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

In summary, the conversation discussed the rotation and angular speed of a uniform ring released from rest. The maximum angular velocity was found using the energy conservation equation, and the minimum initial angular velocity for a full 360° rotation was determined using the parallel axis theorem. The conversation also touched on when and how to use the parallel axis theorem in problem-solving.
  • #1
maniacp08
115
0
Rotation -- Anuglar speed

339698281.jpg


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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  • #2


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.)
 
  • #3


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.
 
  • #4


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?
 
  • #5


Icm = MR^2
distance between the cm and parallel axis would be R right?
 
  • #6


maniacp08 said:
Icm = MR^2
distance between the cm and parallel axis would be R right?
Right!
 
  • #7


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


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.)
 
  • #9


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?
 
Last edited:
  • #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!
 

What is rotation?

Rotation is the circular movement of an object around a fixed point or axis.

What is angular speed?

Angular speed is the rate at which an object rotates around a fixed point or axis. It is measured in radians per second or degrees per second.

How is angular speed different from linear speed?

Angular speed is the rate of change of angular displacement, while linear speed is the rate of change of linear displacement. In other words, angular speed measures how fast an object is rotating, while linear speed measures how fast an object is moving in a straight line.

What factors affect angular speed?

The factors that affect angular speed include the size and shape of the object, the distance from the axis of rotation, and any external forces or torques acting on the object.

How is angular speed measured?

Angular speed can be measured using a variety of instruments, such as an angular velocity sensor or a tachometer. It can also be calculated by measuring the angular displacement over a certain time period.

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