Thick ring period formula (pendulum)

[xzibition8612]

1. Homework Statement

Derive and show that the period for a thick ring would be

T=2π√[d/g+(ΔR)^2/4Rg]


2. Homework Equations

"the ring pendulum consists of a thin metal ring that can be suspended from a knife edge. The ring has inner radius Ri and outer radius Ro. A notch is cut into the ring so that it can be suspended without slipping. The center of the notch is a distance r from the center of mass of the ring. The average of Ri and Ro is equal to r. The moment of inertia of the ring about its center of mass is

I=1/2[M(Ro^2+Ri^2)]

At this point you can make an approximation. If the ring is thin relative to its radius, that is if Ri and Ro are nearly equal and the difference between them is small compared to Ri, the moment of inertia of the ring about its center of gravity can be approximated by I=MR^2. The parallel axis theorem says that the moment of inertia about an axis parallel to that going through the center of mass is Itotal=Icm + MD^2 where D is the distance from the center of mass to the axis of rotation. In the case if the rings in this experiment D equals R. You can use the parallel axis theorem to find the moment of inertia of the ring about the pivot point. The period of a thin ring should be T=(2π√d/g) where d is the average diameter of the ring. It equals 2R or 1/2(do+di). di and do are the inner and outer diameters of each ring and d is the average of the two."



3. The Attempt at a Solution

Obviously, delta R means the difference between Ri and Ro would be considerable. So ΔR=Ro-Ri. Then Ro=R+ΔR/2 and Ri=R-ΔR/2
I also know that T=2π√(I/mgr) and Itotal=Icenter of mass + MR^2...
Can you guys help me? I'm lost here. Thanks a lot.

 

[Eynstone]

Let J = I + Mr^2 be the moment of inertia about the nock. If q denotes the angular displacement, you have
torque = Jq'' = -Mg(rsinq) ~= -Mgrq.
So, T = 2pi*sqrt(J/Mgr). Just simplify.