Newton's laws

Newton’s second law for 1D motion¶

Let’s first derive Newton’s second law for a set of particles, and then we’ll specialize it to 1D motion. Each particle comprising the rigid body satisfies the standard version of Newton’s second law for a particle:

where $\F^\textsf{ext}_i$ is the total external force applied to particle $i$, and $\F_{ij}$ is the internal force applied to particle $i$ from particle $j$.

Define the center of mass as $\r_\textsf{cm} := \frac{1}{m}\sum_i m_i \r_i$, where $m := \sum_i m_i$ is the total mass. differentiating twice and plugging in (1), we obtain

The internal forces cancel because of Newton’s third law: $\F_{ij} = -\F_{ji}$. We therefore obtain Newton’s second law for a system of particles:

In other words, no matter what the particles are doing (translating, rotating, moving apart), the center of mass behaves like a particle of mass $m$ with all external forces acting at the center of mass. This is particularly useful when we have gravity acting on a rigid body, for example; we can just assume the gravity is acting at the center of mass with a force equal to the total gravity force $mg$.

Special case: 1D rigid body motion with no rotation. In this case, all particles of the body have the same velocity and acceleration, and it is a scalar quantity due to the 1D nature of the motion. So if $x(t)$ is the position of any point on the rigid body, we have:

Newton’s second law for planar rotation¶

Let’s first derive a more general result for rotational motion of rigid bodies, and then we’ll specialize it to rotation about a fixed axis. As with translational motion, we can think of the rigid body as a collection of particles with positions $\r_i$ and masses $m_i$. Each particle satisfies Newton’s second law:

where $\F^\textsf{ext}_i$ is the external force on particle $i$ and $\F_{ij}$ is the internal force on particle $i$ from particle $j$. To extract rotational dynamics, we take the moment (torque) of each particle’s equation of motion about a reference point $\r_\textsf{ref}$. We will assume an arbitrary (possibly moving) reference point for now. Taking the cross product with $(\r_i - \r_\textsf{ref})$, position with respect to the reference point, we obtain

If we sum over all particles $i$, the internal forces cancel as in the linear case. However, the reason is more subtle this time. Newton’s third law says that $\F_{ji} = -\F_{ij}$, so we have:

The last step is true because we assume the force between particles $i$ and $j$ acts along the vector joining particles $i$ and $j$ (also known as a central force). Therefore the vector $\r_i-\r_j$ is parallel to the vector $\F_{ij}$ and so their cross product is zero. This assumption is valid for ordinary rigid bodies and classical force interactions. Systems that transmit internal moments (e.g., via friction or actuation) require additional torque terms.

Summing over all particles $i$ in Eq. (6) therefore gives

where we defined $\T_i^\textsf{ext}$ as the torque about the reference point caused by the external force $\F_i^\textsf{ext}$. Let us also define the angular momentum of a particle measured with respect to the reference point:

Taking the time derivative and applying the product rule, we have:

Rearrange and use the fact that the cross-product of parallel vectors is zero:

Substitute (11) into (8) and obtain

where we defined the total angular momentum $\L := \sum_i\L_i$. Recall our definition of the center of mass $\r_\textsf{cm} := \frac{1}{m} \sum_i m_i \r_i$, so we can rewrite (12) as

The extra term $\dot\r_\textsf{ref} \times m \dot\r_\textsf{cm}$ will vanish if we choose our reference $\r_\textsf{ref}$ to be either a fixed point (zero velocity), or the center of mass (so the cross product is zero). In either of these cases, we obtain Newton’s second law for rotational motion of a rigid body:

So far, we assumed $\r_\textsf{ref}$ is either fixed in space or it’s the center of mass of the body. We want the moment of inertia of the body with respect to our reference to be constant, so let’s further assume that our reference is fixed to the body. So we are now assuming that

1. $\r_\textsf{ref}$ is fixed to the body and fixed in space (for example, if the object is pinned at a point and rotates about that point), or

2. $\r_\textsf{ref}$ is the center of mass of the object.

Since the body is rigid, the distance between any pair of points fixed to the body is constant. This means that any point $\r_i$ can only rotate with respect to $\r_\textsf{ref}$, so there must exist some vector $\bomega$ such that for any particle $i$,

substituting this into (14) with the definition (9), we obtain

The rightmost term is zero because either $\dot\r_\textsf{ref}=0$ (fixed reference), or $\r_\textsf{ref}=\r_\textsf{cm}$. The remainder of the expression is a linear function of $\bomega$, so we can write it as $\L = \I \bomega$, where $\I$ is called the inertia tensor.

In general, $\I$ can be a function of time, so when we plug in $\L = \I\bomega$ into (14), we need to use the product rule. This doesn’t happen in our particular setting, because we have a rigid body rotating around a fixed axis.

Suppose the axis of rotation is $\khat$ and the body rotates with angular speed $\dot\theta(t)$. We therefore have $\bomega = \dot\theta\,\khat$. We only care about the component of $\L$ in the $\khat$ direction, so we have:

where we used the cyclic property of the triple product to simplify the expression:

So we have $\khat \cdot \L = J\dot\theta$, where $J$ is the moment of inertia of our body about an axis parallel to our axis of rotation and passing through $\r_\textsf{ref}$. Substituting into (14), we obtain Newton’s second law for a rigid body rotating about a fixed axis, where the reference point is either the axis or the center of mass.