{2} I (\omega_f^2 - \omega_i^2)$ Given: Work done, $W = 484 \text{ J}$ Initial angular speed, $\omega_i = 60 \text{ rpm} = \frac{60 \times 2\pi}{60} \text{ rad/s} = 2\pi \text{ rad/s}$ Final angular speed, $\omega_f = 360 \text{ rpm} = \frac{360 \times 2\pi}{60} \text{ rad/s} = 12\pi \text{ rad/s}$ Substitute the values in the equation: $484 = \frac{1}{2} I ((12\pi)^2 - (2\pi)^2)$ $484 = \frac{1}{2} I (144\pi^2 - 4\pi^2)$ $484 = \frac{1}{2} I (140\pi^2) = 70 I \pi^2$ Taking $\pi^2 \approx \left(\frac{22}{7}\right)^2 = \frac{484}{49}$: $484 = 70 \times I \times \frac{484}{49}$ $1 = I \times \frac{70}{49} = I \times \frac{10}{7}$ $I = \frac{7}{10} = 0.7 \text{ kg-m}^2$