What factors affect the motion of a bead on a spinning hoop?

In summary: N is the force pulling to the center of the hoop. Would it's x component have the force m*arad?x = m*aradthe weight obviously has no x component so the only x force is from...the weight?the weight is the only x force
  • #36
Well I already have the equation for N right? (N=mg/cosθ) And can I just use friction≤μN here?
 
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  • #37
zaper said:
Well I already have the equation for N right? (N=mg/cosθ)

no, because you got that by resolving F = ma vertically, in the non-friction case

now, there's a friction force F, which of course will have both horizontal and vertical components, so that equation needs changing

(you may find it more convenient to use R-radial and tangential components instead of horizontal and vertical … I'm not sure, i haven't checked)
 
  • #38
Ok so with friction for there to be no vertical acceleration then Ny+Fy=mg to stop it from moving down but also mg+Fy=Ny to stop from moving up.

For no horizontal acceleration won't it simply be Nx=Fx?
 
  • #39
zaper said:
Ok so with friction for there to be no vertical acceleration then Ny+Fy=mg to stop it from moving down but also mg+Fy=Ny to stop from moving up.

nooo, it'll drive you mad if you try doing the two friction cases separately :redface:

the only way to stay sane is to just call it F, and find out later whether it's positive or negative :smile:
For no horizontal acceleration won't it simply be Nx=Fx?

but there isn't no horizontal acceleration, is there?
 
  • #40
tiny-tim said:
the only way to stay sane is to just call it F, and find out later whether it's positive or negative :smile:

So Ny+F-mg=ma with no vertical acceleration so it's just simply Ny+F=mg

Horizontally then Nx-Fx=ma but where does this acceleration come from?
 

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