A disk with a rotational inertia of 5.0 kg .m² and a radius of 0.25 m rotates on a fixed axis perpendicular to the disk and through its center. A force of 2.0 N is applied tangentially to the rim. As the disk turns through half a revolution the work done by the force is: A) 1.6 J B) 2.5 J C) 6.3 J D) 10 J E) 40 J

Question

Quizplus · Accepted Answer

The work done by the force is equal to the change in kinetic energy of the disk. The initial kinetic energy of the disk is zero, and the final kinetic energy is:$$K_f = \frac{1}{2}I\omega^2 = \frac{1}{2}(5.0 	ext{ kg m}^2)(2\pi 	ext{ rad/s})^2 = 6.3 	ext{ J}$$Therefore, the work done by the force is 6.3 J.

A Disk with a Rotational Inertia of 5