# Bead on a hoop, angular momentum

1. Aug 10, 2011

### CyberShot

In the above problem, the hoop will gain angular momentum, L, in the +y direction. However, initially the bead is on the bottom of the circle where the slope is exactly zero. The angular momentum will therefore point exactly upward (thus the net force acting on the bead is only in the y direction) , and the bead should not move in the x direction, not even the y direction since it's confined to the hoop.

But how is it that the bead angle increases (i.e. the bead slides up the hoop) when the hoop starts to spin, even though L is always in the +y direction and the instantaneous slope is zero initially?

If the bead were given a slight nudge in the +x direction so it's not longer exactly at the bottom, and thus is not on a surface with slope zero, then the bead should start to slide up.

Last edited by a moderator: May 5, 2017
2. Aug 11, 2011

### Pythagorean

this is not quite correct. While the vector of angular momentum is defined as up here via the right hand rule, the force the bead feels will be radially outward, away from the axis of rotation. No matter how hard you spin the hoop, the bead shouldn't go past the horizontal position at the extrema of +/- x.

The bead only moves along y because it is coupled to y through the hoop, horizontal force being transferred to vertical force.