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Derive the Period of an Oscillating Sphere

  1. Nov 24, 2009 #1
    1. The problem statement, all variables and given/known data
    Derive the period of a sphere oscillating in a hemispherical concave surface at small angles.


    2. Relevant equations
    Net Torque = I x alpha
    Delta E_m = 0


    3. The attempt at a solution
    The thing is, tried using the conservation of mechanical energy to get an expression and isolate for omega (7/10 M r^2 omega^2 = 0). Then I differentiated in order to get omega and alpha, and substituted alpha from the net torque equation (in this case, gravity = - Mg (R - r) sin theta) with the moment of inertia of the sphere. But then it gets really messy and I can't solve for omega without a theta. The bottom line is I don't think I quite understand whether circular motion plays a role here, and whether for the kinetic energy, the velocity in translational kinetic energy is (R-r) x omega (due to circular motion) or just r x omega, and whether the omega in rotational kinetic energy is v/(R-r) or just v/r.

    Thanks
     
  2. jcsd
  3. Nov 24, 2009 #2

    tiny-tim

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    Hi SpringPhysics! :smile:

    (have an alpha: α and an omega: ω and a theta: θ and try using the X2 tag just above the Reply box :wink:)
    Yes, conservation of mechanical energy should give you θ'2 + f(θ) = constant …

    the question says "at small angles", so now replace cos or sin by polynomials, and integrate. :wink:

    (btw, it isn't (R-r)sinθ, it's (R-r)(1-cosθ))
     
  4. Nov 24, 2009 #3
    Oh, thanks.

    So torque is completely irrelevant in this question? I attempted to solve using the conservation law, but I am unsure about the tangential velocity of the sphere - is it based on the radius of the bowl or just the sphere? Also, if I use the small-angle approximation, the gravity portion of the mechanical energy completely disappears. And why would we integrate? Shouldn't we differentiate in order to get an equation in terms of α and θ, in order to relate it to the SHM equation?

    Thanks.
     
  5. Nov 24, 2009 #4

    tiny-tim

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    Hi SpringPhysics! :smile:
    It's irrelevant if you use conservation of energy, it's very relevant if you dont! :wink:
    The speed of the point of contact is zero.
    No it doesn't.
    Yeah, ok, differentiate. :smile:
     
  6. Nov 24, 2009 #5
    Sorry, I'm still very confused.

    Suppose that I use the conservation of energy. Then if I set the zero of potential at the base of the bowl, the change in potential energy would be
    mg[r - (R-r)(1 - cos \theta].
    If I use the small angle approximation, then cos \theta becomes 1 and the change in potential energy becomes just mgr. Then the kinetic energy is the sum of translational energy and rotational energy
    1/2 mv2 + 1/2 I\omega2.
    Then the above becomes
    1/2 m(r\omega)2 + 1/2 (2/5 mr2)\omega2
    Even without solving for omega, there is no R variable here!

    Suppose I use torque. Then
    \alpha = -mg(R-r)sin\theta / (2/5 mr2) = -5/2 g(R-r)/r2.
    When I integrate this from a small angle \theta to 0, I get
    -5/2 g(R-r)/a2 (cos \theta - 1).
    If I use the small angle approximation, cos \theta = 1 and then the entire thing becomes zero.

    What am I doing wrong?
     
  7. Nov 24, 2009 #6

    tiny-tim

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    Hi SpringPhysics! :smile:

    (what happened to that θ ω and α i gave you? :confused:)
    No, cosθ becomes 1 - θ2/2. :smile:
    No, v depends on R also. :wink:
     
  8. Nov 24, 2009 #7
    Sorry, I couldn't make it work (more like, I don't know how to make it work).

    According to my textbook though, cos theta = 1 :confused: It doesn't mention the second derivative approximation (is there a way to derive it?).

    I'll try it out first, thanks.
     
  9. Nov 24, 2009 #8

    tiny-tim

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    Yes, cosθ = √(1 - sin2θ) ~ √(1 - θ2) ~ 1 - θ2/2.

    Alternatively:
    cosθ = 1 - θ2/2 + θ4/24 - … = ∑ (-1)nθ2n/(2n)!
    sinθ = θ - θ3/6 + θ5/120 - … = ∑ (-1)nθ2n+1/(2n+1)!
     
  10. Nov 24, 2009 #9
    Thanks.

    So I used the conservation of mechanical energy.

