# Kinetics of point masses

## Newton's equations

Newton's equations relate the acceleration $\vec{a}$ of a point mass with mass $m$ to the total applied force $\vec{F}$ on the mass (sum of all applied forces). They are:

Newton's equations.

\[\vec{F} = m \vec{a}\]

There is no derivation for Newton's equations, because they are an assumed model for dynamics. We can only verify them by comparing with experimental evidence, which confirms that Newtonian dynamics are accurate for non-relativistic and non-quantum systems.

Click and hold on the figure to apply a constant-magnitude force.

A point mass moving in the plane with an applied force. You can try to made the mass move in a circle and then see what happens when the force is suddenly removed, which will demonstrate Newton's first law (no net force implies motion at constant speed in a constant direction). Also observe which force directions cause the speed to increase or decrease.

If we have objects which are either very massive, very small, or moving very fast, then Newton's equations do not provide a good model of their motion. Instead we must use Einstein's equations of general relativity (for massive and fast objects) or the equations of quantum mechanics (for very small objects). Unfortunately, these two theories cannot be used together, so we currently have no good models for objects which are simultaneously very small and very massive, such as micro black holes or the universe shortly after the big bang. Physicists are currently trying to reconcile general relativity with quantum mechanics by devising a new set of equations (sometimes called quantum gravity or a theory of everything). Current possibilities for new equations include string theory and loop quantum gravity, but none of these are generally accepted yet.

It is important to remember that all of these different equations are only models of reality and are not actually real:

“All models are wrong. Some models are useful.”

— George Box

## Method of assumed forces and method of assumed motion

Newton's equations can be used in two main ways. Either we know the forces and we use this to compute the acceleration of a mass, or we know the acceleration and use this to compute the forces.

Method of assumed forces: | Know $\vec{F}$ | $\Longrightarrow$ | Compute $\vec{a}$ |
---|---|---|---|

Method of assumed motion: | Know $\vec{a}$ | $\Longrightarrow$ | Compute $\vec{F}$ |

Of course there are other possibilities from these two, such as knowing the vertical component of the force and the horizontal component of acceleration, and then computing the missing components of each.

Example Problem: Method of assumed forces

A cannonball of mass $m$ is fired from the origin with initial velocity $\vec{v}_0$ and then experiences a force $\vec{F}_{\rm g} = -m g \hat{\jmath}$ due to gravity and a wind force $\vec{F}_{\rm w} = -C_{\rm w} \hat{\imath}$. What is the motion of the cannonball as a function of time?

We can use Newton's equations to find the acceleration of the cannonball, using the method of assumed forces: \[\begin{aligned} m \vec{a} &= \vec{F}_{\rm w} + \vec{F}_{\rm g} \\ \vec{a} &= \frac{1}{m} \left( - C_{\rm w} \hat{\imath} - m g \hat{\jmath} \right) \end{aligned}\]

If $\vec{r}$ is the position vector of the cannonball then we have the differential equation: \[\begin{aligned} \ddot{\vec{r}}(t) &= - \frac{C_{\rm w}}{m} \hat{\imath} - g \hat{\jmath} & \vec{r}(0) &= \vec{0} & \dot{\vec{r}}(0) &= \vec{v}_0. \end{aligned}\] Because the acceleration is constant we can use the elementary motion solution to find the trajectory of the cannonball: \[ \vec{r}(t) = \vec{v}_0 t - \frac{1}{2} t^2 \left(\frac{C_{\rm w}}{m} \hat{\imath} + g \hat{\jmath}\right). \]

Example Problem: Method of assumed motion

A car of mass $m$ is observed driving on a sinusoidal road at a constant horizontal speed $v_0$. The road surface has equation $y = A \sin(k x)$, where $A$ is the amplitude and $k$ is the wavenumber. What is the force of the road on the car? Gravity $g$ acts vertically and assume no air resistance.

