# Homework Help: Dynamics - spinning skater

1. May 18, 2008

### gupster

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...

1. The problem statement, all variables and given/known data

A skater spins with angular velocity $$\omega = 6 rad/s$$ 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 $$R_{disk} = 2.5 m$$ and mass $$M_{disk} = 500 kg$$. 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?

2. Relevant 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 $$I + Md^{2}$$

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

3. 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

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

$$E = \frac{1}{2}I_{1}\omega_{1}^{2} = \frac{1}{2}I_{2}\omega_{1}^{2}$$

$$\omega_{2} = 13.33 rad/s$$

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

$$v = r\omega$$ and using conservation of momentum

$$v_{disk} = \frac{m_{s}v_{s}}{m_{disk}} = 0.316 m/s$$

$$\omega_{s} = \frac{v_{s}}{r} = 0.83 rad/s$$

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

2. May 18, 2008

### Staff: Mentor

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?

Energy is not conserved, but something else is.

3. May 18, 2008

### gupster

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. May 18, 2008

### Staff: Mentor

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. May 18, 2008

### gupster

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. May 18, 2008

### Staff: Mentor

The arms are modeled as thin rods, not cylinders. Radius is small and irrelevant.