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Period of Thick Ring

  1. Nov 10, 2008 #1
    I can't seem to make the necessary connections in this problem.

    Complete the algebra to show that the period for a thick ring ( the difference between the outer and inner radius is very large) is:

    T = 2π √[(d/g) + ((ΔR)^2/4Rg)]

    d= diameter of the ring


    Ro (outer radius)= R + ΔR/2
    Ri (inner radius) =R - ΔR/2

    I know that Icm(moment of inertia about the center of mass) = 1/2 M (Ro^2 + Ri^2)

    and for a thin ring where Ro & Ri are nearly identical, I=1/2MR^2

    Parallel Axis theorem: I total = Icm + MD^2

    T = 2π √(d/g) = 2π √(I/mgr) for a thin ring

    I tried to find I and substitute it into the equation 2π √(I/mgr), but I did not get the correct answer. Can anyone tell me what the key to this problem is?
  2. jcsd
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