Particle Kinetics

Classical mechanics

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.

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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.

$$ \begin{align}\textbf{Method of assumed forces: } \text{Know } \vec{F} \Longrightarrow \text{Compute } \vec{a} \\ \textbf{Method of assumed motion: } \text{Know } \vec{a} \Longrightarrow \text{Compute } \vec{F} \end{align} $$

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 #rep-xf

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:

$$ \begin{aligned} m \vec{a} &= \vec{F}_{\rm w} + \vec{F}_{\rm g} \\ \vec{a} &= \frac{1}{m} \left( - C_{\rm w} \hat{\imath} - m g \hat{\jmath} \right) \end{aligned} $$

Example Problem: Method of assumed motion #rep-xm

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.

Solution steps

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

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

Example Problem: Pendulum with Newton's equations #rep‑xl

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:

$$ \begin{aligned} \ddot\theta &= - \frac{g}{\ell} \sin\theta \\ T &= mg\cos\theta + m\ell\dot\theta^2. \end{aligned} $$

More information about Free body diagrams included in "Free boy diagrams"

Numerical integration

Add information shown in Fig \ref fig:NumericalIntegration

Fig: NumericalIntegration

Applications

Kiiking

This topic is in L13-Notes, slides 9-10. Include information in Fig \ref and this YouTube link https://www.youtube.com/shorts/qvW0sz4kBLQ

. Application for "Particle kinetics".

Fig: AppKiiking

Accelerating and braking

What happens when we step on the gas or brake in a car? The car pushes against the road to either accelerate (gas pedal) or decelerate (brake pedal). But how are the forces on the car and wheels distributed? What determines whether the wheels grip the road or lose traction and spin or slide?

To study this problem we need a model. Let's start with the simplest model and then gradually consider more complex models.

Fig: red_car.jpg

The classic American Pony Car: a 1965 Ford Mustang Fastback with the Golden Gate Bridge in the background. Image source: flikr image by Nick Ares (CC BY-SA 2.0) (full-sized image).

The simplest model of a car is to treat the entire vehicle as a point mass. On a we have vertical force balance for a stationary car. When the car , there is a horizontal forward force on the car, and a corresponding backwards horizontal force on the ground. As the car picks up speed, air resistance produces a backwards force. On the diagram we have drawn some forces offset from the center of mass, so that the vectors don't overlap. Because we are assuming a point mass model, however, all vectors are really acting at the same point.

While cruising at a constant speed, there is a balance between the horizontal driving force and the drag force due to air resistance. When the car , there is a backwards force that slows the car down to a stop.

We can think of the force vectors (such as the ground force on the car) as either in separate horizontal and vertical or as unified vectors. It may be helpful to pause the during and to consider the forces at work.

Reference material

Extra links

Video of a burnout with a deliberate loss of traction.
Video of a wheelie, where the front wheels lose contact.
V Video of braking in which the rear wheels lose contact.
Another burnout video, taken to the point of destruction.

Banked turns

\label sub:PartKin_turns

Complete in "Banked turns". Just the introduction and the information under "Track geometry" and "Point mass model".

Projectiles with air resistance

Complete in "Projectiles with air resistance".

Consider a spherical object, such as a baseball, moving through the air. The motion of an object though a fluid is one of the most complex problems in all of science, and it is still not completely understood to this day. One of the reasons this problem is so challenging is that, in general, there are many different forces acting on such objects, including:

gravity
drag
lift
thrust
buoyancy
bulk fluid motion, such as wind
inertial forces, such as the centrifugal and Coriolis forces

In most introductory physics and dynamics courses, gravity is the only force that is accounted for (this is equivalent to assuming that the motion takes place in a vacuum). Here we will consider realistic and accurate models of air resistance that are used to model the motion of projectiles like baseballs.

Fig: baseball_field.jpg

Wrigley Field in Chicago, Illinois is the home of the Chicago Cubs baseball team. In baseball, an “out of the park” home run is scored when the ball is hit beyond the green outfield (past the yellow foul pole at the left of the above image). Image source: Flickr image by Mike Bash (CC BY NC ND 2.0) (full-sized image).

Drag forces and drag coefficients

The drag force is always directly opposed to the velocity of the object. In vector notation,

$$ \vec{F}_{\rm D} = -F_{\rm D} \hat{v}, $$

where $ F_{\rm D} $ is the magnitude of the drag force, and $ \hat{v} $ is the unit vector in the direction of the object's velocity.

The magnitude of the drag force is characterized by the dimensionless drag coefficient $ C_{\rm D} $, given by

$$ C_{\rm D} = \frac{F_{\rm D}}{\frac{1}{2} \rho A v^2}, $$

where $ \rho $ is the density of the fluid (in this case, air), $ A=(1/4)\pi D^{2} $ is the cross-sectional area of the object, and $ v $ is the object's speed.

Drag coefficients as a function of Reynolds number

A dimensionless parameter that is very useful in fluid dynamics is the Reynolds number, which is defined as

$$ {\rm Re} = \frac{\rho v L}{\mu}, $$

where $ L $ is a characteristic length for the flow (in this case, the diameter $ D $ of the ball), and $ \mu $ is the dynamic viscosity of the fluid. The Reynolds number gives a ratio between inertial forces and viscous forces in a fluid flow. For very small Reynolds numbers, the viscous forces are much stronger than the inertial forces (think of trying to stir a cup of honey). For very large Reynolds numbers, the viscous forces are negligible, and we refer to the flow as inviscid (think of stirring a cup of coffee).

An important result in fluid dynamics is that the drag coefficient is a function only of the Reynolds number of the fluid flow about the object. That is,

$$ C_{\rm D} = C_{\rm D}({\rm Re}). $$

This functional relationship has no closed form. However, the relationship has been established numerically based on experimental data. See Figure #aft-fd for a schematic diagram of the drag coefficient's dependence on the Reynolds number.

Fig: drag_coefficient

Experimental dependence of drag coefficient on Reynolds number for a sphere. Image credit: NASA via Wikipedia.

Quadratic drag model

Notice from Figure #aft-fd that there is a range of Reynolds numbers ($ 10^3 < {\rm Re} < 10^5 $), characteristic of macroscopic projectiles, for which the drag coefficient is approximately constant at about 1/2 (see the part of the curve labeled “4” in Figure #aft-fd). That the drag coefficient is constant means that, within this region, the magnitude of the drag force is proportional to the square of the object’s speed. Substituting $ C_{\rm D}=1/2$ into the definition of $C_{D} $ above, we obtain

$$ F_{\rm D} = \frac{1}{4} \rho A v^2 = c v^2, $$

where we have defined the quadratic drag parameter $c$ as

$$ c = \frac{1}{4} \rho A = \frac{pi}{16} \rho D^2, $$

which is a constant for a given projectile. This is referred to as quadratic drag, because the drag force magnitude is proportional to the square of the speed. Using this result, we can write the drag force vector as

$$ \vec{F}_{\rm D} = - c v^2 \hat{v}. $$

In the presence of gravity and quadratic drag alone, the net force on an object is given by

$$ \sum \vec{F} = m \vec{g} - c v^2 \hat{v}, $$

where $ m $ is the mass of the object and $ \vec{g} $ is the local acceleration due to gravity. Below is an interactive graphic showing the trajectory of an object with gravity and quadratic drag. You can change the various parameters to see their effects on the object’s motion.