Rotational Frequency [figure skater]

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SUMMARY

The discussion centers on calculating the new rotational frequency of a figure skater who changes her arm position while spinning. The skater, weighing 45 kg, initially spins at 1.0 rev/s with outstretched arms modeled as a cylindrical torso and two rod-like arms. Using the conservation of angular momentum and the Parallel Axis Theorem, participants calculated the moment of inertia for both the body and arms. The correct new rotation frequency, as derived through proper application of the equations, is 5.804 rev/s, not the initially assumed 3.5 rev/s.

PREREQUISITES
  • Understanding of angular momentum conservation
  • Familiarity with the Parallel Axis Theorem
  • Knowledge of moment of inertia calculations for cylinders and rods
  • Basic physics principles related to rotational motion
NEXT STEPS
  • Study the application of the Parallel Axis Theorem in rotational dynamics
  • Learn how to calculate moment of inertia for composite bodies
  • Explore angular momentum conservation in various physical systems
  • Review examples of rotational motion problems in 'Physics for Scientists and Engineers' by Randall D. Knight
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Physics students, educators, and anyone interested in understanding rotational dynamics and angular momentum in real-world applications, particularly in sports science and biomechanics.

otonh210
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Homework Statement



A 45kg figure skater is spinning on the toes of her skates at 1.0 rev/s. Her arms are outstretched as far as they will go/ In this orientation, the skater can be modeled as a cylindrical torso (40kg, 20cm average diameter, 160cm tall) plus two rod-like arms (2.5kg each, 66cm long) attached to the outside of the torso. The skater then raises her arms straight above her head, where she appears to be a 45kg, 20cm diameter, 200cm tall cylinder. What is her new rotation frequency, in revolutions per second?

Homework Equations



L=Iw
Li=Lf
Ii = Ibody + 2 Iarm

The Attempt at a Solution



I did conservation of L
(Ibody + 2 Iarm)(1 rev/s) = Iwholebody (wf)
Ibody = 1/2 Mr^2 = 1/2(40kg)(.1m)^2
Iarm = 1/3Mr^2 (?)
Iwholebody = 1/2Mr^2 = 1/2(45kg)(.1m)^2

When I did it this way, I got ~ 5.7 rev/s
The answer I am supposed to be getting is 3.5 rev/s
All the resources I am using to help me figure this out are not offering any new help.
Am I supposed to be using her height? Where?

Thank you very much.
 
Last edited:
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otonh210 said:
I did conservation of L
(Ibody + 2 Iarm)(1 rev/s) = Iwholebody (wf)
Ibody = 1/2 Mr^2 = 1/2(40kg)(.1m)^2
Good.
Iarm = 1/3Mr^2 (?)
This is the tricky part. The arms attach to the outside of the torso, not to the axis of rotation. You'll need the parallel axis theorem.
Iwholebody = 1/2Mr^2 = 1/2(45kg)(.1m)^2
Good.

Show the details of your calculations.
 
I'm unsure of how to apply the Parallel Axis Theorem, because I've never heard of it until now.
Do I just add the distance from the outside of the torso [attachment point] to the axis of rotation?

Iarm = 1/3(m)(l+r)^2?
Iarm = 1/3(2.5kg)(.66m+.1m)^2
Iarm = 0.4813
Ibody = 1/2(40kg)(.1m^2)
Ibody = 0.2

Iwholebody = 1/2(45kg)(.1m^2)
Iwholebody = 0.225It was mentioned to me that I should pretend there is only one arm?
Also, that I need to find the diameter of the arms?
Do I need the height in there? Or is that extraneous information?

Thank you for the reply!
 
Last edited:
I tried the math above, and it does not work.

Do I need to use a different equation for the arms?
Ibody+MD^2 (?)

If so, what do I use for D? The distance from the edge of the torso to the axis of rotation? The distance from the centre of mass of the arm to the axis of rotation? The distance from the far end of the arm to the axis of rotation?
 
otonh210 said:
I'm unsure of how to apply the Parallel Axis Theorem, because I've never heard of it until now.
Do I just add the distance from the outside of the torso [attachment point] to the axis of rotation?

Iarm = 1/3(m)(l+r)^2?
Iarm = 1/3(2.5kg)(.66m+.1m)^2
Iarm = 0.4813
No, that would be true if the arms were 0.76 long and were directly attached to the axis.

Ibody = 1/2(40kg)(.1m^2)
Ibody = 0.2

Iwholebody = 1/2(45kg)(.1m^2)
Iwholebody = 0.225
OK.

