How Do You Calculate Maximum Angular Velocity of a Rotating Ring?

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To calculate the maximum angular velocity of a rotating ring, energy conservation principles are applied, where potential energy converts to kinetic energy at the lowest point of the swing. The moment of inertia for the ring pivoted at its perimeter is derived using the parallel axis theorem, resulting in an expression of 2MR^2. The maximum angular velocity is found to be approximately 3.0 rad/s when substituting the radius of 1.1 m. For the minimum initial angular velocity required to complete a full 360°, the energy conservation equation is adjusted to account for the potential energy at the top of the motion. Understanding when to use the parallel axis theorem is crucial for calculating rotational inertia about non-central axes.
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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?
 
Last edited:
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maniacp08 said:
R = .75m
But the diameter = 2.2 m.
Moment of inertia for the ring is MR^2
Good. (About an axis through the center.)
Moment of the inertia of the pivot = 1/2 MR^2 + MR^2 = 3MR^2/2
How did you get this? (You made an error in applying the parallel axis theorem.)
 


Oh, Doh!
I was doing a same problem but with different numbers so please ignore the R = .75m, suppose to be R = 1.1m
Moment of the inertia of the pivot = 1/2 MR^2 + MR^2 = 3MR^2/2
How did you get this? (You made an error in applying the parallel axis theorem.)

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.
 


maniacp08 said:
I went to tutoring and this is what the tutor did, I didn't really understood it.
For this very problem? If so, he goofed.
Parallel Axis Theorem is I = Icm + Mh^2
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?
 


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


maniacp08 said:
Icm = MR^2
distance between the cm and parallel axis would be R right?
Right!
 


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?
 


maniacp08 said:
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?
Perfect!


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
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.)
 


Oh, your right, I am 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?
 
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  • #10


maniacp08 said:
Oh, your right, I am 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
Looks good.

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


Thanks a lot Doc!
 
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