Dissonance in E said:
This is somewha related so i didnt want to make a new thread. Could someone explain the mechanical principles and math behind the classic example of a figure skater pulling her arms close to her body and thus increasing her spin velocity?
It has to do with conservation of angular momentum. Just as momentum "quantifies" translational motion, so does angular momentum quantify rotational motion. The angular momentum (L) of a particle is given by:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
where
r is a vector from the centre of rotation to the particle, and
p is its momentum. For the simple case of motion in a circle (edit: at a constant speed),
r is always perpendicular to
p, and so the magnitude of the cross product is simply given by
|L| = L = |r||p| = rp = rmv
where |
r| = r is the radial distance to the particle from the centre of rotation.
You can see that as r decreases, v must increase for L to be conserved. For extended bodies such as figure skaters, the actual calculation of the angular momentum is more complicated, but the principle still applies, because we can consider the figure skater as an aggregate of a large number of such particles, each orbiting about the centre at a different distance (depending on its position in the figure skater's body).
If you're wondering intuitively why a large rotating body has MORE angular momentum than a small one rotating at the same speed, it has to do with the fact that it's harder (it takes more work) to get the larger body rotating at that speed in the first place. For translational motion, an object's inertia is quantified solely by its mass. But an object's
rotational inertia (how much it resists a change in its state of rotational motion) depends not only on its mass, but also on
how that mass is distributed in space. (It is quantified by the moment of inertia). This ties in with the idea of torque...that it's easier to make something spin if you push on it at a point farther away from the spin axis (e.g. it's easier to push open a door using the handle than it would be if you pushed on it at a point close to the hinges). In fact, just as force is the rate of change of momentum (dp/dt) so too is torque the rate of change of angular momentum (dL/dt). That's why, for a system like the figure skater, if no external torques are applied, angular momentum must be constant. There are many such analogies between dynamical quantities describing translational motion, and those describing rotational motion...if you spend some time looking at the area of classical mechanics known as rotational dynamics, you will see how everything ties together.
