Can Gravity Make a Hoop Rise Off Its Support When Beads Slide Down?

[UnterKo]

Hello, I've got a problem and I have no idea how to start. I'll be happy for any hint. Thanks

Homework Statement

Two beads each of mass m are at the top (Z) of a frictionless hoop of mass M and radius R which lies in the vertical plane. The hoop is supported by a frictionless vertical support. The beads are given a tiny impulse and due to gravity they slide down the hoop: one clockwise and one anti-clockwise.
a) Determine the minimum value of X = m/M, X_min, for which the hoop will rise up off the support before the beads reach the bottom of the hoop (Y).
b) If X < X_min and the beads collide inelastically at the bottom of the hoop with a coefficient of restitution e_R = 0.98, determine the minimum angle (θ) to the nearest degree with respect to the initial position (θ = 0) achieved by the clockwise moving bead.
c) How many oscillations are required so that the maximum height achieved by the beads is less than 0.01R?

2. The attempt at a solution
a) To use energy conservation law?
b) Tangens of the angle θ (with respect to the initial position) is going to be reduced by the e?
c) Do I get the height by the energy conservation law (PE = KE)? But how to determine the number of collisions?

 

[haruspex]

For a), yes, use energy conservation. What can you deduce from that?
You will also need to think about forces on the hoop, obviously. What forces will there be?

For b), again, use energy. If the coefficient is e, what fraction of energy is lost in the collision?

What fraction of energy is lost in n collisions?

 

[UnterKo]

a) Each bead experiences its weight and the normal reaction force (because the hoop is frictionless, it acts normal to the surface, right?). If I take θ as the angle of it's position with respect to the horizontal, the normal force may be expressed as:
F = Nr - mgk = (N - mg sin θ)r - (mg cos θ) θ.
Thus the radial eq. is: $N - mg sin θ = - ma θ' ^2$, and the angular eg. is:$- mg cos θ = ma θ''$. So, when the bead reaches the bottom of the hoop, the potential energy is: $V(θ) = mga sin θ$.
And the speed of the bead is aθ', so the K.E. is: $K = ½ ma^2 θ'^2$, right?
So, for the law of conservation of energy I get: K + V = K + mga sin θ = mga ?
But now I don't know precisely what's happening with the hoop at this moment..
b) The lost of KE is: $ΔK = ½ mv_i^2 (e^2 - 1) (1 - sin^2 θ) < 0$, since for the velocity of each bead after the collistion, $v_f applies: v_f^2 = v_i^2 (sin^2 θ + a^2 cos^2 θ) = v_i^2 + v_i^2 (e^2 - 1) cos^2 θ$?
c) Is the fraction: 1/n ΔE ?

 

[haruspex]

take θ as the angle of it's position with respect to the horizontal,

... and measured upwards from the horizontal, yes?

So, when the bead reaches the bottom of the hoop, the potential energy is: V(θ) = mga sin θ.

What is theta here? It has gone from top to bottom.

K + mga sin θ = mga

Right. So plug that in your equation for N.

ΔK=½mv²_i(e²−1)(1−sin²θ)<0

Again, what is theta here?

 

[UnterKo]

a) Yes, it's upwards to the horizontal.
When I substitute, it gets:
$$ ½ ma^2 θ'^2 + mga \sin θ = mga$$
$$ ½ a θ'^2 + g \sin θ = g $$
$$ ½ a θ'^2 = g(1 - \sin θ)$$
$$ θ'^2 = \frac{2g(1 - \sin θ)}{a}$$
and in radial eg.:
$$N - mg \sin θ = -2 mg (1 - \sin θ)$$
$$N = mg(3 \sin θ - 2)$$
But how do I get the fraction of masses, X?
b) And then after the collision (sorry, I used the same notation for the angle as before), so there is φ as measured from vertical of the place of the collision to the right (for the clockwise moving bead). But I don't know how to derive it afterwards. Do I need to use the radius (or height) difference, Δr, from the bottom position to the position after the collision?

