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Please help - mechanics of a bead on a hoop

  1. May 29, 2008 #1
    1. The problem statement, all variables and given/known data

    Consider a bead of mass m that can slide freely on a circular hoop of radius a which lies in a vertical plane. The hoop rotates about a vertical axis through its centre at fixed angular velocity [tex]\omega[/tex]. The angle [tex]\theta[/tex] is the polar angle of the bead measured from the downward vertical.

    1. Find the kinetic and potential energy of the bead.

    2. Find the bead's equation of motion.

    3. Show that for [tex]\omega^{2}a < g [/tex] the bead can remain in a stable equilibrium at the bottom of the hoop, otherwise this position is unstable.

    4. In the case [tex]\omega^{2}a > g [/tex] find the new position [tex]\theta^{*}[/tex] of stable equilibrium and calculate the angular frequancy [tex]\Omega[/tex] of small oscillations about this position.

    2. Relevant equations

    I worked out the answers for the first 2 questions myself (see below)

    3. The attempt at a solution

    1. The kinetic energy is [tex]\stackrel{1}{2}ma^{2}(\omega^{2}sin^{2}(\theta)+ \stackrel{.}{\theta}^{2})[/tex]

    The potential energy is mga[tex](1-cos(\theta))[/tex]

    The above are just from the geometry of the situation

    2. The equation of motion of the bead is [tex]\stackrel{..}{\theta} = (\omega^{2}cos(\theta) - g/a)sin(\theta)[/tex]

    This just comes from using the Lagrangian L=T+V and treating it as a function of [tex]\theta[/tex] and [tex]\stackrel{.}{\theta}[/tex]

    3 and 4. This is where I am getting stuck! When I set to zero [tex]dv/d\theta[/tex] I get [tex]mgasin(\theta)=0[/tex] which does not relate to [tex]\omega[/tex]! Would it be necessary to use minimise an effective potential instead? I've tried this but the algebra is not going anywhere and a little insight would be helpful.
    Last edited: May 29, 2008
  2. jcsd
  3. May 30, 2008 #2
    I don't think your kinetic energy is right. You should have three terms since the kinetic energy ought to be 1/2*m*v.v, where v velocity of the bead.
  4. May 30, 2008 #3
    I agree with his Kinetic energy expression - the velocities are confined to different planes- this should mean the dot product between is zero. Or have I screwed my geometry up there something rotten?
  5. May 30, 2008 #4


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    The KE is correct. It's the sum of two orthogonal velocities, the part along the hoop and the part related to the rotation of the hoop.
  6. May 30, 2008 #5


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    As for the rest of the problem, you can think of the right side of the theta eom as a 'force'. The equiibrium points are where it vanishes. Take the theta=0 point, if you expand to first order in theta, you get
    \stackrel{..}{\theta} = (\omega^{2} - g/a) \theta

    Now you can see that if the factor in front of the theta is positive, then if theta moves away from zero it will keep accelerating away. If it's negative, it will accelerate back. It must be obvious where the other equilibrium point is, right? Play the same game there and write an expansion in terms of dtheta, where dtheta is the difference from equilibrium. To get the oscillation frequency, notice these expansions look like harmonic oscillators.
  7. May 30, 2008 #6
    Yes, sorry. I read the question wrong!
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