    PEi = mg[R-(R-r)cosθ]
    PEf = mgr

    KEi = 0
    KEf = 1/2 mv2 + 1/2 I ω2

    Then by equating the negative change in PE to the change in KE, I eventually end up with

    g(R-r)(1-cosθ) = 1/2[(R-r)ω1]2 + 1/5 r2 ω22

    I know that somehow I have to relate the two terms for kinetic energy, but I do not know how.

    EDIT: I finally got something better

    1 - cos θ= 7/10 (R-r)/g ω2

    Now do I differentiate? I have to somehow get

    T = 2 pi [7/5 (R-r)/g]1/2
     
    Last edited: Nov 24, 2009
  11. Nov 25, 2009 #10

    tiny-tim

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    (just got up :zzz: …)

    Now put 1 - cos θ ~ θ2/2. :wink:

    hmm … that gives θ' proportional to θ, which isn't SHM :redface:

    ah, I think you're missing a constant, you've got ω= 0 when θ = 0, which can't be right.
     
  12. Nov 25, 2009 #11
    Good morning :smile:

    A constant? If I differentiate, I get [tex]\alpha[/tex] = 5g/7(R-r) sin [tex]\theta[/tex]. Using the small angle approximation, I get [tex]\alpha[/tex] = 5g/7(R-r) [tex]\theta[/tex], which mirrors the SHM equation right?

    How can there be a constant? (Although I agree that it can't be true that when [tex]\theta[/tex] is zero, [tex]\omega[/tex] is zero as well.)

    Can someone please help me?
     
    Last edited: Nov 25, 2009
  13. Nov 25, 2009 #12

    tiny-tim

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    Good afternooooon. :wink:

    No, you get α = a positive constant times θ, which is no good (it's exponential, ie positive feedback) …

    you need α = a negative constant times θ for SHM (so it's negative feedback, or retarding).

    I'm not sure, but I think you must have got your PE negative when you did KE + PE = constant.

    (btw, it didn't occur to me before, but you can also use one of the standard trigonometric identities before approximating … 1 - cosθ = 2sin2(θ/2) :wink:)

    (now i'm off to watch sean the sheep :tongue2:)
     
  14. Nov 25, 2009 #13
    I actually didn't start with KE + PE = constant, I started with the change in KE + change in PE = 0. So then the change in KE is just the negative of the change in PE.

    EDIT:

    It worked when I placed the zero of potential at the top of the concave surface (as opposed to at the bottom). But I don't know why it doesn't work if it's at the bottom.
     
    Last edited: Nov 25, 2009
  15. Nov 26, 2009 #14
    Sorry for the double post, but I just found out that I used a relation incorrectly.

    So I have that the displacement of the circumference of the ball is equal to [tex]\phi[/tex] times r, and also equal to R times [tex]\theta[/tex]. So I differentiate and get a relation between the angular velocities, and isolate for the angular velocity for the ball. I substitute that into the PE-KE equation, and then I'm stuck. What do I do next?
     
  16. Nov 26, 2009 #15

    tiny-tim

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    Hi SpringPhysics! :smile:
    Sorry, you've lost me …

    what is the PE-KE equation you got stuck with?
     
  17. Nov 26, 2009 #16
    Hey, I have to derive the exact same equation for my lab too O_O
     
  18. Nov 26, 2009 #17
    =D

    I placed the zero of gravity at the top of the concave surface at the centre.
    PEi = -mg(R-r)cos[tex]\theta[/tex]
    PEf = -mg(R-r)
    KEi = 0
    KEf = 1/5 mv2 + 1/2 I[tex]\omega[/tex]2

    So I = mr2
    and [tex]\omega[/tex]2 = (R-r)/r [tex]\omega[/tex]
    and v = (R-r)[tex]\omega[/tex]

    When you substitute them into [tex]\Delta[/tex]PE + [tex]\Delta[/tex]KE = 0, you get g(1-cos[tex]\theta[/tex]) = 7/10 (R-r)[tex]\omega[/tex]2

    When you differentiate with respect to time, you get gsin[tex]\theta[/tex] = [tex]\alpha[/tex] 7/5 (R-r)

    Obviously, alpha is not a negative number, so what am I doing wrong?
     
  19. Nov 26, 2009 #18
  20. Nov 26, 2009 #19
    Thanks, but I would still like to know what went wrong with my solution.
     
  21. Nov 26, 2009 #20
    You were trying to use trigonometry and torques, which overcomplicated matters and thus got you confused.

    It's ok, I've had my fair share of experiences where I try to apply unneeded concepts to a problem and make things harder for myself :(
     
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