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Bead on a hoop

  • #1
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Homework Statement



A circular wire hoop rotates with constant angular velocity ! about a vertical diameter. A small bead moves, without friction, along the hoop. Find the equilibrium position of the particle and calculate the frequency of small oscillations about this position.

Homework Equations




The Attempt at a Solution


Let's take our reference inertial frame as a spherical co-ordinate system whose axis is along the axis of rotation of the who and whose origin is at the centre of the hoop.
Then position of M can be given by r, ## \theta, \phi ##.
Constraint: r= R,
## \dot \phi = \omega ## , constant.
There are two generalised coordinates ## \phi ## and ## \theta ##.

L = T - U

## T = \frac 1 2 mR^2 ( {\dot \theta}^2 + \sin ^2 \theta {\dot \phi}^2) ##
Taking U = 0 at the origin, U = mgR ## \cos \theta ##
So, L = ## \frac 1 2 mR^2 ( {\dot \theta}^2 + \sin ^2 \theta {\dot \phi}^2 ) - mgR ## ## \cos \theta ##

Lagrange's equation of motion gives,
## \ddot \theta = \frac g R \sin {\theta} + \sin {\theta}~ \cos { \theta}~{ \dot \phi}^2 ##

At eqbm. ## \ddot \theta = 0 ##
## \theta = 0, \Pi , \cos {\theta} = \frac {-g} { \omega^2 R} ##

What to do next?
 
Last edited:

Answers and Replies

  • #2
kuruman
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What to do next?
Rewrite your equation of motion in terms of a small angular departure from equilibrium and see if you can bring the equation into the form of a harmonic oscillator.
Let ##\theta \rightarrow \theta_0+\alpha##, where ##\theta_0 = \arccos(-g/(\omega^2 R))## and expand the EOM for small values of ##\alpha##. Don't forget that ##\sin \theta \cos \theta = \sin(2\theta)/2##.
 
  • #3
vela
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Where are you measuring ##\theta## from? If ##\theta=0## is supposed to be when the bead is at the bottom of the loop, then the sign of ##U## is wrong.
 
  • #4
kuruman
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Where are you measuring ##\theta## from? If ##\theta=0## is supposed to be when the bead is at the bottom of the loop, then the sign of ##U## is wrong.
It all makes sense if ##\theta## is measured from the top. Then the equilibrium angle is (as it should) below the horizontal diameter and its cosine is negative. U is zero when ##\theta = \pi/2##.
 
  • #5
vela
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Yeah, I know. It's just that most people use the convention that ##\theta=0## is the (stable) equilibrium position when the hoop isn't spinning. Thought it was worth making sure.
 
  • #6
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I solved it. Thank you all for the guidance.
 

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