 

[haruspex]

But how do I get the fraction of masses, X?

What is the net force they exert on the hoop?

For b), what is the KE at the bottom?

 

[UnterKo]

$$ N = mg (3 \sin \theta - 2)$$<- it isn't this multiplied by two? I'm not actually sure about what even physically means 'to rise up off the support'. Does it mean that the hoop is not fixed?
b) KE before the collision is:
$$ K = 1/2 ma^2 \dot{\theta}^2$$

 

[BvU]

Haru may have gone to bed; I liked following this thread and will chip in carefully, without giving away too much (that would spoil the fun and the ultimate satidfaction).

So: The $N$ you calculated represents the radial force each bead experiences from the hoop, right ? So yes, you now have to look at the sum of the two forces the hoop experiences from the beads. That's not a factor of two ! (make a sketch).

The hoop will rise up from the support if the sum of the two forces it experiences from the beads offsets $M\vec g$. The hoop appears to be not fixed to the support.

 

[haruspex]

I confirm all of BvU's observations.

b) KE before the collision is:

Yes, but you also know what it is in terms of the given variables, R, m etc. (it would be less confusing if you were to use the given radius R instead of introducing a, which is often used for acceleration.)

 

[UnterKo]

So: The $N$ you calculated represents the radial force each bead experiences from the hoop, right ? So yes, you now have to look at the sum of the two forces the hoop experiences from the beads. That's not a factor of two ! (make a sketch).
The hoop will rise up from the support if the sum of the two forces it experiences from the beads offsets $M\vec g$. The hoop appears to be not fixed to the support.

So the force experienced by beads on the hoop is a sum of this force (N) of both beads? Then: 2mg(3sin θ - 2) > Mg. And then it is: m/M > (3 sin θ - 2)/(2)?

Yes, but you also know what it is in terms of the given variables, R, m etc. (it would be less confusing if you were to use the given radius R instead of introducing a, which is often used for acceleration.)

So: K = ½ mR θ'². But if I substitute, for example, m = M (3 sin θ - 2)/(2), I'm sorry, but I don't see how it's going to help me.

 

[BvU]

force experienced by beads on the hoop is a sum of this force (N) of both beads

Yes. But they are not pointing in the same direction, so it's not a factor of 2 (make a sketch).

 

[UnterKo]

Yes. But they are not pointing in the same direction, so it's not a factor of 2 (make a sketch).

But then they are at the bottom of the hoop, no? And for this position the fraction m/M has to be determined, right?

 

[haruspex]

But then they are at the bottom of the hoop, no? And for this position the fraction m/M has to be determined, right?

No, they will not be at the bottom of the hoop when they apply the greatest upward force on it. Indeed, at the bottom it would be the maximum downward force.
Find the component of the normal force which acts upwards on the hoop, then find the angle at which this is maximised.

 

[haruspex]

So: K = ½ mR θ'². But if I substitute, for example, m = M (3 sin θ - 2)/(2), I'm sorry, but I don't see how it's going to help me.

You dropped an exponent in the KE expression.
For this part of the question, forget m = M (3 sin θ - 2)/(2). That is not relevant now. What is the lost PE when a bead reaches the bottom?

 

[UnterKo]

No, they will not be at the bottom of the hoop when they apply the greatest upward force on it. Indeed, at the bottom it would be the maximum downward force.
Find the component of the normal force which acts upwards on the hoop, then find the angle at which this is maximised.

I see, I have not understood this till now! :) So it's going to be a centripetal force? And the angle may be found by some derivation?
Is the loss of PE going to be like: $PE = mgh$, so for one of the beads at the bottom: $PE = mg(\sin \frac{\pi}{2})$, so the loss is: $PE = mg(2R - 1)$?

 

[haruspex]

I see, I have not understood this till now! :) So it's going to be a centripetal force? And the angle may be found by some derivation?

For part a), in post #5 you already found the normal force each bead exerts on the hoop when at angle theta. What part of that force tends to raise the hoop up? For what theta is it maximised?

