Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Bead on a hoop+accelerating elevator

  1. Aug 25, 2010 #1
    1. The problem statement, all variables and given/known data
    http://img839.imageshack.us/img839/8904/bead.jpg [Broken]

    Uploaded with ImageShack.us

    The picture speaks for itself. I'm asked to find the minimum velocty that this bead should be given so that it will reach point B.

    2. The attempt at a solution

    I've solved this question using energy conservation with a "new gravity": [tex] g_{eff}=g+a [/tex] , so that:

    [tex] 0.5mV^{2}=m(g+a)(2R) \to V=2 \sqrt{(g+a)R} [/tex]

    and this is the right answer.

    However, supposing I want to solve this question using Newton's 2nd law in an accelerating frame, so:

    [tex] m \vec {a}_{rel} = \vec{F}_{real} - m \vec{a}_{frame} [/tex]

    If i take the positive values to be in the direction of the relative acceleration, i'll obtain:

    [tex] m \frac{V^{2}}{R} = mg + ma \to V= \sqrt {(g+a) R } [/tex] ?!?!?!?!

    Why is that? Am I missing something?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 25, 2010 #2
    What does this equation mean? Is it the correct condition for the bead to reach B? :wink:
  4. Aug 25, 2010 #3
    Obviously it's not correct because the answer is messed up :D , but assuming i didn't know the correct answer i'll say yes, it the the correct one. I don't see where's the problem.

    What does this equation means? hmmm ... Isn't it simply the modified 2nd law of newton in accelerating frames?
  5. Aug 25, 2010 #4
    That equation describes the motion of the bead in the lift frame at the lowest point, given that the bead doesn't experience normal force from the hoop. But that doesn't guarantee the bead can climb up to B :wink:

    EDIT: I see something deadly wrong here. At the lowest point, in the lift frame, gravitational force and fictitious force point down, but the centripetal force points up. Normal force cannot be zero.
    Last edited: Aug 25, 2010
  6. Aug 25, 2010 #5
    Reagarding the normal force, I really forgot all about it (in addition, I'm given that the hoop is frictionless) ... so the equation should be:

    m \frac{V^{2}}{R} = mg -N + ma

    and at point B we want it to be: [tex] N \neq 0 [/tex] ... but I still have 2 unknows :/
  7. Aug 25, 2010 #6
    You got the wrong signs! It should be:
    m \frac{V^{2}}{R} = -mg +N - ma
    Anyway, this doesn't show any possibility that the bead can go up to B. You may disregard this force method; otherwise, if you continue with it, the equation you get is eventually the same as the energy equation.

    P.S.: Some more comments on this. You may repeat the idea that [tex]N \neq 0[/tex] is the condition taken from some other problems, but it does not apply to this problem. We always have [tex]N \neq 0[/tex], as the bead is forced to follow a circular motion.
  8. Aug 25, 2010 #7
    yeah, I used the energy equation at first and got it right but I just wanted to see whether I get the same answer form the force method. So are you saying it's insolvable this way?
  9. Aug 25, 2010 #8
    It's solvable, but it's actually the same as energy method. Here it is:
    1 - Because the bead is forced to follow a circular path, the normal force must always satisfy that. In other words, the normal force is determined by the circular motion.
    2 - The only thing that matter is that if the component of other forces (i.e. weight & fictitious force) along the hoop can pull the bead down before the bead reaches B. Let [tex]\phi[/tex] denote the angle coordinate of the bead to the horizontal. We have:

    [tex]-m(g+a)cos(\phi) = mRd\omega/dt=mR\omega d\omega/d\phi[/tex]

    Therefore: [tex]\int_{V/R}^{0}R\omega d\omega = -\int_{-\pi/2}^{\pi/2}(g+a)cos(\phi)d\phi[/tex]

    And that is the same as the energy equation.
    Last edited: Aug 25, 2010
  10. Jun 10, 2012 #9
    Hello could you look at ? the post /showthread.php?t=612910
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook