# How does the kinetic energy of a ballerina increase? [duplicate]

When a ballerina pulls her arms in, her rotational kinetic energy increases because angular momentum is conserved. That means that work must have been done on her. I saw somewhere that there is work done because she has to pull her arms in against a centrifugal force.

Actually, the particular example is from the Feynman lectures volume I, chapter 19 section 19-4. Here is the problematic part: "When we move a weight horizontally we do not do any work. If we hold a thing out and pull it in, we do not do any work. But that is when we are not rotating! When we are rotating there is centrifugal force on the weights. They are trying to fly out, so when we are going around we have to pull the weights in against the centrifugal force. So the work we do against the centrifugal force ought to agree with the difference in rotational energy."

I have two questions: what is this centrifugal force in the quote above, and how is it that by pulling her arms in against this force, she has done work? A centrifugal force is a fictitious force from what I know of, so the kinetic energy increase cannot be accounted for with this.

The second question is explained here: Suppose you have some bar that you are holding at arm's length, perfectly horizontally (meaning ignore gravity). Because the bar is still, there is no net force. You now pull the bar towards you. The bar must have accelerated, so you did some positive work on the bar. However, because at the end of the process, when the bar is next to your body, it is at rest, you must have done negative work, decelerating the bar to make it come to rest. Therefore,the total work done on the bar must be zero. Note that this property does not depend on whether or not other forces are acting upon the bar, as it is only the net force that does work. So how is it, that by saying that there some force pulling the bar away from you (the centrifugal force for the ballerina), you have done work? It must be the same thing, because if the bar is at rest initially, and at rest when near your body, there must be no net force, so it is the same situation as if there was only one force applied to the bar.

We could say that you accelerated the bar and let the other force decelerate it, meaning that in all, you did work on the bar. However, the net work on the bar would be zero, because the other force did negative work on it by decelerating it. And we are interested in the net work done on the arms of the ballerina, not the work she does herself. So what part am I missing? Thanks for any answers and sorry for the long question.

To clarify, the bar in question is not rotating. So the net force on the bar is zero.

How does the kinetic energy of a ballerina increase?

Conservation of angular momentum:

$$L_1=L_2 \implies I_1\omega_1=I_2\omega_2\quad\quad (1)$$

Pulling in your arms reduces moment of inertia $I$, since the same mass is now distributed over a volume closer to the spin centre, $I=\sum mr^2$. As you say, reducing $I$, so $I_2<I_1$, implies increasing angular speed $\omega_2>\omega_1$.

Conservation of energy:

$$E_1=E_2 \implies K_1+W=K_2 \implies \frac{1}{2}I_1\omega_1^2+W=\frac{1}{2}I_2\omega_2^2\quad\quad (2)$$

Kinetic energy will also be affected by a reduction in moment of inertia. Theoretically, with no added work, $W=0$, reducing $I$, so $I_2<I_1$, will as before have to imply $\omega_2^2>\omega_1^2$. Your angular speed will increase.

But there's a clinch: For $W=0$, these two equations obviously do not result in the same angular speed $\omega_2$. So $W>0$ must be the case.

Combining the equations (inserting $\omega_1=\frac{I_2}{I_1}\omega_2$ from $(1)$ in $(2)$):

$$\frac{1}{2}I_1\left(\frac{I_2}{I_1}\omega_2\right)^2+W=\frac{1}{2}I_2\omega_2^2\implies\\ \frac{1}{2}\frac{I_2^2}{I_1}\omega_2^2+W=\frac{1}{2}I_2\omega_2^2\implies\\ W=\frac{1}{2}I_2\omega_2^2-\frac{1}{2}\frac{I_2^2}{I_1}\omega_2^2=\frac{1}{2}I_2\omega_2^2\left(1-\frac{I_2}{I_1}\right)$$

This amount of work $W$ must be added. The work comes from the ballerina herself pulling in her arms. This requires energy from within the system (from energy comsumption in the muscles), in other words (from the 1st law of thermodynamics):

$$\Delta U=Q-W\implies \Delta U=-W$$

where $U$ is internal energy, and there is no heat addition $Q$. The change in internal energy $U$ to perform this work must be within the body of the person.

what is this centrifugal force

Centrifugal force is the tendency of particles to not stay in circular motion. Spreading out her arms in a spin, her hands are as far out as possible.

The hands have a velocity at some instantaneous point (if you cut off the hands (sorry for the brutallity) at that moment, they would fly straight away, not in a circular motion anymore), but to change the direction of that velocity - that is, acceleration - a force must be applied, $F=ma$. Force must be applied to keep particles in circular motion. This is the reason for "centrifugal force", which is not a force itself, but rather the phenomenon that force is required to keep the circle.

and how is it that by pulling her arms in against this force, she has done work?

Changing the radius of the circular motion for some particles makes them move along with the direction of the force (inwards towards the centre). $W=\int F dx$, where $dx$ is the change in position. This is so to speak where the work results from.

Suppose you have some bar that you are holding at arm's length, perfectly horizontally (meaning ignore gravity). Because the bar is still, there is no net force.

Not intirely correct (although not important for the question). To clear this one out, as said before, to keep particles in circular motion they must constantly change velocity direction. That means, they must constantly accelerate (because a velocity change is acceleration). For a circular motion, this acceleration is towards the center (this will not change the velocity magnitude but only the direction, and so the circle is kept constant). Called radial acceleration $a_{rad}$. A force must constantly be applied to keep this acceleration:

$$F=m a_{rad}$$

This force is what she feels pulling in her arms during the spin.

You now pull the bar towards you. The bar must have accelerated, so you did some positive work on the bar. However, because at the end of the process, when the bar is next to your body, it is at rest, you must have done negative work, decelerating the bar to make it come to rest. Therefore,the total work done on the bar must be zero.

No. It is wrong to say when the bar is next to your body, it is at rest, since the angular speed is still present. The objects in your hands have much more kinetic energy than before.

I understand why you would think this: you are considering the velocity towards the center that changes the radius. But remember that kinetic energy is not a directional property. It is a value of the object regardless of the direction.

There is no difference... Doing work is by definition: applying a force to do a displacement, $W=\int F dx$. Other force working at the same time do not have influence on this (except for helping out with the displacement of course).