For part b):

for one of the beads at the bottom: $PE = mg(\sin \frac{\pi}{2}),$

You need a distance in there.

 

[UnterKo]

For part a), in post #5 you already found the normal force each bead exerts on the hoop when at angle theta. What part of that force tends to raise the hoop up? For what theta is it maximised?
For part b):You need a distance in there.

It is maximised for $\theta = \frac{\pi}{2}$. Is it centripetal force?
So is it going to be part b): $PE = mgR \sin \theta$?

 

[CWatters]

Can I suggest you draw a bead on a hoop at several different positions. Mark on it the direction of the centripetal force and the corresponding reaction force on the hoop. It should then be obvious that there is a range of angles for which the reaction force on the hoop has an upwards component (lifting the hoop) and a range of angles for which it's downwards (not lifting the hoop).

I don't think you can determine angle at which the force is a maximum simply by inspection because the velocity and centripetal force changes with the angle. You will have to write equations, some of which you have already done.

 

[UnterKo]

Can I suggest you draw a bead on a hoop at several different positions. Mark on it the direction of the centripetal force and the corresponding reaction force on the hoop. It should then be obvious that there is a range of angles for which the reaction force on the hoop has an upwards component (lifting the hoop) and a range of angles for which it's downwards (not lifting the hoop).

I don't think you can determine angle at which the force is a maximum simply by inspection because the velocity and centripetal force changes with the angle. You will have to write equations, some of which you have already done.

I've done it and I see what you mean. But there can not be found any max. lifting force at some angle? Eg. $\theta = \frac{\pi}{2}$(as measured from horizontal in downwards direction)?. So how can I find the value of $X = \frac{m}{M}$?
And could you, please, explicitly write which ones of mine equations are correct and which ones I am missing?
Thank you

 

[BvU]

check numbers 8,11,13,16,18 , all of us want you to replace the factor 2 by something that depends on theta...

 

[haruspex]

It is maximised for

You skipped this question, which must be answered first:

What part of that force tends to raise the hoop up?

The normal force acts at angle theta above the horizontal. What part of that tends to lift the hoop up?

 

[UnterKo]

BvU, I really don't know what is it... could you please answer to my questions in 19? I am trying to figure out this problem for more than a week so that's the reason why I've asked you for help.

 

[CWatters]

Earlier you wrote...

N=mg(3sinθ−2)

where N is the Normal force.

Then you appear to have multiplied by 2 because there are two beads...

So the force experienced by beads on the hoop is a sum of this force (N) of both beads? Then: 2mg(3sin θ - 2)...

That's not correct. The normal force on each bead is a vector so you have to do vector addition, not just multiply by 2. You will remember that one way to do vector addition is to break each vector down into vertical and horizontal components and sum these. However in some problems you aren't interested in one of those sums.

At this point I'll refer you to what Haruspex has been asking...

What part of that force tends to raise the hoop up?
The normal force acts at angle theta above the horizontal. What part of that tends to lift the hoop up?

 

[johnnyhgrace]

Hopefully someone sees this. I am solving this problem as well. I don't understand how the beads can exert an upwards force on the hoop. If gravity exerts a downwards force on the beads, and the beads are always accelerating downward by less than g, then the normal force (from the hoop on the beads) never has a downward component. Thus the normal force (from the beads on the hoop) never has an upward component.

Can someone please explain.

 

[TSny]

The comments below are from a different thread that deals with a similar problem.

At the beginning of the motion, the normal force from the hoop acting on a bead is radially outward. Thus, the force of the bead on the hoop is radially inward.

However, as the bead continues to slide there will be a special point where the normal force between the bead and the hoop goes to zero while the bead is still on the upper half of the hoop. (Recall the problem of someone sliding down a hemispherical dome.)

What is the direction of the normal force on the bead once the bead is past this special point?

 

[haruspex]

If … the beads are always accelerating downward by less than g

Yes, if.
Think about the total acceleration of the beads.