# I Why an integral vanishes? Angular momentum of a rigid body

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1. May 5, 2017

### Alexanddros81

Hi.
I am revising my Mechanics: Dynamics by reading the Beer 10th edition textbook and Pytel 2nd edition
In Pytel pg 358 art. 17.3 the angular momentum about the mass center of a rigid body in general motion is being calculated.
Can someone explain to me why (∫vr' dm) x v- = 0
Text says this "happens according to the definition of mass center". I need further explanation

Thanks

2. May 5, 2017

### Sturk200

Isn't it just that the center of mass is the weighted sum of positions with weighting factor equal to the mass at said positions, which is precisely what that integral term is, so that if you define the center of mass to be at the zero of coordinates, it will be zero?

3. May 9, 2017

### Alexanddros81

Can you provide a graph and the equations in order to understand it better?

4. May 9, 2017

### Sturk200

Let's see if I can get this right.

You want to analyze the motion of a rigid body. To do so we will do the usual thing and define two systems of coordinates: a body system and a stationary system. The body system is rigidly fixed in the body, meaning it translates and rotates with the body. The stationary system is fixed in you, the observer, or the ground, or something like that.

We shall defined three vectors by specifying their endpoints.
The vector $\mathbf{R}$ points from the origin of the stationary system to the origin of the body system.
The vector $\mathbf{r}$ points from the origin of the body system to some point on the body.
The vector $\mathbf{a}$ points from the origin of the stationary system to that same point on the body.

If you are confused at this point, stop and draw everything I just described. If not, good -- you shouldn't be because it's simple. It should be evident that $\mathbf{a}=\mathbf{R}+\mathbf{r}$.

Now we would like to find an expression for the velocity of a point on the body with respect to the stationary frame, i.e., $\dot{\mathbf{a}} = \dot{\mathbf{R}} + \dot{\mathbf{r}}$. Of course, the only thing that can cause $\dot{\mathbf{r}}$ to be nonzero is a rotation of the body (since the body is rigid and by definition points on it cannot translate with respect to one another). Let the axis of rotation pass through the origin of the body frame, and let the angular velocity of rotation be $\mathbf{\Omega}$, then $\dot{\mathbf{r}}=\mathbf{\Omega}\times\mathbf{r}$.

Defining $\dot{\mathbf{a}}=\mathbf{v}$ and $\dot{\mathbf{R}}=\mathbf{V}$, we have derived the necessary expression for the velocity as $$\mathbf{v} = \mathbf{V} + \mathbf{\Omega}\times\mathbf{r}.$$

Fine. Now the angular momentum of a point mass is given by $\mathbf{L} = m\mathbf{r}\times\mathbf{v}$. We shall treat our rigid body as a collection of point masses, so that its angular momentum will be given by a sum over all mass elements on the body: $\mathbf{L} = \Sigma m \mathbf{a}\times\mathbf{v}$ (I am using the vector $\mathbf{a}$ because I want the angular momentum relative to the stationary system). Substituting for the velocity:

$$\mathbf{L} = \Sigma m \big(\mathbf{a}\times\mathbf{V}+\mathbf{a}\times\mathbf{\Omega}\times\mathbf{r}\big)$$

I guess we probably want to write $\mathbf{a} = \mathbf{R} + \mathbf{r}$. Then

$$\mathbf{L} = \Sigma m \big(\mathbf{R}\times\mathbf{V}+\mathbf{r}\times\mathbf{V}+\mathbf{R}\times\mathbf{\Omega}\times\mathbf{r}+\mathbf{r}\times\mathbf{\Omega}\times\mathbf{r}\big)$$

Remember that the sum here is taken over all points of the body, so that any quantity which does not vary from point to point (such as, say, $\mathbf{V}$ or $\mathbf{R}$ or $\mathbf{\Omega}$ can be taken out of the sum:

$$\mathbf{L} =( \Sigma m) \mathbf{R}\times\mathbf{V}+(\Sigma m\mathbf{r})\times\mathbf{V}+\mathbf{R}\times\mathbf{\Omega}\times(\Sigma m\mathbf{r})+\Sigma m(\mathbf{r}\times\mathbf{\Omega}\times\mathbf{r})$$.

Notice that we have not yet assumed anything about the center of mass. At this point we can simplify our formula by stating that the center of mass shall be taken to coincide with the origin of coordinates in the body frame. As a consequence, any sum of the form $\sum m\mathbf{r}$ must by definition be zero (where the vector $\mathbf{r}$ is defined relative to the origin of coordinates in the body frame -- i.e. relative to the center of mass). Saying that that sum is zero and saying that the center of mass is at the origin of coordinates are equivalent statements.

Now the formula for the angular momentum simplifies to

$$\mathbf{L} =( \Sigma m) \mathbf{R}\times\mathbf{V}+\Sigma m(\mathbf{r}\times\mathbf{\Omega}\times\mathbf{r})$$.

Or even better:

$$\mathbf{L} =( \Sigma m) \mathbf{R}\times\mathbf{V}+\Sigma m(\mathbf{r}\times\mathbf{v})$$.

This, as it should be, is the sum of the angular momentum with respect to the stationary origin and the angular momentum with respect to the body origin, sometimes called the orbital and spin angular momenta respectively.

I probably complicated the notation in unnecessary ways at points, but maybe this will help. Also, if you prefer integrals to sums, just change them. All of the logic still holds.

5. May 10, 2017

### zwierz

By definition radius-vector of the center of mass equals
$$\boldsymbol r_C=\frac{\int_V\boldsymbol rdm}{\int_Vdm}$$ If the center of mass is at the origin then $\boldsymbol r_C=0$ and thus $\int_V\boldsymbol rdm=0$