Newton's Second Law for Composite Systems

Newton's three laws of motion form the basis for classical mechanics. Central to the modern discussion of mechanics is the concept of a particle, i.e. a body that is small enough that its motion is described sufficiently by giving its position as a function of time. Newton's first law defines how one recognizes the presence of a net force on a particle by describing how the particle behaves in the absence of a net force. The second law expresses the relation between a net force on a particle of mass, $m$, and the acceleration produced by that force. Using bold face letters for vectors and regular type for scalars, that relationship is expressed by the formula

$$\mathbf{F} = m \mathbf{a},$$

where $\mathbf{F}$ is the net force on the particle of mass, $m$, and $\mathbf{a}$ is the particle's acceleration.

For extended bodies, where rotational motion becomes relevant, we must consider where the various forces that contribute to the net force are applied. This consideration leads to the concept of torque. Before considering torque, we will show that a system of particles obeys Newton's second law, for the net external force and the acceleration of the center of mass.

An extended body will be treated as a collection of point particles. Since the particles in the system may exert forces on each other, the net force $\mathbf{F}_\textrm{i,net}$ on the $i^\text{th}$ particle is the sum of the net external force $\mathbf{F}_\textrm{i,ext}$ and the net internal force $\mathbf{F}_\textrm{i,int}$. It is shown below that, because of Newton's third law, the forces that the particles of a system exert on each other can be ignored when considering translational motion of the system as a whole:

$$\mathbf{F}_\textrm{i,net} = \mathbf{F}_\textrm{i,ext} + \mathbf{F}_\textrm{i,int}. \qquad (1)$$

The net force on the system is therefore

$$\mathbf{F}_\textrm{net} = \sum_i \mathbf{F}_\textrm{i,ext} + \sum_i \mathbf{F}_\textrm{i,int} . \qquad (2)$$

The first summation on the right side of (2) is the net external force on the system $\mathbf{F}_\textrm{net,ext}$, and the second sum will be seen to be zero when Newton's third law is used. Let $\mathbf{F}_\textrm{i,j}$ be the force on the $i^\text{th}$ particle due to the $j^\text{th}$ particle. Newton's third law is

$$\mathbf{F}_\textrm{i,j} = - \mathbf{F}_\textrm{j,i}. \qquad (3)$$

The second term on the right side of (2) is

$$\sum_i \mathbf{F}_\textrm{i,int} = \sum_i \sum_j \mathbf{F}_\textrm{i,j} = 0, \qquad (4)$$

since the internal forces cancel in pairs. Therefore,

$$\mathbf{F}_\textrm{net} = \mathbf{F}_\textrm{net,ext}. \qquad (5)$$

It will now be shown, by introducing the concept of center of mass of the system of particles, that for translational motion, one can treat the entire system as a "particle" obeying Newton's second law, where the applied force is the net external force and the acceleration of the "particle" is the acceleration of the center of mass. This is a considerable simplification when considering a composite system.

The position vector $\mathbf{r}_\textrm{cm}$ of the center of mass of a system of particles is a "weighted average" of the position vectors $\mathbf{r}_\textrm{i}$ of all particles:

$$\mathbf{r}_\textrm{cm} = \sum_\textrm{i} \frac{m_i \mathbf{r}_i}{M}, \qquad (6)$$

where $M$ is the total mass $M = \sum_i m_i$. The velocity and acceleration of the center of mass are, respectively,

$$\mathbf{v}_\textrm{cm} = \frac{d \mathbf{r}_\textrm{cm}}{dt},\quad \mathbf{a}_\textrm{cm} = \frac{d^2 \mathbf{r}_\textrm{cm}}{dt^2}. \qquad (7)$$

Apply Newton's second law to each particle:

$$\begin{aligned} \mathbf{F}_\textrm{i} &= m_\textrm{i} \mathbf{a}_\textrm{i} \\&= m_\textrm{i}\frac{d^2 \mathbf{r}_\textrm{i}}{dt^2}\\\\ \mathbf{F}_\textrm{net,ext} &= \frac{d^2}{dt^2} \sum_\textrm{i} m_\textrm{i} \mathbf{r}_\textrm{i} \\&= M \mathbf{a}_\textrm{cm}. \qquad (8) \end{aligned}$$

This means that the center of mass accelerates under the influence of the net external force just as a particle accelerates under the influence of the net force acting on it.

An artillery shell is fired at an angle of projection of $\theta_{0} = 30^\circ$ and a speed of $v_{0} = 400\text{ m/s}$. At the apex of its trajectory the shell explodes into two pieces with the larger piece three times as massive as the smaller one. The smaller piece drops straight down below the point of the explosion. How far from the launch point does the larger piece land? Assume that there is no air resistance, that the ground is level, and that the acceleration of gravity is $g = 9.8\text{ m/s}^2$.

