# Homework Help: Rotation - Anuglar speed

1. Oct 22, 2008

### maniacp08

Rotation -- Anuglar speed

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

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?

Last edited: Oct 22, 2008
2. Oct 22, 2008

### Staff: Mentor

Re: Rotation -- Anuglar speed

But the diameter = 2.2 m.
Good. (About an axis through the center.)
How did you get this? (You made an error in applying the parallel axis theorem.)

3. Oct 22, 2008

### maniacp08

Re: Rotation -- Anuglar speed

Oh, Doh!
I was doing a same problem but with different numbers so please ignore the R = .75m, suppose to be R = 1.1m

I went to tutoring and this is what the tutor did, I didn't really understood it.

Parallel Axis Theorem is I = Icm + Mh^2
Im not sure what to do.

4. Oct 22, 2008

### Staff: Mentor

Re: Rotation -- Anuglar speed

For this very problem? If so, he goofed.
Right. You know Icm. h is just the distance between the cm and the parallel axis that you want to use. What's that distance for this problem?

5. Oct 22, 2008

### maniacp08

Re: Rotation -- Anuglar speed

Icm = MR^2
distance between the cm and parallel axis would be R right?

6. Oct 22, 2008

### Staff: Mentor

Re: Rotation -- Anuglar speed

Right!

7. Oct 22, 2008

### maniacp08

Re: Rotation -- Anuglar speed

So Parallel Axis Theorem is I = Icm + Mh^2
I = MR^2 + MR^2
I = 2MR^2

So I can still apply the energy of conservation equation?
Ui + Ki = Uf + Kf
MgR + 0 = 0 + 1/2 (2MR^2)*omega ^2
MgR = 1/2 (2MR^2)*omega ^2

omega = square root of g/R
this is the maximum angular velocity
root of 9.81m/s/1.1m = 2.986331203 approx 3.0 rad/s?

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 (2MR^2)*omega ^2

and solve for angular velocity as I did above?

8. Oct 22, 2008

### Staff: Mentor

Re: Rotation -- Anuglar speed

Perfect!

Careful. You're trying to find the required initial angular speed--you have it as zero. The speed at the top of the motion is what? (It just barely makes it over.)

9. Oct 22, 2008

### maniacp08

Re: Rotation -- Anuglar speed

Oh, your right, Im solving for initial angular speed. So would Kf be 0? Since all we care if it gets barely over?

Ui + Ki = Uf + kf
MgR + 1/2(2MR^2)*omega initial ^2 = 2MgR + 0

Also Doc, when would I use the parallel axis theorem on problems? Like how would I know I should use it?

Last edited: Oct 22, 2008
10. Oct 23, 2008

### Staff: Mentor

Re: Rotation -- Anuglar speed

Looks good.

Use the parallel axis theorem whenever you need to find the rotational inertia of an object about an axis that doesn't pass through the center of mass. Usually you can look up or calculate the Icm of an object. But for a given problem the needed axis of rotation could be anywhere.

11. Oct 23, 2008

### maniacp08

Re: Rotation -- Anuglar speed

Thanks a lot Doc!