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

  • Thread starter Thread starter zaper
  • Start date Start date
  • Tags Tags
    Bead Spinning
Click For Summary

Homework Help Overview

The discussion revolves around the motion of a bead on a frictionless circular hoop that spins around a vertical axis. Participants are exploring how the bead's position, represented by the angle θ, is influenced by various factors such as angular velocity (ω), mass (m), radius (R), and gravitational acceleration (g). The problem also introduces friction and seeks to understand the conditions under which the angular velocity can change without affecting the angle θ.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between centripetal force and the forces acting on the bead, including weight and normal force. There are attempts to derive equations for the forces in both vertical and horizontal directions, with some questioning the components of acceleration and the role of friction.

Discussion Status

The discussion is active, with participants providing guidance on how to approach the problem. There is a mix of confusion and clarification regarding the application of forces and the correct interpretation of acceleration components. Some participants are exploring the implications of friction on the forces involved, while others are attempting to derive the relationship between ω and θ.

Contextual Notes

Participants are navigating through the complexities of the problem, including the effects of friction and the need to resolve forces accurately. There is an emphasis on ensuring that the equations reflect the conditions of equilibrium for the bead on the hoop.

  • #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:)
 
Physics news on Phys.org
  • #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?
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 44 ·
2
Replies
44
Views
10K
Replies
25
Views
6K
  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 5 ·
Replies
5
Views
5K
Replies
13
Views
7K
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
Replies
2
Views
2K
Replies
14
Views
3K