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Let $R$ be the range of an identical shell which does not explode and is launched at the same angle of projection and with the same speed. Choose the origin at the launch point and let $m_1$ be the mass of the smaller piece and $m_2$ be the mass of the larger piece. We are given that $m_2 = 3m_1$ and that $x_1 = \frac{1}{2}R,$ and are asked to find $x_2$.

The net external force on the system, after the shell is launched, is the weight of the shell. Therefore, the center of mass of the exploded shell follows the same trajectory as the unexploded shell and has, therefore, the same range. For $x_\text{cm}$ after the pieces have landed, we have $x_\text{cm} = R$. Using the information we are given, we have

$$\begin{aligned} R &= \frac{0.5Rm_1 +3m_1 x_2}{4m_1}\\ &= \frac{0.5R + 3x_2}{4}\\ \Rightarrow x_2 &= \frac{7}{6}R. \qquad (9) \end{aligned}$$

The range of a projectile is

$$R = \frac{v_{0}^2 \sin (2\theta_0)}{g}. \qquad (10)$$

Calculation gives $x_2 = 19 \text{ km}$. $_\square$

The torque $\mathbf{\tau}$ about the origin produced by a force $\mathbf{F}$ is defined in terms of the position vector $\mathbf{r}$ of a particle and the force as

$$\mathbf{\tau} = \mathbf{r}\times \mathbf{F}. \qquad (11)$$

For a system of $N$ particles of masses $m_\textrm{i}$ and position vectors $\mathbf{r}_\textrm{i}$, the torque on the $i^\text{th}$ particle produced by the net force on it, $\mathbf{F}_\textrm{i,net}$, is

$$\mathbf{\tau}_\textrm{i} = \mathbf{r}_\textrm{i}\times \mathbf{F}_\textrm{i,net}. \qquad (12)$$

The net or total torque on the system about the chosen origin is

$$\mathbf{\tau}_\textrm{net} = \sum_i \mathbf{r}_i \times \mathbf{F}_\textrm{i,net}. \qquad (13)$$

Using equation (1), the net torque on the system is

$$\mathbf{ \tau}_\textrm{net} = \sum_i \mathbf{r}_i \times \mathbf{F}_\textrm{i,ext} + \sum_i \mathbf{r}_i \times \mathbf{F}_\textrm{i,int}. \qquad (14)$$

The first term on the right side of the equation is the net external torque $\mathbf{ \tau}_\textrm{net,ext}$, and the second term is the sum of internal torques, which can be shown to be zero for purely mechanical forces. (In electromagnetic theory, this is not the case.)

The displacement vector from $m_i$ to $m_j$ is $\mathbf{r}_\textrm{i,j}$ = $\mathbf{r}_j$ - $\mathbf{r}_i$. (See the picture below.)

For mechanical forces one has, in addition to Newton's third law $\mathbf{F}_\textrm{i,j}$ = - $\mathbf{F}_\textrm{j,i}$, the so-called strong form of Newton's third law, i.e. that the internal forces $\mathbf{F}_\textrm{i,j}$ are parallel to $\mathbf{r}_\textrm{i,j}$. Substituting $\sum_i \mathbf{F}_\textrm{i,int}$ = $\sum_i \sum_j \mathbf{F}_\textrm{i,j}$, the sum of internal torque is

$$\sum_i \mathbf{r}_i \times \mathbf{F}_\textrm{i,int} = \sum_i \sum_j \mathbf{r}_\textrm{i} \times \mathbf{F}_\textrm{i,j}. \qquad (15)$$

The summation on the right in equation (15) can be written as

$$\sum_i \sum_j \mathbf{r}_\textrm{i} \times \mathbf{F}_\textrm{i,j} = \frac{1}{2} \sum_i \sum_j \mathbf{r}_\textrm{i} \times \mathbf{F}_\textrm{i,j} +\frac{1}{2} \sum_j \sum_i \mathbf{r}_\textrm{j} \times \mathbf{F}_\textrm{j,i}. \qquad (16)$$

Collecting terms and using Newton's third law gives

$$\sum_i \sum_j \mathbf{r}_\textrm{i} \times \mathbf{F}_\textrm{i,j} = \frac{1}{2} \sum_i \sum_j ( \mathbf{r}_i - \mathbf{r}_j ) \times \mathbf{F}_\textrm{i,j} = \frac{1}{2} \sum_i \sum_j \mathbf{r}_{j.i} \times \mathbf{F}_\textrm{i,j}. \qquad (17)$$

Using the strong form of Newton's third law, the cross product in each term is seen to be zero. Therefore,

$$\sum_i \mathbf{r}_i \times \mathbf{F}_\textrm{i,int} = 0, \qquad (18)$$

and

$$\mathbf{ \tau}_\textrm{net} = \sum_i \mathbf{r}_i \times \mathbf{F}_\textrm{i,ext} = \mathbf{ \tau }_\textrm{ext}. \qquad (19)$$

