# Mechanics - Motion

 Content Motion

## Motion

There exists different type of motions: translational motion (motion along a straight line), rotational motion (motion around a internal axis of rotation) and oscillatory motion (also known as vibrational or periodic motion, is the movement of an object between two points continuously).

We start our study of motion with particle kinematics. Particle kinematics is the study of the motion of a single particle. It is an idealization of a real world scenario of an object whose mass is neglectible compared to its environment.

The results obtained in particle kinematics are used to study the kinematics of a collection of particles by aggregation of the behaviour of the individual particles. These composite objects are assumed to be rigid bodies, meaning that the relative positions of the particles of which the object is composed remain constant over time. Relaxing this rigidness assumption allows for a further refinement of the study of mechanics to include deformable bodies.

The position of a point in space is the most fundamental idea in particle kinematics. To specify the position of a point, one must specify three things: the reference point (often called the origin), distance from the reference point and the direction in space of the straight line from the reference point to the particle. Position is usually described by mathematical quantities: vectors. For measurement of distances and directions, usually three dimensional coordinate systems are used with the origin coinciding with the reference point. A three-dimensional coordinate system (whose origin coincides with the reference point) with some provision for time measurement is called a reference frame. All observations in physics are incomplete without the reference frame being specified.

If the position vector of the particle (relative to a given reference frame) changes with time, then the particle is said to be in motion with respect to the chosen reference frame. However, if the position vector of the particle (relative to a given reference frame) remains the same with time, then the particle is said to be at rest with respect to the chosen frame. Note that rest and motion are relative to the reference frame chosen. It is quite possible that a particle at rest relative to a particular reference frame is in motion relative to the other. Hence, rest and motion aren't absolute terms, rather they are dependent on reference frame.

The position of a particle at any time can be specified by its rectangular coordinates $x_1,x_2,x_3$, its cylindric coordinates $R,\theta,z$ or its spherical coordinates $R,\theta, \phi$. Furthermore the motion of a particle along its path can also be measured by its path variables, the tangent and normal to the path.

The path of motion can be along a straight line, (one-dimensional motion), along a curve in a plane (two-dimensional motion), or along a curve in space (three-dimensional motion).

Displacement is a vector describing the difference in position between two points, i.e. it is the change in position the particle undergoes during the time interval. Geometrically, displacement is the shortest distance between the points A and B. Displacement, distinct from position vector, is independent of the reference frame.

A particle's path is the locus between its beginning and end points which is reference-frame dependent. The path of a particle may be rectilinear (straight line) in one frame, and curved in another. Distance is a scalar quantity, describing the length of the path between two points along which a particle has travelled.

In kinematics motion is described by three quantities:

• position $\vec{r}$ (unit: meter, $m$)
• velocity $\vec{v}$ (unit: meter/second, $m.s^{-1}$)
• acceleration $\vec{a}$ (unit: meter/(second x second), $m.s^{-2}$)

These variables are mathematical objects called vectors. Normally they are described as vector functions of time: $\vec{r}(t),\vec{v}(t),\vec{a}(t)$. We denote vectors with bold letters and scalars with normal letters.

The relation between the kinematic quantities is based on calculus: \begin{align} \vec{v}(t)&=\lim_{\Delta t\to 0}\frac{\Delta \vec{r}}{\Delta t}=\frac{d\vec{r}}{dt}=\dot{\vec{r}}(t) \\ \vec{a}(t)&=\lim_{\Delta t\to 0}\frac{\Delta \vec{v}}{\Delta t}=\frac{d\vec{v}}{dt}=\ddot{\vec{r}}(t) \end{align}
$\Delta \vec{r}$ is the displacement vector. For infinitesimally small segments the increase of the distance (arc length), $s$, we have $\norm{d\vec{r}}=ds$ from which follows: $\frac{ds}{dt}=\norm{\vec{v}}$

The direction of a velocity vector is always tangential to the path of motion. This follows from the definition of the velocity vector as being the first derivative of the position vector.

Uniform motion is defined as a motion with constant speed.In case of uniform motion there is either no acceleration or there is a constant acceleration whose direction is perpendicular (normal) to the path of motion. If an acceleration is not normal to the path of motion there exists a component tangential to the path of motion and this component changes the speed. This is excluded by the assumption of uniform motion.

