Olympian Midori Ito
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.
Angular Momentum Conservation
Angular momentum characterizes an object's resistance to change in
rotation. The basic idea is the same as with
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.
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.
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.
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
We need to
figure out the moment of inertia Iout when her
arms and a leg are out (and she's spinning slowly) and
Iin when her arms and leg are in (and she's
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 Itorso of her torso is the same in both
cases, and it's given by
factor of ½ comes in because not all her torso mass is a
distance rtorso 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 Itorso = 0.2 kg m².
Now, when her arms and extra leg are in, she just has that extra mass
at a distance rtorso 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 Iin:
Iin = Itorso + 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
rarm 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
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 mleg
= 5 kg and marm = 2.5 kg. With these estimates,
the arms contribute 0.9 kg m² and the legs contribute 2.5 kg
m² to Iout. So, we wind up with
Iout = Itorso +
½mleg = 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 Lin = Lout, or
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.
Iout/Iin = 12
- rotational inertia of mass at distance r: I = mr²
- angular momentum: L = I
- 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.