1
$\begingroup$

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.

  2. 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

$\endgroup$
0
$\begingroup$

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

$$p=mv$$

We have

$$L=I\omega$$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 Dec 1 '19 at 19:45
  • $\begingroup$ @Gert good enough? I never heard of conservation of kinetic energy :) $\endgroup$ – user192234 Dec 1 '19 at 20:01
0
$\begingroup$
  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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Im just confused how are we going to consider r in this case $\endgroup$ – VVNV Dec 1 '19 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? $\endgroup$ – aditya_stack Dec 1 '19 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 Dec 2 '19 at 19:32
  • $\begingroup$ Okay, and what do you mean by 'r'. Is it the radius of rotation? $\endgroup$ – aditya_stack Dec 3 '19 at 15:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.