Figure Skater Spins

Everyone has seen the classic "scratch spin" in figure skating, where the skater draws her arms and a leg in and speeds up tremendously. This is the result of conservation of angular momentum: as the skater reduces her rotational inertia by pulling her arms and leg in, her rotation speed must increase to maintain constant angular momentum. Angular momentum conservation plays a VERY important role in all figure skating routines.

Here's a 564 kB AVI movie of Paul Wylie doing a scratch spin, at 30 frames/sec.

Here's an 849 kB AVI movie of Scott Davis doing a scratch spin, at 25 frames/sec. These AVIs were taken from Kevin Anderson's Technical Figure Skating page.

Angular Momentum Conservation

Angular momentum characterizes an object's resistance to change in rotation. The basic idea is the same as with linear momentum: moving things like to keep moving, and to change their motion we have to apply a force. If no force is present, then momentum doesn't change, ie. it is conserved. In the case of rotation, the force is called torque: when, for example, you pull the string on a top, you are applying a torque to make it speed up. Its angular momentum increases. It slows down after being released due to frictional torque. A spinning figure skater has a nearly frictionless contact with the ice, and there is little net torque on her body. Her angular momentum is nearly conserved.

Rotational Inertia

For straight-line motion, inertia is mass. For rotational motion, it's a bit more involved. It's harder to make a given mass rotate around an axis that it's far from than one that it's close to. The rotational inertia, or moment of inertia, I, of a single mass m rotating a distance r around an axis (like a planet around the Sun or a rock on a string) is given by

I = mr²

Note that rotational inertia increases as the square of the distance from the axis: if you double the distance of a mass from the axis of rotation, you quadruple the rotational inertia. This is why such a minor change such as a skater's leg position has such a huge effect on her rotational speed.

Rotational Speed

The other parameter of rotational motion is rotational speed, or angular velocity,

. This is the rate of rotation, expressed in radians/sec, revolutions/minute (RPM) and other units. A complete rotation is 2

radians, so one revolution per second is an angular velocity of 2

rad/s.

Angular Momentum

Armed with rotational inertia and angular velocity, we can write the expression for angular momentum, L:

L = I

So, if angular momentum is conserved, and one factor like I changes, the other factor (

in this case) must change to compensate.

Example: Figure Skater Spins

When a figure skater draws her arms and a leg inward, she reduces the distance between the axis of rotation and some of her mass, reducing her moment of inertia. Since angular momentum is conserved, her rotational velocity must increase to compensate. Let's estimate how much she speeds up by estimating the change in her rotational inertia.

We need to figure out the moment of inertia I_out when her arms and a leg are out (and she's spinning slowly) and I_in when her arms and leg are in (and she's spinning fast). A crude approximation of the the skater's shape, good enough for the purpose here, says that she is a solid cylinder made up of most of her mass plus three rods representing her arms and a leg. The moment of inertia I_torso of her torso is the same in both cases, and it's given by ½M_torsor²_torso (the factor of ½ comes in because not all her torso mass is a distance r_torso away from the axis, it's only halfway out on the average).

A typical female skater has mass of around 50 kg, I'd guess. I'd also guess that about 40 kg is in her torso plus one leg. Finally, I'll guess that the appropriate radius of our figure skater cylinder is 0.1 m. That means that her torso moment of inertia I_torso = 0.2 kg m². Now, when her arms and extra leg are in, she just has that extra mass at a distance r_torso away from the axis. So, let's just add mr² = 0.1 kg m², with m = 10 kg and r = 0.1 m, to get I_in:

I_in = I_torso + mr² = 0.3 kg m²

Now when her arms and a leg are out, they are further from her rotation axis. If her arms are straight out they have moment of inertia ½(2m_arm)r²_arm, where r_arm is the distance from the axis to her fingertips. I'll guess that's about 0.6 m. If her leg is straight out, it contributes ½m_legr²_leg. I'll guess her leg is 1.0 m long. All that's left is to decide how to divide her 10 kg non-torso mass into her arms and leg. I'll guess that one leg is about equal to two arms, so m_leg = 5 kg and m_arm = 2.5 kg. With these estimates, the arms contribute 0.9 kg m² and the legs contribute 2.5 kg m² to I_out. So, we wind up with

I_out = I_torso + ½(2m_arm)r²_arm + ½m_leg = 3.6 kg m²

From this estimate, the skater's moment of inertia is much larger when her arms and one leg are out, all due to the r² dependence of I. We can now estimate how much she speeds up by pulling her arms and leg in by applying conservation of angular momentum, which says L_in = L_out, or

I_in_in = I_out_out => _in /_out = I_out/I_in = 12

A typical rotation speed with arms and a leg out is 2 revolutions per second; the above estimate says she'll spin up to 24 revs/sec with her arms and leg pulled in! If you look carefully at the AVI movies linked at the top of this page, you'll see that the increase isn't quite that drastic, but this calculation is just an estimate, anyway.

Equations

rotational inertia of mass at distance r: I = mr²
angular momentum: L = I

Summary

Rotational inertia characterizes the resistance to change in rotation.
Torque is the type of force which makes something rotate.
Angular momentum is conserved if there is no net torque on an object. A change in rotational inertia is compensated by a change in rotation speed.

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Sam Hokin / samh@fusion.kth.se