We can determine the acceleration of the car, and then use Newton's equations and method of assumed motion to find the total force and thus the road force. First, we find the acceleration: \[\begin{aligned} x(t) &= x_0 + v_0 t \\ \vec{r}(t) &= x(t)\,\hat{\imath} + y(x(t))\,\hat{\jmath} \\ &= (x_0 + v_0 t)\,\hat{\imath} + A \sin(k x_0 + k v_0 t)\,\hat{\jmath} \\ \vec{a}(t) = \ddot{\vec{r}}(t) &= - A (k v_0)^2 \sin(k x_0 + k v_0 t)\,\hat{\jmath} \\ &= - A (k v_0)^2 \sin(k x)\,\hat{\jmath}. \end{aligned}\] Given the force of gravity $\vec{F}_{\rm g} = - m g \,\hat{\jmath}$ and the road force $\vec{F}_{\rm r}$, Newton's equations give the forces as: \[\begin{aligned} \vec{F}_{\rm r} + \vec{F}_{\rm g} &= m \vec{a} \\ \vec{F}_{\rm r} &= - m A (k v_0)^2 \sin(k x)\,\hat{\jmath} + m g \,\hat{\jmath} \\ &= m \Big(g - A (k v_0)^2 \sin(k x)\Big) \,\hat{\jmath}, \end{aligned}\] where we have solved for the road force on the car.

The steps involved in analyzing a mechanical system with Newton's equations are as follows.

Solution procedure with Newton's equations.

\[\begin{aligned} &\text{1. FBD: draw a Free Body Diagram.} \\ &\text{2. Kinematics: determine $\vec{a}$.} \\ &\text{3. Newton: use $\vec{F} = m\vec{a}$.} \\ &\text{4. Algebra: rearrange and solve as needed.} \end{aligned}\]

Example Problem: Pendulum with Newton's equations

Consider the 2D pendulum with a massless rigid rod of length $\ell$ and a point mass $m$. What is the equation of motion for $\ddot\theta$ and the tension $T$ in the rod?

It is helpful in this problem to use both Cartesian and polar bases:

**1. FBD:** The free body diagram for the point mass is:

The forces on the free body diagram are: \[\begin{aligned} \vec{F}_g &= - mg \,\hat\jmath \\ \vec{F}_T &= - T \,\hat{e}_r. \end{aligned}\]

**2. Kinematics:** Using the polar
basis acceleration equation #rkv-ep gives:
\[\begin{aligned}
\vec{a} &= (\ddot{r} - r\dot\theta^2) \,\hat{e}_r
+ (r\ddot\theta + 2\dot{r}\dot\theta) \,\hat{e}_\theta \\
&= -\ell \dot\theta^2 \,\hat{e}_r
+ \ell\ddot\theta \,\hat{e}_\theta.
\end{aligned}\]

**3. Newton:** Using #rep-en gives:
\[\begin{aligned}
\vec{F} &= m\vec{a} \\
\vec{F}_T + \vec{F}_g
&= m(-\ell \dot\theta^2 \,\hat{e}_r
+ \ell\ddot\theta \,\hat{e}_\theta) \\
-T\,\hat{e}_r - mg\,\hat\jmath
&= -m\ell \dot\theta^2 \,\hat{e}_r
+ m\ell\ddot\theta \,\hat{e}_\theta.
\end{aligned}\]

**4. Algebra:** To compare components
in the above equation we need to switch to a single
basis. We will convert to $\hat{e}_r,\hat{e}_\theta$
using:
\[
\hat\jmath = -\cos\theta \,\hat{e}_r
+ \sin\theta \,\hat{e}_\theta,
\]
which gives:
\[
(-T + mg\cos\theta)\,\hat{e}_r
- mg\sin\theta\,\hat{e}_\theta
= -m\ell \dot\theta^2 \,\hat{e}_r
+ m\ell\ddot\theta \,\hat{e}_\theta.
\]
Equating the $\hat{e}_\theta$ and $\hat{e}_r$ terms
gives $\ddot\theta$ and $T$ by:
\[\begin{aligned}
\ddot\theta &= - \frac{g}{\ell} \sin\theta \\
T &= mg\cos\theta + m\ell\dot\theta^2.
\end{aligned}\]

Newton's equations $\vec{F} = m\vec{a}$ are an algebraic
expression of *Newton's second law*:

“Mutationem‑mutation/change
motûs‑motion
proportionalem‑proportional
esse‑is
vi‑force
motrici‑motive
impressæ‑impressed/applied,
et‑and
fieri‑is
secundum‑along
lineam‑line
rectam‑right/straight
quâ‑which
vis‑force
illa‑that
imprimitur‑imprints/acts.”

