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A uniform ring 2.2 M in diameter is pivoted at one point on its perimeter so that it is free to rotate about a horizontal axis. Initially, the line joining the support point and center is horizontal.

(a) If the ring is released from rest, what is its maximum angular velocity?

rad/s

(b) What minimum initial angular velocity must it be given if it is to rotate a full 360°?

rad/s

What I did:

I used energy of conservation on this.

The maximum velocity would be when the potential energy is at 0 since it is all converted into kinetic energy.

R = 1.1m

Moment of inertia for the ring is MR^2

Moment of the inertia of the pivot = 1/2 MR^2 + MR^2 = 3MR^2/2

Ui + Ki = Uf + Kf

MgR + 0 = 0 + 1/2 (3MR^2/2)*omega ^2

MgR = 1/2 (3MR^2/2)*omega ^2

omega = square root of 2(g/3R)

V = omega * R = square root of 2(gR/3R)

It is at the bottom when it is 2R so

V= omega * 2R = square root of 4(gR/3)

What did I do wrong?

For Part B:

To find the minimum velocity to do a full 360, can I do this:

Ui + Ki = Uf + kf

MgR + 0 = 2MgR + 1/2 (3MR^2/2)*omega ^2

and solve for velocity as I did above?

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# Homework Help: Rotation - Anuglar speed

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