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Rotating bead

  1. Aug 3, 2011 #1
    1. The problem statement, all variables and given/known data

    A small bead of mass m is free to slide along a long, thin rod without any friction. The rod rotates in a horizontal plane about a vertical axis passing through its end at a constant rate of f revolutions per second. Show that the displacement of the bead as a function of time is given by r(t)=A1ebt +A2e–bt , where r is measured from the axis of rotation. Find the expression for the constant b. Also, how would you determine the constants A1 and A2?

    2. Relevant equations

    L=rp


    3. The attempt at a solution

    I know this has to do with Angular momentum and torque but I don't know where to start
     
  2. jcsd
  3. Aug 3, 2011 #2

    kuruman

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    This problem is done quite easily if you transform to the (non-inertial) rotating frame of the bead. Suppose you are sitting on the bead. What you write down for Newton's 2nd Law?
     
  4. Aug 3, 2011 #3
    F = ma = mw^2r
    where w = 2pif ?????
     
    Last edited: Aug 3, 2011
  5. Aug 4, 2011 #4

    kuruman

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    Correct. Now write Newton's 2nd law as a differential equation and solve it.
     
  6. Aug 4, 2011 #5
    let u = tangential velocity
    let v = radial velocity = [itex]\omega[/itex] r

    dv/dt = [itex]\omega[/itex] dr/dt

    at = du/dt = d/dt([itex]\omega[/itex]r) = [itex]\omega[/itex] dr/dt
    ar = d^2r/dt^2 = dv/dt= ([itex]\omega[/itex]^2)r

    so
    [itex]\omega[/itex] dr/dt = [itex]\omega[/itex]^2 r
    dr/dt = [itex]\omega[/itex] r
    or
    dr/r = [itex]\omega[/itex] dt
    ln r = [itex]\omega[/itex] dt + c'
    r = e^([itex]\omega[/itex]t + e^c') = C e^[itex]\omega[/itex]t
    that is the radius versus time and C is the initial radius at t = 0

    theta = theta at t =0 +[itex]\omega[/itex]t = To + [itex]\omega[/itex] t
    for simplicity call To = 0
    so
    theta = [itex]\omega[/itex] t9
    or R = C e^[itex]\omega[/itex]t (A1 cos [itex]\omega[/itex]t + A2 sin [itex]\omega[/itex]t)
    where A1 and A2 depend on the theta at t = 0

    Is that correct?
     
  7. Aug 4, 2011 #6

    kuruman

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    It is not correct. You are supposed to get exponentials, not sines and cosines. Why don't you start with Newton's Second Law as you wrote it down, cancel the masses on each side, move everything to the left side so that you have in the radial direction
    [itex]\frac{d^2r}{dt^2}-\omega^2r=0[/itex]
    That's the diff. eq. to solve.
     
  8. Aug 4, 2011 #7
    ok so
    [itex]\omega[/itex] = sqrt(a/r)

    d^2r/dt^2 = a^2/r

    ???
     
  9. Aug 4, 2011 #8

    kuruman

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    I guess you have not taken a course on differential equations. So time for plan B. What if you take the expression for r(t) as given by the problem and substitute it in the differential equation. If r(t) is indeed a solution, you should end up with 0 = 0.
     
  10. Aug 4, 2011 #9
    Right you are i am still unaware of differential equations, i dont get what you're trying to get me to do here you want me to plug in r(t)=A1ebt +A2e–bt into d^2r/dt^2 = [itex]\omega[/itex]^2r? Do i have to take the double derivative of r(t) first?
     
  11. Aug 4, 2011 #10

    kuruman

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    Yes, take the double derivative first and then see if it is equal to ω2r. If it isn't, what must be true so that it is? This may sound confusing, but do take the double derivative and, if you don't see what is going on, we'll take it from there.
     
  12. Aug 4, 2011 #11
    ok so i took the double derivative and it equals
    r"(t) = A1b2ebt+A2b2e-bt

    whats next boss?
     
  13. Aug 4, 2011 #12

    kuruman

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    Like I said, is this equal to ω2r? If it isn't, what must be true so that it is?
     
  14. Aug 4, 2011 #13
    I don't get how those equal zero, none of the values cancel out, wouldn't i need another equation that relates some of the values in order to be able to cancel them out?
     
  15. Aug 4, 2011 #14

    kuruman

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    Suppose you factor b2 that appears in both terms on the right. What does the equation become then?
     
  16. Aug 4, 2011 #15
    b2(A1ebt+A2e-bt)= [itex]\omega[/itex]2r

    b2(A1ebt+A2e-bt)= [itex]\omega[/itex]2(A1ebt+A2e-bt)

    b2= [itex]\omega[/itex]2

    this is where i am stuck how do i prove 0 = 0?
     
    Last edited: Aug 4, 2011
  17. Aug 5, 2011 #16

    ehild

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    As Kuruman is temporary absent I take his role over :smile:.

    You arrived to b22. That means the constant b has to be equal to the angular frequency ω of the rotation. If it is so the function r(t) =A1 ebt+A2e-bt is the displacement as function of time along the length of the rod.
    You need to write up ω in terms of f, revolutions per second.

    ehild
     
  18. Aug 5, 2011 #17
    Thank you very very much!

    so [itex]\omega[/itex]= 2[itex]\pi[/itex]f? Do i plug that back into r(t)?
     
  19. Aug 5, 2011 #18

    ehild

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    The problem asked to show that the function is really the displacement. You have proved that it is, with the condition that b=2πf. It is enough.

    You need to say something about the constants A1 and A2. What do you think?

    ehild
     
  20. Aug 5, 2011 #19
    I am not sure why this is...why is b so important that we had to relate it to [itex]\omega[/itex] just to show that is the position function

    Any ways...ummm good question would these be something like this?

    2[itex]\pi[/itex]ft(A1-A2)?
     
  21. Aug 5, 2011 #20

    ehild

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    What are those "these" which are equal to 2πft(A1-A2)?

    ehild
     
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