Suppose that a rigid body is moving under its own inertia. Show that its center of mass moves linearly and uniformly. If the center of mass is at rest, then the angular momentum with respect to it is conserved.

Let's consider an isolated rigid body consisting of particles and let the total mass of the rigid body be M. Given that there are no external forces acting on the system, we can apply Newton's second law to find the center of mass: 1. Calculate the center of mass position vector, \(\textbf{R}\), using the formula: \[\textbf{R} = \frac{1}{M}\sum_{i=1}^{n} m_i\textbf{r}_i\] Here, \(m_i\) is the mass of the \(i\)th particle, \(\textbf{r}_i\) is its position vector, and the sum is from \(i = 1\) to \(n\) particles. 2. Calculate the total momentum of the system, \(\textbf{P}\): \[\textbf{P} = \sum_{i=1}^{n} m_i\textbf{v}_i\] Here, \(\textbf{v}_i\) is the velocity of the \(i\)th particle. 3. Since there are no external forces acting on the rigid body, the total momentum of the system remains constant, i.e., \(\textbf{P}\) is constant. 4. Now we differentiate the center of mass position vector, \(\textbf{R}\), with respect to time: \[\frac{d\textbf{R}}{dt} = \frac{1}{M}\sum_{i=1}^{n} m_i\frac{d\textbf{r}_i}{dt}\] \[\Rightarrow \frac{d\textbf{R}}{dt} = \frac{1}{M}\sum_{i=1}^{n} m_i\textbf{v}_i = \frac{1}{M}\textbf{P}\] Since \(\textbf{P}\) is constant, the derivative of \(\textbf{R}\) with respect to time is also constant, which implies that the center of mass moves linearly and uniformly.

Now, we will show that when the center of mass is at rest, the angular momentum with respect to the center of mass is conserved. 1. Since the center of mass is at rest, \(\textbf{R}\) is a fixed point, and we can calculate the angular momentum, \(\textbf{L}\), of each particle about the center of mass using: \[\textbf{L}_i = m_i(\textbf{r}_i - \textbf{R}) \times \textbf{v}_i\] Here, \(\times\) denotes the cross product. 2. Calculate the total angular momentum of the system, \(\textbf{L}\): \[\textbf{L} = \sum_{i=1}^{n} \textbf{L}_i\] \[= \sum_{i=1}^{n} m_i(\textbf{r}_i - \textbf{R}) \times \textbf{v}_i\] 3. Since the total momentum of the system is conserved, the rate of change of the total angular momentum with respect to time is given by: \[\frac{d\textbf{L}}{dt} = \sum_{i=1}^{n} m_i(\textbf{r}_i - \textbf{R}) \times \frac{d\textbf{v}_i}{dt}\] 4. As there are no external forces acting, Newton's second law states that the rate of change of momentum is zero: \[\frac{d\textbf{v}_i}{dt} = 0\] 5. Substituting this result into the expression for the rate of change of the total angular momentum, we get: \[\frac{d\textbf{L}}{dt} = \sum_{i=1}^{n} m_i(\textbf{r}_i - \textbf{R}) \times 0 = 0\] This result implies that the angular momentum with respect to the center of mass is conserved when the center of mass is at rest.