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

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The discussion focuses on analyzing the motion of a bead on a spinning hoop, emphasizing the relationship between the bead's angle θ, angular velocity ω, radius R, mass m, and gravitational force g. Participants explore the equations of motion, particularly using F = ma to derive the normal force and centripetal acceleration. They clarify the need to consider both horizontal and vertical components of forces acting on the bead, leading to the conclusion that the bead's equilibrium requires balancing these forces. The conversation also touches on the effects of friction and how it influences the bead's motion without altering the angle θ. Ultimately, the thread illustrates the complexities of applying physics principles to dynamic systems.
  • #31
zaper said:
Not sure if this is what you want or not, but subbing Rω2=arad into Fx I get mRω2=N*sinθ.

but you're not told what N is!

(and that's not right anyway :redface:)
 
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  • #32
Ok so since we don't know what N is and the question doesn't ask for it we need to get rid of it. You said earlier that there is no vertical acceleration so that means N*cosθ=mg or N=mg/cosθ
 
  • #33
(just got up :zzz: …)

that's right :smile:

now put everything together!
 
  • #34
So then plugging N into mrω2=N*sinθ (again r is the horizontal circle's radius) I get:

mrω2=mg*sinθ/cosθ

Simplified this is rω2=g*tanθ

Now I believe that r is R*sinθ so subbing again:

R*sinθ*ω2=g*tanθ which simplifies again to

R*ω2=g/cosθ

Hopefully I did all that right and from here it's easy to get θ by itself
 
  • #35
zaper said:
R*ω2=g/cosθ

yes! :smile:

ok, now you have ω as a function of θ, find both the normal force and the friction (tangential) force as a function of θ, and check whether µ is sufficient :wink:
 
  • #36
Well I already have the equation for N right? (N=mg/cosθ) And can I just use friction≤μN here?
 
  • #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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