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!

Derive an Equation for Period of Ring Pendulum

  1. Dec 7, 2012 #1
    1. The problem statement, all variables and given/known data
    Apply the physical pendulum equation to a ring pivoted on its edge to derive the equation for the period of a ring pendulum for small oscillations about the pivot point. Include a diagram showing the restoring torque acting on a ring pendulum displaced from equilibrium.


    2. Relevant equations
    T=2∏√(I/mgd) physical pendulum equation
    I=I(COM)+mR^(2) Parallel Axis Thm


    3. The attempt at a solution
     
  2. jcsd
  3. Dec 8, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi Arrhenius7991! Welcome to PF! :wink:

    What is the torque?

    Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
     
  4. Dec 8, 2012 #3
    Ʃ(Torque)=Iα
    From Newton's Second Law: (Torque)=mgsinθ
    And I is given by Parallel Axis Thm: I=I(COM)+MR^(2), and I(COM)=MR^(2), the moment of inertia for a thin ring.
    α=ω^(2)x(max)
    ω=√(κ/I)=√(κ/2MR^(2))
    So, Mgsinθ=(2MR^(2))(ω^(2)x(max))
    Mgsinθ=(2MR^(2))((κ/(2MR^(2))x(max))
    Mgsinθ=(κ/x(max))

    And now I'm stuck.

    I know the equation ofr the period of a ring pendulum is: T=2∏√(I/κ)
     
  5. Dec 8, 2012 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    No, that's a force. What do you need to multiply it by to make it the torque?
    I don't understand that equation. What exactly are x and x(max) there? It appears to be saying [itex]\ddot\theta = constant ×\dot\theta^2[/itex], which doesn't look like SHM to me.
     
  6. Dec 8, 2012 #5
    T(Torque) = -κθ.

    And α(alpha)=angular acceleration=ω^(2)x(max), ω=angular speed, and x(max) is the amplitude.
     
  7. Dec 8, 2012 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I have no idea where you would get such an equation from.
    In SHM, θ=Asin(ct), say. So ## ω(t) = \dot\theta = Ac{cos(ct)} ## and ## α(t) = \ddot\theta = -Ac^2 sin(ct) = -c^2θ(t)##. I see no way that this will satisfy ## α(t) = B ω^2(t) = BA^2c^2 cos^2(ct)##
     
  8. Dec 8, 2012 #7
    Ok. So now what?
     
  9. Dec 9, 2012 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Let's try the torque once more. You wrote Torque=mgsinθ. I explained that was the force, and you needed to multiply it by something to get torque. Instead of doing that you responded "(Torque) = -κθ".
    If you want to know the torque about some point due to a force, what do you multiply the force by?
     
  10. Dec 9, 2012 #9
    You'd multiple the force by the distance the point is from the center of mass. Given by r. So, Torque=vector(r) x(Cross-Product) F(Force).
     
  11. Dec 9, 2012 #10

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    If you do that as a vector cross product (as you wrote next) then fine. But as scalars, it would be the distance from the axis to the line of force (i.e. measured perpendicularly to the force). Equivalently, you can use the whole distance but take the component of the force perpendicular to that.
    So what do you get for the torque?
     
  12. Dec 9, 2012 #11
    Torque would be mgrsinθ, r being the radius of the ring.
     
  13. Dec 9, 2012 #12

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    ok, and you can call that mgrθ (for small θ)

    now write the τ = Iθ'' equation …

    what do you get? :smile:

    (and remember that your axis of rotation is parallel to a diameter of the ring, not the axis of the ring)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook