1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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: 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.

  2. jcsd
  3. Nov 24, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper

    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?

  5. Nov 24, 2009 #4


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

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

    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

    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


    User Avatar
    Science Advisor
    Homework Helper

    (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


    User Avatar
    Science Advisor
    Homework Helper

    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.


    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


    User Avatar
    Science Advisor
    Homework Helper

    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

    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 :(
  22. Nov 26, 2009 #21
    Torque o.0...

    Nevertheless, I finally got the answer lol.

    Thanks for your help guys :smile:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook