Understanding Centripetal Forces on a Rotating Bead Hoop

[Loro]

We have a bead confined on a circular hoop. The hoop is rotating around an axis tangential to it. Suppose the bead is intially at the point, farthest away from the axis, and has got some intial velocity.

I have a question - in the frame of the hoop, there is a Coriolis force perpendicular to the plane of the hoop, and a balancing reaction force. Then there is a centrifugal force, which has got a component normal to the hoop.

How big is the force from the hoop, balancing this one? Is it of the same magnitude as this normal component, or is it bigger so that it provides a net centripetal force associated with the motion of the bead around the hoop?

 

[tiny-tim]

Hi Loro!

We have a bead confined on a circular hoop. The hoop is rotating around an axis tangential to it. Suppose the bead is intially at the point, farthest away from the axis, and has got some intial velocity.

If I'm understanding it correctly, the initial relative velocity (which is tangential) is parallel to the axis of rotation, so the initial Coriolis force is zero.

 

[Loro]

Yes, that's what I mean. Sorry that I didn't make a picture.

 

[tiny-tim]

I'm not sure what you're asking.

In the frame of the hoop, there's a centrifugal force away from the axis, there's a Coriolis force "vertically" out of the plane of the hoop, there's a centripetal acceleration towards the centre of the hoop, and there's a tangential acceleration …

when you put them all together, what equations did you get?

 

[Loro]

So if I were to compute the reaction forces of the hoop, there would be:

- one "vertically" out - balancing the Coriolis force
- and one towards the centre of the hoop

But would the latter be equal in magnitude to the component of the centrifugal force, normal to the loop? Or would it be that, + the centripetal force?

 

[tiny-tim]

So if I were to compute the reaction forces of the hoop, there would be:

- one "vertically" out - balancing the Coriolis force
- and one towards the centre of the hoop

Yes, the reaction force (for a frictionless hoop) must be perpendicular to the tangent, so it will have a "vertical" component and a radial component.

But would the latter be equal in magnitude to the component of the centrifugal force, normal to the loop? Or would it be that, + the centripetal force?

F_total = ma …

F_total is the centrifugal force plus the reaction force

a is the centripetal acceleration plus the tangential acceleration

So, in the radial direction, the component of the centrifugal force plus the component of the reaction force must equal the centripetal acceleration times the mass.

 

[Loro]

Thanks, that answers my question! :)