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A bead rests at the top of a fixed, frictionless hoop.

  1. Mar 5, 2017 #1
    • Member warned to use the homework template for posts in the homework sections of PF.
    1. The problem statement, all variables and given/known data
    A bead rests at the top of a fixed, frictionless hoop of radius R that lies in a vertical plane. The bead is given a tiny push so that it slides down and around the hoop. At what points is the bead's acceleration vertical? What is the acceleration at these points? This problem is from David Morin's "Introductory Classical Mechanics with problems and solutions"

    2. Relevant equations
    $$v=\sqrt{2gh}$$ from conservation of energy, $$a_R=-mR\dot\phi^2$$ and $$a_T=mR\ddot\phi$$

    3. The attempt at a solution
    I started off using the conservation of energy to find that $$v=\sqrt{2gh}$$ where h is the distance travelled from the top of the hoop. I also wrote down the tangential acceleration Rφ'' and radial acceleration aR-v^2/R. I then got aR=-2gh/R. But now I have no idea what to do.
     
    Last edited: Mar 5, 2017
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  3. Mar 5, 2017 #2

    haruspex

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    At what points is the bead's acceleration what?
     
  4. Mar 5, 2017 #3
    *At what points is the bead's acceleration vertical* oops forgot to add that.
     
  5. Mar 5, 2017 #4

    haruspex

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    At angle Φ from the top, what is h? What are the forces on the bead there?
     
  6. Mar 5, 2017 #5
    $$h=R(1-cos\phi)$$ right? and the forces would be normal force N in the radial direction and gravitational force mg downwards?
     
    Last edited: Mar 5, 2017
  7. Mar 5, 2017 #6
    Ok so I managed to get two equations: $$N=mg(3-2cos\phi)$$ and $$mg\sin\phi=R
    \ddot{\phi}$$ by equating the radial directions together and the tangential directions together. I'm not sure where to go from here though. Are these correct btw?
     
    Last edited: Mar 5, 2017
  8. Mar 5, 2017 #7

    kuruman

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    The second equation is equation is correct. The first equation for the normal force cannot be correct. That's because, at some point between the top and the bottom of the loop, the normal force has to change direction from radially out to radially in. At that point the normal force has to go through zero. Your expression goes through zero when cosφ = 3/2. Not possible.
     
  9. Mar 5, 2017 #8
    How could I go about resolving this problem?
     
  10. Mar 5, 2017 #9

    kuruman

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    Check your FBD and your derivation for the normal force. There must be an error somewhere. If you cannot find the error, post your solution and someone else will.
     
  11. Mar 5, 2017 #10
    I think I found my error. I had the directions wrong. I started off with, $$N-mg\cos\phi=\frac{mv^2}{R}$$ $$\implies N=mg\cos\phi+2mg(1-\cos\phi)$$ since $$v^2=2gR(1-\cos\phi)$$ from conservation of energy, $$\implies N=mg(2-\cos\phi)$$ Is it correct this time? If so where do I go next.

    EDIT: Hang on this doesn't make any sense either since cosφ=2 is impossible.
     
  12. Mar 5, 2017 #11

    kuruman

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    In what direction is the centripetal acceleration? Radially out or radially in?
     
  13. Mar 5, 2017 #12
    Radially in. Oh god I made an typo in my 6th post. I meant to say $$N=mg(3\cos\phi-2)$$ Hence all the confusion. Ok what is the next step now?
     
  14. Mar 5, 2017 #13

    kuruman

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    There is also an error in post #10 where you show the centripetal acceleration radially outwards. Anyway, the expression for the normal force is now correct.
    What do you think it should be? The question is
    Think critically. If you add all the forces on the bead, what must be true for the acceleration to be vertical?
     
  15. Mar 5, 2017 #14
    Yes there is. I neglected to mention this one for some reason.
    Would it simply be where the normal force is zero? Because then that only leaves gravity, which is vertical? In which case you'd have $$\phi=\cos^{-1}\frac{2}{3}, 2\pi-\cos^{-1}\frac{2}{3}$$
     
  16. Mar 5, 2017 #15

    kuruman

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    That is correct, good thinking. However, is that the only angle? What about other angles after the normal force goes through zero and changes direction?
     
  17. Mar 5, 2017 #16
    Why does it change direction exactly? Also, would I have to start the problem again but put a negative sign on the normal force for the other case?

    EDIT: The question said there were only two other points aside from the trivial ones at φ=0,π (this was given in a hint that I should've mentioned earlier). So I'm not sure how my two solutions can't be the only ones, according to the question.
     
    Last edited: Mar 5, 2017
  18. Mar 5, 2017 #17
    Oh and for the acceleration at those two points I gave above, is it just simply g?

    EDIT: I used the second equation I found earlier and got answers of $$a=\frac{\sqrt{5}g}{3}$$ downwards at the point $$\phi=\cos^{-1}2/3$$ and $$a=\frac{\sqrt{5}g}{3}$$ upwards at the point $$\phi=2\pi-\cos^{-1}2/3$$
     
    Last edited: Mar 5, 2017
  19. Mar 5, 2017 #18

    kuruman

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    Mathematically, because the function that describes it goes through zero and becomes negative. Physically, because the force that is needed to keep the bead in the circular path has to start pushing inwards instead of outwards. It is the point where the radius of curvature of the circular path matches the radius of curvature of the parabolic path in free fall at that velocity. This problem is a variant of the problem where you have a small block at the top of a frictionless hemisphere and you are asked to find at what angle the block flies off the hemisphere if you give it a slight push.
    Why the question mark? Are you asking me or are you telling me?

    There are still other angles where the acceleration has no horizontal component.
     
  20. Mar 5, 2017 #19
    I'm asking you. Also, just to be clear is there more than 4 angles where the acceleration is vertical?
     
  21. Mar 5, 2017 #20

    kuruman

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    Fair enough. I will guide your thinkng with two questions of my own. If the normal force is zero, what other force(s) act(s) on the bead? What is the acceleration in that case?
     
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