How come ballerina rotate so fast when there is no external force?

Question

The question is where are they getting this torque ($\tau$) thats causing them to increase their angular acceleration ($\alpha$) and therefore increasing their angular velocity ($\omega)?$
Assuming it's frictionless ground(like on ice)

Answer 1

They initially kick the ground and receive an equal and opposite force from it (Newton III), that's where the initial torque comes from. They would not be able to get this from a frictionless surface.

Then, to spin even faster, they usually move their arms close to their chest. This decreases their moment of intertia ($I$) and hence increases their angular velocity ($\omega$), following conservation of angular momentum $L = I\omega$.

Answer 2

For those who are equation-minded, I think one of the comments in this thread gave away the key Newtonian trick here, and I think the question stems from a misconception about Newton's second law.

Usually, we phrase Newton's second law(s), ignoring the nitty-gritty of vector stuff, as $$ F_\text{net} = ma = m\frac{dv}{dt}\qquad \text{and}\qquad \tau_\text{net} = I\alpha = I\frac{d\omega}{dt} $$ ... which, for rigid bodies, apply to the centre of mass of the object, or for systems of point masses, apply to their centre of mass. By "net", we mean the net sum over all interactions with masses not part of the body or system. Clearly, when the ballerina (or ice skater) goes to spin faster, her $\omega$ changes, so $d\omega/dt\neq 0$, and since $I\neq 0$, you would suspect $\tau_\text{net}\neq0$ by some magical ground interaction.

But actually, the above equations are only true in a limited sense: when mass and moment of inertia don't change in time (which is a safe assumption made in your typical high-school mechanics riddle). In Newton's full second law, the derivative is taken for the entire right-hand side, not just (angular) velocity. Hence: $$ F_\text{net} = \frac{d(mv)}{dt} = \frac{dp}{dt} \qquad \text{and}\qquad \tau_\text{net} = \frac{d(I\omega)}{dt} = \frac{dL}{dt} $$ Aha! We can now live with $\tau_\text{net} = 0$, because that simply means $$ 0 =\frac{dL}{dt}\quad\iff\quad 0 =\frac{d(I\omega)}{dt}\quad\iff\quad 0 =\frac{dI}{dt}\omega + I\frac{d\omega}{dt} = \frac{dI}{dt}\omega + I\alpha $$ which yields $$ I\alpha = -\,\frac{dI}{dt}\omega. $$ Let's assume we're spinning in the direction that $\omega>0$ (so, to speed up, $\alpha>0$). Also, just like mass, $I >0$. This equation tells us the full story that we see in the ballerina or ice skater: when she speeds up, $\alpha>0$, so the left-hand side is completely positive. What must she do to achieve this? Make the right-hand side completely positive, which only happens when $-dI/dt > 0$, or in other words, when she decreases her moment of inertia, e.g. by pulling in her arms and legs. No external torques or forces needed when already spinning ($\omega>0$)!

Answer 3

They push off the ground to get their angular momentum.

The ground is actually far from frictionless in this regard - it's only (mostly) frictionless in the direction parallel to the blades of their skates. They'd have a much harder time spinning if they were in street shoes.

Answer 4

We assume no friction once the object is rotating.

When you rotate with stretched arms (kicked off by the ground, say), you have a rotational momentum. Without friction, that value stays constant, even if you bring the arms close. However, the rotational energy goes up during this process.

To bring the arms close you have to use force against the centrifugal force over the distance you need to cover bringing the arms closer to your body. This work is injected into the rotational energy of the body and causes this to go up.

If you compute the amount of work required to bring the outer masses closer to the rotational axis, you will find that it precisely matches the gain in rotational energy.

The torque required is the effect of a Coriolis-type force exerted on the masses when you move them closer to the rotational axes.

Disclaimer: I have computed that the rotational energy difference and confirm that it matches precisely the work exerted against the centrifugal force; however, I have not yet confirmed computationally that the Coriolis forces match the torques required to speed up the rotation.

Answer 5

Torque increases an object's angular momentum, not its angular velocity. If an object has a constant moment of inertia, then increasing one means increasing the other, but in this case, the ballerina drawing their arms in is decreasing their moment of inertia, and so their angular velocity can increase without increasing their angular momentum.