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Dynamics - spinning skater

  • Thread starter gupster
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  • #1
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Practice question for a general physics exam - I'm a 4th year undergraduate but have mostly taken astrophysics courses this year so am a bit stuck going back to general physics...

Homework Statement



A skater spins with angular velocity [tex]\omega = 6 rad/s[/tex] with his arms extended. How fast will he spin with his arms by his sides? Treat the skater's body as a uniform cylinder of radius R = 20cm, approximate his arms as thin uniform rods of length L = 70 cm and mass m = 4.5 kg. His total mass excluding arms is M = 70 kg.

After ceasing to spin the skater now steps onto the outer edge of a large uniform disk of radius [tex]R_{disk} = 2.5 m[/tex] and mass [tex]M_{disk} = 500 kg[/tex]. Assume that the disk is mounted on a frictionless bearing with a vertical axis of rotation and is initially at rest. If the skater begins skaing around the edge of the disk at a speed of 2 m/s how fast does the disk turn and what is its angular momentum?

Homework Equations



For a body of mass M with a moment of inertia I about an axis through its ventre of mass the
moment of inertia about a parallel axis a distance d from the first is [tex]I + Md^{2}[/tex]

The moment of inertia of a uniform cylinder of mass M, radius R and length L rotating about its axis is [tex]\frac{1}{2}ML^{2}[/tex]

The Attempt at a Solution



I think I've done the first part. I called the situation with the arms outstretched 1 and arms by his side 2 so

[tex]I_{1} = \frac{1}{2}MR^{2} = \frac{1}{2}(79)(0.2)^{2} = 1.58[/tex]
[tex]I_{2} = \frac{1}{2}MR^{2} + 2(\frac{1}{12}ML^{2} +Md^{2})= \frac{1}{2}(70)(0.2)^{2} + 2(\frac{1}{12}(45)(0.7)^{2} + (4.5)(0.55)^{2})= 7.798[/tex]

[tex]E = \frac{1}{2}I_{1}\omega_{1}^{2} = \frac{1}{2}I_{2}\omega_{1}^{2}[/tex]

[tex]\omega_{2} = 13.33 rad/s[/tex]

Its the second part I'm struggling with. I've said

[tex]v = r\omega[/tex] and using conservation of momentum

[tex]v_{disk} = \frac{m_{s}v_{s}}{m_{disk}} = 0.316 m/s [/tex]

[tex]\omega_{s} = \frac{v_{s}}{r} = 0.83 rad/s [/tex]

Is this right because I'm not convinced...
 

Answers and Replies

  • #2
Doc Al
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I think I've done the first part. I called the situation with the arms outstretched 1 and arms by his side 2 so

[tex]I_{1} = \frac{1}{2}MR^{2} = \frac{1}{2}(79)(0.2)^{2} = 1.58[/tex]
[tex]I_{2} = \frac{1}{2}MR^{2} + 2(\frac{1}{12}ML^{2} +Md^{2})= \frac{1}{2}(70)(0.2)^{2} + 2(\frac{1}{12}(45)(0.7)^{2} + (4.5)(0.55)^{2})= 7.798[/tex]
Looks like you reversed your notation. I_1 looks like an attempt at I for arms by his side.

What's the I for the torso? That should not change.

What's the I for the arms in each case?

[tex]E = \frac{1}{2}I_{1}\omega_{1}^{2} = \frac{1}{2}I_{2}\omega_{1}^{2}[/tex]
Energy is not conserved, but something else is.
 
  • #3
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Oops yeah that notation is the wrong way round. I can't seem to edit my original post though.

So I do the moment of inertia for body+arms for each case and then assume momentum is conserved? That's the only other thing I can think of that is conserved! I'll have a go at that now then. I feel like I'm being really stupid not being able to do this but I was taught this stuff 3 and a half years ago and I don't think I ever fully grasped it!
 
  • #4
Doc Al
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So I do the moment of inertia for body+arms for each case and then assume momentum is conserved?
Yes. And it's angular momentum that's conserved. Assuming no friction, then there's no external torque and thus angular momentum is conserved.
 
  • #5
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I've hit a snag when trying to calcuate the moment of inertia for when his arms are by his side because I don't know the radius of his arms. I've said

I_arm = I + Md^2 where I = MR^2 and d is the distance from the axis i.e. R_body+R_arm
 
  • #6
Doc Al
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The arms are modeled as thin rods, not cylinders. Radius is small and irrelevant.
 

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