# Homework Help: Bead on a Spinning Hoop

1. Feb 9, 2012

### zaper

A bead of mass m can slide on a frictionless circular hoop in the vertical plane with radius R. When the hoop spins around the vertical axis with a rate of ω, the bead moves up the hoop by an angle θ which depends on the angular velocity of the hoop. (Pic attached)

a) Find θ in terms of ω, R, m and g.

b) If there is friction between the bead and hoop given by μ, within what range can ω change without changing θ?

For part a what I have so far is that arad=R*ω2. The centripetal force (F) then is m*arad and the normal force is F/sinθ.

The weight needs to cancel out the normal force then so the weight in the direction of the normal force is mg*cosθ = F/sinθ

Am I anywhere close to being right?

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2. Feb 9, 2012

### tiny-tim

hi zaper!

no

keep it simple

do F = ma vertically to find N

then do F= ma horizontally to find ω

(and remember that the centripetal acceleration is horizontal, so use the correct r )

3. Feb 9, 2012

### zaper

Hey there tiny tim!

Sorry, but I'm really struggling to keep up with this problem so forgive me for anything dumb that I do here.

So you said to find N as F=ma vertically so that means that N=mg*cosθ

By you're second part are you saying m*arad=m*ω2*r (r being R*sinθ)

4. Feb 9, 2012

### tiny-tim

hey there zaper!
that's not in the vertical direction, is it?
no, that's ma = ma, you need F = ma

5. Feb 9, 2012

### zaper

Ok so then N=mg*sinθ I assume. Could you please explain why this is and what exactly you want for the second equation. I'm not really following you here.

6. Feb 9, 2012

### tiny-tim

no

start again …

Fx = max

Fy = may

what are ax and ay ?​

7. Feb 9, 2012

### zaper

I'm really really sorry. I guess I'm way overthinking this or something. Is ax the radial acceleration?

8. Feb 9, 2012

### tiny-tim

a is the acceleration

ax and ay are its x and y components

(i'm assuming you're using y for up and x for radial)

9. Feb 9, 2012

### zaper

Ok so ax is the radial acceleration and the total acceleration is pointing up and to the left. The triangle formed by the accelerations has top angle θ and ay=ax/tanθ. Is this correct? Also does ay actually describe something here or is it just a component of the total a?

10. Feb 9, 2012

### tiny-tim

what??

why?

11. Feb 9, 2012

### zaper

Wait, I mean to the right

12. Feb 10, 2012

### tiny-tim

do you mean up and to the right, or only to the right?

13. Feb 10, 2012

### abhinav111

hi zapper i have an easily understandable solution, if u have not come to the answer i may help.

14. Feb 10, 2012

### zaper

Up and to the right tiny tim and what's your idea abhinav?

15. Feb 10, 2012

### tiny-tim

(along the normal?)

nooo

i think you're confusing physics and maths

force is physics

acceleration is maths (geometry)

if you know how the object is moving, then the acceleration follows automatically as a matter of maths

it doesn't matter what forces are producing that motion, the acceleration depends only on the motion, not on the forces

in this case, in equilibrium, the bead is moving uniformly in a horizontal circle …

so what is its acceleration?

16. Feb 10, 2012

### zaper

Since there's no angular acceleration then the only acceleration is radial right?

17. Feb 10, 2012

### tiny-tim

yes, but you're looking at the wrong radius …

the circle to look at is the circle of the bead's motion, which is horizontal, isn't it?

18. Feb 10, 2012

### zaper

Yes. That last post was supposed to be in terms of the horizontal circle. Sorry that I forgot to explain that

19. Feb 10, 2012

### tiny-tim

ok, so now do Ftotal = ma in both the horizontal and vertical directions

20. Feb 10, 2012

### zaper

Ok I think we're getting back to my point of confusion from before. We've established that the only acceleration in the horizontal circle is the radial acceleration so I'm not quite seeing where this vertical force can come from unless you mean the weight

21. Feb 10, 2012

### tiny-tim

yes, the weight is vertical

but the bead has no vertical acceleration

so what's the other vertical force (that makes F=ma)?

22. Feb 10, 2012

### zaper

The normal force? That's about all that's left

23. Feb 10, 2012

### tiny-tim

yes, N

N and mg are the only two forces on the bead

so their components in the vertical direction have to equal the component of ma in the vertical direction,

and their components in the horizontal direction have to equal the component of ma in the horizontal direction

24. Feb 10, 2012

### zaper

Ok so N is the force pulling to the center of the hoop. Would it's x component have the force m*arad? Also the weight obviously has no x component so the only x force is from N.

25. Feb 10, 2012

### tiny-tim

that's right!

so the x and y equations for F = ma are … ?​