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## 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...