—
Isaac Newton,
Principia, 1687.

“The change of motion is proportional to the applied force, and takes place in the direction of the straight line in which the force acts.”

Although Newton was English and lived in England, he wrote the Principia in Latin because that was the standard language for scientific communication at the time. By the late 1600s the use of Latin was in decline, however, and Newton's later works, such as Opticks in 1704, were written in English.

## Linear momentum

The linear momentum of a point mass is the product of its mass and velocity:

Linear momentum of a point mass.

\[\vec{p} = m\vec{v}\]

Using the linear momentum, Newton's equations are:

Newton's equations in momentum form.

\[\vec{F} = \dot{\vec{p}}\]

If we differentiate the linear momentum with respect to time then we obtain \[\frac{d}{dt} \vec{p} = \frac{d}{dt} (m \vec{v}) = m \dot{\vec{v}} = m \vec{a} = \vec{F}. \]

Warning: Variable mass problems can't simply use $\vec{F} = \dot{\vec{p}}$.

For a system with variable mass, such as a rocket burning up its fuel, it is tempting to write \[\vec{F} = \dot{\vec{p}} = \frac{d}{dt} (m \vec{v}) = \dot{m} \vec{v} + m \dot{\vec{v}}\] and to then rearrange to find the acceleration: \[m \vec{a} = \vec{F} - \dot{m} \vec{v}.\] But this is wrong! The acceleration must depend not only on the rate of mass loss, but also on the speed with which the lost mass is ejected.

The right way to calculate the acceleration of a rocket is to consider the center of mass of the total system (rocket plus ejected mass) and to use the equation for the center of mass of a many-body system. This results in the Tsiolkovsky rocket equation: \[m \vec{a} = \vec{v}_{\rm e} \dot{m},\] where the acceleration $\vec{a}$ of the rocket is determined by the rate of mass loss $\dot{m}$ and the velocity $\vec{v}_{\rm e}$ of the ejected mass relative to the rocket.

## Angular momentum

Angular momentum is defined with respect to a given base point $O$. For a point mass, the angular momentum is the cross product between the position vector and the linear momentum:

Angular momentum about fixed base point $O$ of a point mass at $P$.

\[\vec{H}_O = \vec{r}_{OP} \times \vec{p}_P = \vec{r}_{OP} \times m\vec{v}_P\]

When a force is applied to a body it produces a
*moment* about any given fixed base point $O$:

Moment about fixed point $O$ of a force acting at $P$.

\[\vec{M}_O = \vec{r}_{OP} \times \vec{F}_P\]

The angular momentum and applied moment are related by:

Moment equation about fixed base point $O$.

\[\vec{M}_O = \dot{\vec{H}}_O\]

If we compute the derivative of $\vec{H}_O$ and use Newton's equations #rep-en, then we obtain: \[\begin{aligned} \dot{\vec{H}}_O &= \frac{d}{dt} \Big( \vec{r}_{OP} \times m \vec{v}_P \Big) \\ &= \dot{\vec{r}}_{OP} \times m \vec{v}_P + \vec{r}_{OP} \times m \dot{\vec{v}}_P \\ &= m \vec{v}_P \times \vec{v}_P + \vec{r}_{OP} \times m \vec{a}_P \\ &= 0 + \vec{r}_{OP} \times \vec{F} \\ &= \vec{M}_O. \end{aligned}\] Here we used the fact that, for a fixed point $O$, the derivative $\dot{\vec{r}}_{OP}$ is simply the velocity $\vec{v}_P$ of point $P$, and the cross product of the velocity with itself is zero.

Warning: Always indicate the base point for angular momentum and moments.

In the moment equation #rep-et it
is crucial that $\vec{M}_O$ and $\vec{H}_O$ are both
computed about the same base point $O$. If we have
$\vec{M}_O$ about $O$ and $\vec{H}_C$ about $C$, then it
is *not true* that $\vec{M}_O =
\dot{\vec{H}}_C$. For this reason we always write the
base point $O$ as a subscript on $M$ and $H$, otherwise
it's easy to forget which point we've used as the base
point.