# Linear Momentum/Impulse for a Rigid Body

Problem

Consider a system of $N$ point-masses $m_i$, each travelling at a velocity $\mathbf{v}_i$. Then, the total linear momentum / impulse $\mathbf{p}$ of the system can be calculated as (see Smith, J.O. Physical Audio Signal Processing)

$$\mathbf{p} := \sum_{i=1}^N \mathbf{p}_i = \sum_{i=1}^N m_i \mathbf{v}_i = \sum_{i=1}^N m_i \dot{\mathbf{x}}_i = M \frac{d}{dt} \left(\frac{1}{M}\sum_{i=1}^N m_i \mathbf{x}_i \right) = M \frac{d}{dt} ( \mathbf{x}_c ) = M \mathbf{v}_c$$

where $M = \sum_{i=1}^N m_i$ denotes the total mass and $\mathbf{v}_c$ the velocity of the center of mass.

Similarly, the total linear momentum / impulse $\mathbf{p}$ for a rigid body $K$ can be calculated using points $\mathbf{x}_P$ on the body having the infinitesimal mass $dm$ as

$$\mathbf{p} := \int_{K} d\mathbf{p} = \int_{K} \mathbf{v}_P \: dm = \int_{K} \dot{\mathbf{x}}_P \: dm = M \frac{d}{dt} \left(\frac{1}{M}\int_{K} \mathbf{x}_P \: dm \right) = M \frac{d}{dt} ( \mathbf{x}_c ) = M \mathbf{v}_c$$

Question

Why is it possible to write $$\int_{K} \dot{\mathbf{x}}_P dm = \frac{d}{dt} \left(\int_{K} \mathbf{x}_P dm \right)~?$$
Is it due to the Leibniz Integral Rule or due to the Reynolds Transport Theorem for a constant volume $K$?

• Isn't the latter just a generalization of the former? – probably_someone Aug 3 '17 at 8:25
• @probably_someone As I understand it, it is. However, I think that it is not possible to apply Leibniz Rule for each dimension separately, as the integral constants depend on each other, since a body is in all generality not a cuboid. – Discbrake Aug 3 '17 at 8:38
• An integral is a linear operator, meaning that $$\frac{\partial}{\partial t} \int f(t,x) {\rm d}x = \int \frac{\partial}{\partial t} f(t,x)\,{\rm d}x$$ – ja72 Aug 4 '17 at 21:29
• @ja72 That is only true for constant integration limits, which is not the case for a moving body. – Discbrake Aug 5 '17 at 16:11

$$\frac{d}{dt}\int_{K(t)} x_P(t) \: dm = \lim_{dt \: \rightarrow \: 0} \left( \frac{1}{dt}\int_{K(t+dt)} x_P(t+dt) \: dm - \int_{K(t)} x_P(t) \: dm \right)$$
The mass $K(t+dt)$ has changed during $dt$ only at the surface $\partial K$. The contribution is $\int_{\partial K(t)} d\sigma dS$ with the surface element $dS$ and the change of the mass per surface element $d\sigma$. But since no mass does flows out, it holds $K(t+dt) = K(t)$ and the only contribution left in above equation is the differential of $x_P(t)$.