Ring oscillation formula derivation

[xzibition8612]

Homework Statement

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

T=2π√[d/g+(ΔR)^2/4Rg]

Homework Equations

I'm not sure...

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.

 

[anathema]

Hello,

To derive the period for a thick ring, we first need to understand its motion. The ring will be rotating about its center of mass, with a radius of R and a thickness of ΔR. We can also assume that the ring is rotating in a uniform gravitational field, with a strength of g.

To begin, let's consider the motion of a single point on the ring. This point will have a tangential velocity of v = ωr, where ω is the angular velocity of the ring. As the ring rotates, this point will experience a centripetal acceleration, given by a = v^2/r = ω^2r.

Now, let's consider the forces acting on this point. We know that there is a gravitational force pulling the point towards the center of the ring, with a magnitude of mg. There is also a normal force acting on the point, perpendicular to the surface of the ring, which is responsible for providing the centripetal acceleration. This normal force can be found using Newton's second law: ΣF = ma, where ΣF is the sum of all forces acting on the point.

Since the point is moving in a circular path, we can use the radial component of this equation: N - mgcosθ = mω^2r, where θ is the angle between the normal force and the gravitational force. We can rewrite this as N = m(ω^2r + gcosθ).

Now, let's consider the moment of inertia of the ring. Since the ring has a thickness, we need to use the parallel axis theorem to find the moment of inertia about the center of mass. This gives us I = Icm + md^2, where Icm is the moment of inertia about the center of mass and d is the distance from the center of mass to the point on the ring.

Plugging in our expression for the normal force and our moment of inertia into the equation T = 2π√(I/mgr), we get T = 2π√[(Icm + md^2)/mgr]. We can simplify this by dividing both the numerator and denominator by mr, giving us T = 2π√[Icm/mr + d^2/g].

Now, let's consider the moment of inertia about the center of mass. We can use the parallel axis theorem again to find that Icm = Icm + MR^