# Kinetic energy of a cylinder

It is a long cylinder (you can approx $R=0$), and it has a fixed point in one os its ending points, it's rotating on a plane and I have to calculate the kinetic energy from reference systems situated in the center of mass, and both extremes. If I calculate it in the fixed point, it has only rotational energy: $$\frac{1}{6}ML^2\dot{\varphi}^2$$ If I choose the center of mass, the moment of inertia changes but I have to add to that the energy of that moving point $1/2Mv_{CM}^2$, and the result is the same. My problem is that when I choose the moving extreme, the result changes: The moment of inertia is the same as with the fixed point: $1/3ML^2$, and the speed at that point is $1/2ML^2\dot\varphi ^2$, so adding them gives you something different from the $1/6$ result, and they should be the same, ¿What am I doing wrong?

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Isn't the reference frame attached to the moving end just the same as the fixed end, only the sense of rotation is reversed? – Michael Brown Jan 26 '13 at 15:03
@MichaelBrown But in the moving end, wouldn't I have to add the kinetic energy corresponding to the moving ref. frame? That is: $1/2mv^2$, being $v$ the speed of that point, the same way that I have to add that term when the ref. frame is in the center of mass? – MyUserIsThis Jan 26 '13 at 15:27

The outcomes are not supposed to be the same.

There are two ways to interpret your question:

1. You want to calculate the kinetic energy in different reference frames.

Let's think for example of a point-like body moving in a constant velocity $\mathbf{v}$. It's kinetic energy is $\frac{1}{2}mv^2$, but if we calculate it in a reference frame that is moving with the body, in that frame the body is at rest and we get zero. So we don't expect the same outcome when calculating kinetic energy in different reference frames.

2. You want to stay in the laboratory reference frame, but pick different points as the axis of rotation for your calculations.

Here there is a subtle point we need to be aware of. It's true that some calculations involving the rotation of rigid bodies can be done in several different ways, each time picking a different axis, and still resulting in the correct outcome. For example this works if we want to calculate the linear and angular acceleration of a body when a given force is applied to it.

However, if a rigid body has linear movement and rotation simultaneously, and we want to calculate it's kinetic energy, we need to be careful when using the formula:

$$E_k = \frac{1}{2}M V^2 + \frac{1}{2} I \omega^2,$$

where $\mathbf{V}$ is the velocity of the axis point and $\omega$ is the angular velocity of rotation around that point. This is the formula you used in your calculations, but in fact it is valid only in the following cases (and I give a proof of this below):

1. When we use the center of mass as the axis of rotation.
2. When the axis of rotation is at rest (i.e. $\mathbf{V}=0$).
3. When the velocity is parallel to the line connecting the axis point to the center of mass.

Regarding the calculations you showed in your question:
When you used the fixed point of the cylinder as the axis case 2 applied. When you used the center of mass case 1 applied. When you used the moving extreme none of the cases applied, and you cannot use the above formula in this case.

Proof:
We model the rigid body as a collection of point-like masses $m_i$ with their positions relative to the axis of rotation denoted as $\mathbf{r}_i$. The velocity of mass $i$ is:

$$\mathbf{v}_i = \mathbf{V} + \boldsymbol\omega \times\mathbf{r}_i.$$

The total kinetic energy is then:

$$E_k = \sum_i \frac{1}{2} m_i v_i^2 = \frac{1}{2} MV^2 + \frac{1}{2} I \omega^2 + M \mathbf{V} \cdot (\boldsymbol\omega \times \mathbf{R}_{CM}),$$

where $M=\sum_i m_i$ is the total mass, $I=\sum_i m_i |\hat{\boldsymbol\omega} \times \mathbf{r}_i|^2$ is the moment of inertia and $\mathbf{R}_{CM} = (\sum_i m_i \mathbf{r}_i )/M$ is the center of mass relative to the axis of rotation.
We see that we need the last term to vanish in order to get the formula we want to prove, and we can get this if $\mathbf{R}_{CM}=0$ (case 1), $\mathbf{V}=0$ (case 2) or $\mathbf{V} \cdot (\boldsymbol\omega \times \mathbf{R}_{CM})=0$ (case 3).

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Thank you very much for your very detailed answer, that more general formula at the end is what I was looking for. – MyUserIsThis Jan 27 '13 at 10:42
Yes, in retrospect I could have just said that the general formula is the correct one, and the special cases where the simpler formula holds just follow trivially... – Joe Jan 27 '13 at 12:08

I think I found your mistake. The moment of intertia of a rod fixed about the center is

$$\frac{1}{12} m L^2$$

You can derive this in several ways. I derived it by considering two rods moving separately (but still rigid), so then the moment of intertia is as before, but

$$\frac{2}{3} \frac{m}{2} \bigg(\frac{1}{2} L \bigg)^2 = \frac{1}{12} m L^2$$

You can consider that the angular velocity $\dot{\phi}$ of each rod is the same as before.

When you then take the speed of the center of mass, you should have

$$v_{cm} = \bigg(\frac{1}{2} L \bigg) \dot{\phi}$$

Since the part that is translating is center of mass, at half the total length from the fixed point (think of the rolling wheel problems).

You should be able to solve for the energy now.

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You've done what I've done. I get the same results, the only problem is that the energy from the moving end is not the same as for the other two points – MyUserIsThis Jan 26 '13 at 15:31
You're right, sorry about that. I agree with what Michael Brown posted. There isn't another term added to the energy because now the world looks like it is spinning the other way from where you're standing. It's like the difference between being on Earth and seeing an airplane fly by, and being on the airplane seeing the Earth coast below you. You could say the Earth is more "still", but you also feel the same from the plane (or car, if you prefer). In any case, the Earth's frame is not absolute either. – Angel Jan 26 '13 at 15:40