Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

We first find the moment of inertia of the flywheel. $I = \frac{1}{2}MR^2$ $I = \frac{1}{2}(250~kg)(0.75~m)^2$ $I = 70.3~kg~m^2$ We then find the angular acceleration of the flywheel. $\tau = I~\alpha$ $\alpha = \frac{\tau}{I}$ $\alpha = \frac{50~N~m}{70.3~kg~m^2}$ $\alpha = 0.711~rad/s^2$ Next, we convert the final angular velocity to units of rad/s $\omega_f = (1200~rpm)(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s})$ $\omega_f = 126~rad/s$ We can find the time it takes to reach this final angular velocity. $t = \frac{\omega_f-\omega_0}{\alpha}$ $t = \frac{126~rad/s-0}{0.711~rad/s^2}$ $t = 177~s$ It takes the flywheel 177 seconds to reach the top angular speed of 1200 rpm.