## Essential University Physics: Volume 1 (3rd Edition)

In order to solve this problem, we first need to find the torque: $\tau = Fr =75 \times .45 = 33.75 \ Nm$ We now find the moment of inertia: $I = \frac{1}{2} mr^2 = \frac{1}{2} (120)(.45)^2 =12.15 \ kgm^2$ Thus, we find the final angular velocity: $\omega_f =\sqrt{\frac{2\tau\theta}{I}}=\sqrt{\frac{2(33.75 \ Nm)(.785 \ rads)}{12.15 \ kgm^2}}=\fbox{2.088 radians per second}$