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1. Aug 17, 2016

### UnterKo

Hello, I've got a problem and I have no idea how to start. I'll be happy for any hint. Thanks
1. The problem statement, all variables and given/known data
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, Xmin, for which the hoop will rise up off the support before the beads reach the bottom of the hoop (Y).
b) If X < Xmin and the beads collide inelastically at the bottom of the hoop with a coefficient of restitution eR = 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?

2. Aug 17, 2016

### 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?

3. Aug 19, 2016

### 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 ?

4. Aug 19, 2016

### haruspex

... and measured upwards from the horizontal, yes?
What is theta here? It has gone from top to bottom.
Right. So plug that in your equation for N.
Again, what is theta here?

5. Aug 19, 2016

### UnterKo

a) Yes, it's upwards to the horizontal.
When I substitute, it gets:
$$½ ma^2 θ'^2 + mga \sin θ = mga$$
$$½ a θ'^2 + g \sin θ = g$$
$$½ a θ'^2 = g(1 - \sin θ)$$
$$θ'^2 = \frac{2g(1 - \sin θ)}{a}$$
$$N - mg \sin θ = -2 mg (1 - \sin θ)$$
$$N = mg(3 \sin θ - 2)$$
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?

6. Aug 19, 2016

### haruspex

What is the net force they exert on the hoop?

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

7. Aug 19, 2016

### 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:
$$K = 1/2 ma^2 \dot{\theta}^2$$

8. Aug 19, 2016

### 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.

9. Aug 19, 2016

### haruspex

I confirm all of BvU's observations.
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.)

10. Aug 26, 2016

### UnterKo

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)?
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.

11. Aug 26, 2016

### BvU

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

12. Aug 26, 2016

### UnterKo

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

13. Aug 26, 2016

### haruspex

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.

14. Aug 26, 2016

### haruspex

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?

15. Aug 28, 2016

### UnterKo

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: $PE = mg(2R - 1)$?

16. Aug 28, 2016

### haruspex

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.

17. Aug 29, 2016

### UnterKo

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

18. Aug 29, 2016

### 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.

19. Aug 30, 2016

### UnterKo

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

Last edited: Aug 30, 2016
20. Aug 30, 2016

### BvU

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