Question

A string is wound around the edge of a solid 1.60 kg disk with a 0.130 m radius. The disk is initially at rest when the string is pulled, applying a force of 8.75 N in the plane of the disk and tangent to its edge. If the force is applied for 1.90 seconds, what is the magnitude of its final angular velocity?

Answer 1

Torque of force = 8.75 Newtons * 0.13 meters
alpha = angular acceleration = Torque / moment of inertia
= 8.75 * 0.13 / [ 1.60 * (1/2)(.13)^2]
omega = angular velocity = alpha * t = alpha * 1.90 seconds