Finally, we derive the relationship between torque and angular momentum and arrive at the rotational analogue of Newton's second law. The angular momentum of the $i^\text{th}$ particle in the system is $\mathbf{L}_i$ = $m_i$$\mathbf{r}_i \times \mathbf{v}_i$. The total angular momentum for the system of particles is

$$\mathbf{L} = \sum_i \mathbf{L}_i = \sum_i m_i \mathbf{r}_i \times \mathbf{v}_i. \qquad (20)$$

Differentiating $\mathbf{L}$ with respect to time gives

$$\frac{d \mathbf{L}}{dt} = \sum_i m_i \frac{d \mathbf{r}_i}{dt} \times \mathbf{v}_i + \sum_i m_i \mathbf{r}_i \times\frac{d \mathbf{v}_i}{dt}. \qquad (21)$$

The first sum on the right side of equation (21) is zero because of the cross product, and the second term is the net torque. Thus, we have

$$\mathbf{ \tau}_\textrm{net} = \frac{d \mathbf{L}}{dt}. \qquad (22)$$

APPLICATION: When riding a bicycle, turning the handle bars produces a torque on the turning front wheel. This forces the direction of the angular momentum vector to change. The change is either upward or downward, depending on whether a right turn or left turn, respectively, is made. This is the way a bicycle is kept upright by the rider.

We will now consider the rotation of a rigid body about an arbitrary fixed axis. The particles of the system rotate around the axis in fixed circles and maintain their positions relative to each other. The figure below shows a particle of the system, of mass $m_{i}$, rotating in a circle around an axis through the origin $O$. The axis is perpendicular to the plane of the circle and the direction of motion is counterclockwise as seen from above, as indicated by the curved arrow:

The velocity vector of the particle is the cross product

$$\mathbf{v}_{i} = \mathbf{ \omega}\times \mathbf{r}_{i}, \qquad (23)$$

and the angular momentum is

$$\mathbf{L}_{i} = m_i \mathbf{r}_{i} \times \mathbf{v}_{i} = m_i \mathbf{r}_{i} \times ( \mathbf{ \omega}\times \mathbf{r}_{i}). \qquad (24)$$

Applying an identity for the vector triple product, the so called BAC-CAB identity, gives

$$\mathbf{L}_{i} = m_{i} \big[ r_{i}^2 \mathbf{ \omega } - ( \mathbf{r}_{i} \centerdot \mathbf{ \omega }) \mathbf{r}_{i} \big]. \qquad (25)$$

The total angular momentum of the rigid body about the axis of rotation is, therefore,

$$\mathbf{L} = \sum_i \mathbf{L}_i = \sum_i m_{i} \big[ r_{i}^2 \mathbf{ \omega } - ( \mathbf{r}_{i} \centerdot \mathbf{ \omega }) \mathbf{r}_{i} \big]. \qquad (26)$$

Define the unit second rank tensor $\mathbf{ \iota }$ such that, for each vector $\mathbf{w}$,

$$\mathbf{ \iota } \centerdot \mathbf{w} = \mathbf{w}. \qquad (27)$$

The moment of inertia tensor is defined by

$$\mathbf{I} = \sum_i m_{i} \big[ r_{i}^2 \mathbf{ \iota } - \mathbf{r}_{i} \otimes \mathbf{r}_{i}\big]. \qquad (28)$$

With this definition of the moment of inertia tensor, the angular momentum of the rigid body is

$$\mathbf{L} = \mathbf{I} \centerdot \mathbf{ \omega }. \qquad (29)$$

The components of the moment of inertia tensor produce a symmetric matrix, which consequently has three mutually orthogonal eigenvectors $\mathbf{a}_{k}\, (k = 1, 2, 3)$ with eigenvalues $I_{k}$ such that

$$\mathbf{I} \centerdot \mathbf{a}_{k} = I_{k} \mathbf{a}_{k}. \qquad (30)$$

Axes of rotation through the center of mass, parallel to the eigenvectors of the moment of inertia tensor, are called principle axes of the body, and the eigenvalues are called principal moments of inertia. For a rigid body rotating around a principal axis, the angular momentum is $\mathbf{L}$ = $I$$\mathbf{\omega}$, where $I$ is the moment of inertia about the axis and $\mathbf{ \omega}$ is the angular velocity vector. Since $I$ is a constant, one has

$$\mathbf{ \tau}_\textrm{net} = \mathbf{ \tau }_\textrm{ext} = I\frac{d \mathbf{\omega}}{dt} = I \mathbf{\alpha}. \qquad (31)$$

This is known as Newton's second law for rotational motion.