For context, i've read:

Why does a ballerina speed up when she pulls in her arms?

How do figure skaters get angular momentum for a spin?

  1. If a skater is a rigid body, how can we apply Newton's second law?

I understand how it works if the skater was a pontual body, but i'm not sure how it's going to work if we consider her as an exetended body.

  1. How come the torque of the net force be zero? In Why does a ballerina speed up when she pulls in her arms? it's mention that:

Fr is the centripetal force towards the center of the osculating circle that curves the arm's path, and Ft is a (small) tangential force in the direction of motion that speeds the arm up.

but I still remain clueless about what does that imply because acting on the skater. So what forces are acting on the skater?

  1. If she is actually a rigid body she while be a particle aggregate, which makes sense beacuse we are considering the distribution of her mass. So we know that her global mass will be the Summation of the masses of her particules, but this particules have different radius to the center of mass. So when we are looking at the spin of the skater how do we take in consideration the r? Do we just considere it zero? That would make the torque zero... I really don't know

edit: please don't use polar coordinates, i don't really know how they work


3 Answers 3


Let's ignore friction between skate and ice, and also any horizontal pushing force from the blade biting into the ice. In this situation the only force from the ice is directly upwards on the ice skates (the normal reaction force). The only other force acting is gravity which acts straight down. If the centre of mass of the skater is above the point where the skate meets the ice (if skater is spinning on one skate), then the gravity will not make the skater fall over. In this case there is simply an upwards force and an equal downwards force. Neither of these forces is making the skater rotate any more or any less. This what we mean when we say there is no torque.

Ok so there's no net torque.

Now what happens when the arms come in is that there is a combination of internal forces: the forces provided by muscles, skin and bone in this example. These forces are acting in quite a complicated way if you try to track them all in detail, because as the skater rotates so do the directions of all these internal forces. A nice quick way to do the calculation is first to note (from a general argument) that the internal forces cannot change the total angular momentum. But when the arms come in the only way for the angular momentum to stay constant is if the rotation speeds up. So we deduce that the rotation must speed up, as indeed is observed.

But this calculation leaves us still asking how did the rotation speed up? Ultimately some particles in the skater's body are moving faster than they were to begin with, so an internal force must have acted in a direction so as to speed them up. When the skater pulls his or her arms in, they are indeed providing internal forces which have a non-zero component along the direction of motion of the outer part of their torso. The force is acting in this circular direction. This part is counter-intuitive because you would think the forces only act in the radial direction, but while the rotation is changing there is this more complicated situation. This circular-acting internal force takes angular momentum away from the arms and gives it to the torso. The total $L_{\rm arms} + L_{\rm torso}$ remains constant. Newton's third law is thus maintained because what is taken from one is given to the other.

P.S. actually the blade does bite into the ice and this provides a horizontal force which helps the skater to stay balanced. But this force is acting so close to the axis of rotation that any torque it produces is negligible.


There is a thing called moment of inertia of extended bodies. It is the fundamental leap that allows us to still do mechanics, and we treat it sort of like a mass. Look up the definition, it is a summation or integral of the mass points times their position squared about the axis of rotation.

When the skater has her arms extended then $I$ is much bigger. However in analogy to


We have


Where $I$ is the moment of inertia and $\omega$ is the angular velocity.

So if $I$ becomes smaller $\omega$ becomes bigger, for there is no reason for $L$ to change on ice.. :)

For torque to happen (which is a rate of change of $L$ in time) we need a force to be applied, and in this case the force should be applied at the interface with the ice, in the form of friction. Friction is non conservative and generates heat on the surfaces. Since this is ice, the heat in the the very small interface will turn it to water and deny the friction force to ever happen, ideally. So we can't have force, and therefore we can't have torque, and therefore it is an interplay between $\omega$ and $I$ that keeps $L$ constant in time.

  • $\begingroup$ for there is no reason for L to change on ice. That's true but it won't be clear to the OP why that is so. Please clarify. (Hint: conservation of kinetic energy) $\endgroup$
    – Gert
    Commented Dec 1, 2019 at 19:45
  • $\begingroup$ @Gert good enough? I never heard of conservation of kinetic energy :) $\endgroup$
    – user192234
    Commented Dec 1, 2019 at 20:01
  1. Newton's second Law does in fact apply to rigid bodies. There is no restriction on applying it to only point masses.

  2. The torque of the net force (which is acting on the skater's body minus the arms, and the arms themselves) is not zero. A net non-zero torque will be produced by the tangential forces $F_t$ which will speed up both the skater's armless body and the two arms. The torque produced by the radial force will however be zero as the line of force (line that you get by extending the force vector on both sides) passes through the centre of mass.

  3. I didn't understand this question. If you are talking about the moment of inertia then yes we will calculate that by integrating (adding up) $r^2 dm$ for each particle of mass $dm$ where $r$ is the distance of the particle from the axis of rotation.

  • $\begingroup$ Im just confused how are we going to consider r in this case $\endgroup$
    – VVNV
    Commented Dec 1, 2019 at 20:46
  • $\begingroup$ In the question part 3 what do you mean by "spin". Do you mean angular momentum or angular velocity or something else entirely? $\endgroup$ Commented Dec 1, 2019 at 21:39
  • $\begingroup$ Spin is the rotation that she is describing. Now I kinda get it that im overthinking things in 3. $\endgroup$
    – VVNV
    Commented Dec 2, 2019 at 19:32
  • $\begingroup$ Okay, and what do you mean by 'r'. Is it the radius of rotation? $\endgroup$ Commented Dec 3, 2019 at 15:32

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