Question

A smooth rod of length L rotates in a horizontal plane with a constant angular velocity ? about a vertical axis fixed at one end. A bead of mass m is released from rest relative to the rod at its midpoint. Viewed from an inertial frame and using plane polar coordinates derive an expression for the radial displacement of the bead versus time. Do Not Use fictious forces.

Answer 1

A smooth rod of length l rotates in a plane with a constant angular velocity ? about an axis fixed at the end of the rod and perpendicular to the plane of rotation. A bead of mass m can slide along the rod without friction is initially positioned at the stationary end of the rod and given a slight push such that its initial speed directed down the rod is ? = ?l (draw figure). Calculate how long it takes for the bead to reach the other end of the rod.

Choose a coordinate system that rotates with the rod. Let xhat be along the rod and yhat be perpendicular to the rod. Since the bead is constrained to the rod, the problem becomes one dimensional. The only regular force is the one constraining the bead to the rod, and since this force is perpendicular to the rod, it points in the y direction. There is no regular foce in the x direction.

We need to find the fictitious forces on the bead. To do this, assume the bead is a distance x from the origin, that is, it is located at position (x, 0, 0) in the rotating frame, and assume it has velocity (xdot, 0, 0). (The y and z components of the velocity must be zero in this frame, because the bead can only move along the rod, in x.) Using ? = ?khat = (0, 0, ?), the fictitious forces are

F_cor = 2mv