Force diagram for ring/bead problem

[PhizKid]

Homework Statement

A ring of mass M hangs from a thread, and two beads of mass m slide on it without friction. The beads are released simultaneously from the top of the ring and slide down opposite sides. Show that the ring will start to rise if m > 3M/2, and find the angle at which this occurs.

Homework Equations

F = ma
Kinetic/potential energy

The Attempt at a Solution

The ring is supposed to rise when the Normal force overcomes the gravitational force on the ring, but how is the Normal force supposed to point directly upward? There is no component of the Normal force that goes up.

 

[lewando]

I assume the beads are constrained to travel along the ring and either miss each other or collide in a perfectly elastic manner at the bottom.

 

[PhizKid]

Yes, but for our problem we don't care what happens after the ring rises (assuming the ring rises before the beads collide, in which case here the ring does indeed rise before the rings reach the bottom).

 

[lewando]

I cannot visualize how the ring can rise when the beads are falling. I can, when the beads are rising, though.

 

[PhizKid]

Right, I thought it was unintuitive as well. But apparently the mathematics of the physics work out, which is what I am trying to do with the force diagram, but I can't seem to get it

 

[voko]

The kinetic energy of the beads depends on their angular position. Find it, and find the corresponding velocity.

Then, since the motion if circular you should be able to find out the centripetal acceleration of the beads at any given angle. From that, you should be able to figure out the forces the beads are exerting on the hoop; then find their vertical components.

 

[PhizKid]

I've already found the KE/velocity: KE = mg*(R - Rcos(x)) and I can get the velocity from there. And I know the net force on the bead is N - mgcos(x) which is = m(v^2/R). But I don't know how to make any force diagram that will make the ring go vertically upwards.

 

[voko]

You have two forces exerted by a bead: one related to its curved motion, which acts radially outward; another is due to its weight, acting vertically down. You can find the vertical component of the radial force; the horizontal components of the two beads' radial forces will cancel each other.

 

[PhizKid]

This is what I don't understand. If I decompose the Normal force and the gravitational force based on the axes I set up in my picture, there is a net force pointing southeast. So what cancels out here?

 

[tiny-tim]

Hi PhizKid!

The ring is supposed to rise when the Normal force overcomes the gravitational force on the ring, but how is the Normal force supposed to point directly upward? There is no component of the Normal force that goes up.

I don't understand.

The normal force on/from the beads is at an angle, and always has a component in the vertical direction (except at the ±90° positions).

(but you don't need the forces …

how can the center of mass accelerate faster than g? )

 

[PhizKid]

But then you would need to make a new set of axes:

But I don't think this would be of any help?

 

[tiny-tim]

But then you would need to make a new set of axes …

what's new about vertical and horizontal?

even the ancient greeks knew about them!

get on with it! ​

 

[PhizKid]

I don't know how to solve this without using forces. But the center of mass isn't moving but it would move vertically upwards if the Normal force pointing upwards was greater than the force of gravity

 

[voko]

You have a radial force, at some angle from the vertical. You can surely project the force to the vertical and horizontal axes.

 

[PhizKid]

The radial force is the same thing as the normal force, or is it something different?

 

[voko]

Not exactly. The centripetal acceleration is due to two forces: the reaction force (normal) and gravity. The radial component of this combined force is what I call the radial force (even though its direction is the same as that of the normal force).

 

[PhizKid]

The only force to decompose is the gravitational force, in which one component is parallel to the normal force radially, and the other component is tangential to the ring. I don't understand where this is going

 

[voko]

Again, get the radial force first. That is easy to do, because all you need is the velocity and the radius, which you have. Then get the vertical component of this force, and subtract the force of gravity from it; this gives you the vertical component of the normal force, but mind the signs.

 

[TSny]

There is no component of the Normal force that goes up.

I think you're trying to see conceptually how either bead can exert a normal force on the wire that has a vertical component that is upward. Near the top of the loop, the wire exerts a normal force on the bead that is radially outward, as you have drawn in your first drawing. So, the bead exerts a normal force on the wire that is radially inward (and therefore has a vertical component downward.) But think about what happens as the bead continues farther along the loop. You might be familiar with the problem of something sliding down a frictionless spherical surface (see attachment). At some point the object leaves the surface! What prevents the bead from leaving the wire loop?

 

[PhizKid]

Again, get the radial force first. That is easy to do, because all you need is the velocity and the radius, which you have. Then get the vertical component of this force, and subtract the force of gravity from it; this gives you the vertical component of the normal force, but mind the signs.

I'm not sure what "radial force" means. I thought the radial force was N - mgcos(x) = m(v^2/R)? What other force is there?

I think you're trying to see conceptually how either bead can exert a normal force on the wire that has a vertical component that is upward. Near the top of the loop, the wire exerts a normal force on the bead that is radially outward, as you have drawn in your first drawing. So, the bead exerts a normal force on the wire that is radially inward (and therefore has a vertical component downward.) But think about what happens as the bead continues farther along the loop. You might be familiar with the problem of something sliding down a frictionless spherical surface (see attachment). At some point the object leaves the surface! What prevents the bead from leaving the wire loop?

The object leaves the surface because the Normal force becomes 0 since it loses contact. In this problem, the bead can't because it's attached to the ring.

So do I have to think about the forces on the beads and the forces on the ring separately?

 

[voko]

I'm not sure what "radial force" means. I thought the radial force was N - mgcos(x) = m(v^2/R)? What other force is there?

Again. The net force on the bead is the normal force plus its weight. This net force can be decomposed into two components: one radial and another tangential to the ring. The radial component of the net force (radial force for short) is responsible for the centripetal acceleration; this is the equation you have above. The other force is the tangential force, which you don't care about, because you can obtain the normal force from the equation for the radial force alone.

 

[lewando]

Thanks, TSny--I see what you mean. Facepalm.

 

[PhizKid]

Oh, so once I have the net radial force, I decompose that into its horizontal/vertical components? So it's like decomposing within decomposing or something similar?

 

[voko]

Oh, so once I have the net radial force, I decompose that into its horizontal/vertical components? So it's like decomposing within decomposing or something similar?

Correct. Note the two beads will product twice the vertical component, while their horizontal components cancel - so all that remains is pure vertical force.

 

[PhizKid]

But this vertical force is on the beads from the ring, right? So it's pointing in the same direction as the gravitational force is initially pointing. Where are we going with this?

 

[voko]

The radial force is due to the normal force AND the radial component of the weight. Given the radial force, you can obtain just the normal force (that's equation you got earlier). Find the vertical component of this normal force - that's what lifts the ring up.