It was mentioned to me that I should pretend there is only one arm?
That's OK. (As long as that one arm has a mass of 5 kg.)
Also, that I need to find the diameter of the arms?
Nah.
Do I need the height in there? Or is that extraneous information?
You won't need it. (Note that the formula for rotational inertia of a cylinder does not contain the height.)

What's the initial rotational rate? 1 rev/sec? (You had written 1/0, a typo I presume.)
 
Is Parallel Axis Theorem Ic + MD^2?
I tried the above, with many different Ds, but none seemed to give me the right answer.

Yes, that is a typo. It is supposed to be 1.0 rev/s

Each arm has a mass of 2.5 kg. The person mentioned to me that they got the correct answer by pretending it was one giant arm of 5kg.
 
otonh210 said:
Is Parallel Axis Theorem Ic + MD^2?
Yes.
I tried the above, with many different Ds, but none seemed to give me the right answer.
I don't see how you can get the 3.5 rev/s either. Why do you think it's correct? (If this is a textbook problem, tell me the textbook.)

Yes, that is a typo. It is supposed to be 1.0 rev/s
OK.

Each arm has a mass of 2.5 kg. The person mentioned to me that they got the correct answer by pretending it was one giant arm of 5kg.
That's fine. Since the arms have the same shape, the rotational inertia of both is just twice that of one.
 
For my Physics class, we have sets of problems that we must hand in each week. They tell us the answers, so that we know we are doing it correctly [but not how to get there].

The answer they gave us for this problem was 3.5 rev/s. It appears to be 12.90, but I am unsure of which textbook. Possibly 'Mastering Physics' or 'Physics for Scientists and Engineers'.


For Parallel Axis Theorem, is the Ic just the Iarm [1/3Mr^2] and then I have to add MD^2, where D is the distance from the edge of the torso to the axis of rotation?
And then I add two of those to the Ibody?
 
otonh210 said:
Possibly 'Mastering Physics' or 'Physics for Scientists and Engineers'.
Who's the author?

For Parallel Axis Theorem, is the Ic just the Iarm [1/3Mr^2] and then I have to add MD^2,
Yes.
where D is the distance from the edge of the torso to the axis of rotation?
No, D is the distance between the center of mass and the axis.
And then I add two of those to the Ibody?
Yes.
 
  • #10
Ah, Mastering Physics is a program that comes with 'Physics for Scientists and Engineers' to help people study. Author for Physics for S&E is Randall D Knight, I believe.

Thank you very very much for your assistance!

I am now getting 8.22 rev/s as a final answer. Does this sound more correct to you?
 
  • #11
otonh210 said:
Ah, Mastering Physics is a program that comes with 'Physics for Scientists and Engineers' to help people study. Author for Physics for S&E is Randall D Knight, I believe.
I have his textbook, but not the Mastering Physics program.
I am now getting 8.22 rev/s as a final answer. Does this sound more correct to you?
I get a different answer. (I get an answer very close to the one you first posted.) What did you get for the I of the outstretched arms?
 
  • #12
For Iarm I did,

Iarm = (1/3)(m)(r^2) + (m)(d^2)
Iarm = (1/3(5kg)(.66m^2) + (5kg) (.43m^2)

Iarm = 1.6505


then,
[Ibody + Iarm]/Iwholebody = wf

[0.2 + 1.6505]/0.225 = 8.22 rev/s
 
  • #13
otonh210 said:
For Iarm I did,

Iarm = (1/3)(m)(r^2) + (m)(d^2)
Iarm = (1/3(5kg)(.66m^2) + (5kg) (.43m^2)
For a thin rod about its center of mass, the formula is (1/12)(m)(r^2).
 
  • #14
Okay. I thought you had to use the rod about one end, because they are attached to the torso which is rotating.

Thank you SO much for your help though. I appreciate it a lot!
 
  • #15
otonh210 said:
I thought you had to use the rod about one end, because they are attached to the torso which is rotating.
Whenever you use the parallel axis theorem, you start with the rotational inertia about the center of mass.
 
  • #16
Ah. I have clearly absorbed nothing from the lessons on rotation.

So,

Iarm = 1/12 (5kg)(.66m^2) + (5kg)(.43m^2)
Iarm = 1.106


wf = 5.804 rev/s
 
  • #17
otonh210 said:
So,

Iarm = 1/12 (5kg)(.66m^2) + (5kg)(.43m^2)
Iarm = 1.106


wf = 5.804 rev/s
Good. This is how I would do this problem, based on the description given.
 

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