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)

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    $\begingroup$ Are you sitting in an office chair right now? Make some room around youself, then give a push with your legs and stick them out as far as you can (or get someone to spin you with your legs stuck out)..Now while you're spinning, pull your legs in so your knees are bent and your feet are under the seat, near the post $\endgroup$
    – Caius Jard
    Aug 10 '21 at 14:43
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    $\begingroup$ @CaiusJard ... and then fall down miserably after having been ejected from your seat as my physics teacher once did, showing us exactly that. He intended to go the other way round (starting with arms and legs close to the body and decelerate) but got interrupted, mixed things up and bam. $\endgroup$
    – WoJ
    Aug 10 '21 at 16:52

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 $L = I\omega$.

  • $\begingroup$ "They give themselves an initial spin" this is misleading - the initial spin comes from the ground $\endgroup$
    – Señor O
    Aug 10 '21 at 0:38
  • $\begingroup$ @SeñorO Rephrased. $\endgroup$
    – SuperCiocia
    Aug 10 '21 at 0:40
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    $\begingroup$ It looks like you got an answer, but I thought his question was "where does the $\alpha$ come from that takes the ice skater from one $\omega$ to a larger $\omega$." In that case, since angular momentum is conserved, then during the process where the ice skater draws their arms in, $\dot{I} \omega + I \alpha = 0$ which does not require an external torque. $\endgroup$
    – Evan
    Aug 10 '21 at 1:31
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    $\begingroup$ Unlike an ice skater, a ballerina has to keep kicking the floor every couple of rotations, because there's a decent amount of friction. $\endgroup$
    – Barmar
    Aug 10 '21 at 14:57
  • $\begingroup$ @Barmar: Even ice skating can have a decent amount of friction. You hold the skate blade at an angle to the direction you're pushing, so that the edge digs in a bit. $\endgroup$
    – jamesqf
    Aug 10 '21 at 20:08

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 ($\omega>0$)!

  • $\begingroup$ hold on i have question when she kicks the ground and twists her leg then she gets intial torque she needed, what you did there suggests $\Delta\tau= 0$ on the system, but i'm confused when she kicks the ground she's getting force from ground that means external force is being applied on the system and she's converting it into torque, that means angular momentum is variable?, how is the angular momentum is being conserved? $L=I\omega$ if $L$ is constant then we can say that when her moment of inhertia decreases when she brings her arms close, to account for decrease in $I$, $\omega$ will rise $\endgroup$ Aug 12 '21 at 0:45
  • $\begingroup$ I don't know what you're asking exactly (I never used $\Delta\tau$, e.g.), but here's what I think you need for an answer: in the general case, as specified above, $$\tau_\text{net} =\frac{dL}{dt} = \frac{dI}{dt}\omega + I\alpha.$$ So, when the ballerina interacts with the ground, the size of her angular momentum $L$ increases, because the ground system applies a torque on the ballerina system. Indeed, this can change $\omega$ and $\alpha$. Once she is spinning, she can also change $\omega$ and $\alpha$ (and $I$), but now, there's only one way to do it, namely as I described ($\tau = 0$). $\endgroup$
    – Mew
    Aug 12 '21 at 14:22

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.

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    $\begingroup$ Ballerina do wear shoes. I think you're thinking of figure skaters. $\endgroup$
    – J...
    Aug 10 '21 at 10:32
  • $\begingroup$ @J... same difference though, and the OP did say "on ice".. Probably a hair that didn't need splitting... $\endgroup$
    – Caius Jard
    Aug 10 '21 at 14:44
  • $\begingroup$ @J... I see - so either "thats why ice skaters don't wear street shoes" or "thats why ballerinas don't perform on ice", take your pick ;) $\endgroup$
    – Señor O
    Aug 10 '21 at 17:50
  • $\begingroup$ @SeñorO there is such a thing as Ballet on Ice... $\endgroup$
    – Ruslan
    Aug 10 '21 at 22:03
  • $\begingroup$ @Ruslan they wear skates. Skates = spin on ice. Shoes = hard to spin on ice. $\endgroup$
    – Señor O
    Aug 10 '21 at 22:17

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.


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.


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