The inverse of the derivative relations are the integral relations for the kinematic quantities:

displacement: \begin{align} ds&=\vec{v}dt \\ \int_{s_0}^s{ds}&=\int_0^t{\vec{v}dt} \\ s&=s_0+ \int_0^t{\vec{v}dt} \tag{1} \end{align}

velocity: \begin{align} d\vec{v}&=\vec{a}dt \\ \int_{\vec{v_0}}^\vec{v}{d\vec{v}}&=\int_0^t{\vec{a}dt} \\ \vec{v}&=\vec{v_0}+\int_0^t{\vec{a}dt} \tag{2} \end{align}

velocity with elimination of time: \begin{align} \vec{v}d\vec{v}&=\vec{a}ds \\ \int_{\vec{v_0}}^\vec{v}{\vec{v}d\vec{v}}&=\int_{s_0}^s{\vec{a}d\vec{s}} \\ \vec{v}^2&=2(\vec{v_0}^2+\int_{s_0}^s{\vec{a}ds}) \tag{3} \end{align}

### Translational motion

We derive 4 kinematic equations of translational motion for the case of constant acceleration (uniformally acceleration), $\vec{a}=a$. As the motion is along a straight line we do not need the vector quantities but instead use only scalar variables: distance $s$, velocity $v$, acceleration $a$. With a plus or minus the sense of the direction along the line is specified.

### $v=v_0+at \tag{1}$

#### Displacement

\begin{align} s&=s_0+ \int_0^t{vdt} \\ s&=s_0+ \int_0^t{(v_0+at)dt} \\ \end{align}

### $s=s_0+v_0t+1/2at^2 \tag{2}$

#### Velocity, without time

\begin{align} t&=\frac{v-v_0}{a} \\ s&=s_0+v_0\left(\frac{v_n-v_0}{a}\right)+1/2a\left(\frac{v_n-v_0}{a}\right)^2 \\ 2as&=2as_0+2v_0(v_n-v_0)+(v_n-v_0)^2 \\ \end{align}

### $v^2=v_0^2+2a(s-s_0) \tag{3}$

#### Displacement, without acceleration

\begin{align} a&=\frac{v-v_0}{t} \\ s&=s_0+v_0t+1/2\left(\frac{v-v_0}{t}\right)t^2 \\ s&=s_0+v_0t+1/2(v-v_0)t \\ \end{align}

### Rotational motion

This type of motion is a circular motion around a center of rotation. This type of motion is most conveniently described with angular quantities. We first describe the scalar angular quantities and then analyse their vector representation.

### Angular scalar quantities

#### Angular position

$\theta = \frac{s}{r} \text{ (rad) } \Leftrightarrow s=r\theta$

The angular position $\theta$ is measured in radians (rad), with the positive direction defined counterclockwise from the positive x-axis. A radian is a dimensionless quantity as its is a ratio of two lengths. One radian is the angle subtended by an arc length, $s$ equal to the radius $r$. A full circle is $2\pi \; \text{rad}$. But a full circle is also $360^\circ$ degrees. So we can express a radian also in degrees by: $1 \; \text{rad}=\frac{360^\circ}{2 \pi}\thickapprox 57.3^\circ$

#### Angular velocity

$\omega = \lim_{\Delta t \to 0}\frac{\Delta \theta}{\Delta t} = \dot{\theta}$

The angular velocity, $\omega$ (omega) has a dimension of an inverse time and is measured in units of radians/second $\text{rad}.s^{-1}$ or revolutions/second $rev.s^{-1}$.

#### Angular acceleration

$\alpha = \lim_{\Delta t \to 0}\frac{\Delta \omega}{\Delta t} = \dot{\omega}=\ddot{\theta}$

The angular acceleration, $\alpha$ (alpha) has a dimension of an inverse time squared and is measured in units of radians/second squared $\text{rad}.s^{-2}$ or revolutions/second squared $rev.s^{-2}$.

When a rigid object, (a non-deformable body where the relative locations of its constituents parts remain constant during the motion) is rotating about a fixed axis, every particle on the object rotates through the same angle in a given time interval and has the same angular speed and the same angular acceleration. That is, the quantities $\theta,\omega,\alpha$, characterize the rotational motion of the entire rigid object as well as individual particles in the object. Using these quantities, we can greatly simplify the analysis of rigid-object rotation.

Angular position ($\theta$), angular speed ($\omega$), and angular acceleration ($\alpha$) are analogous to translational position ($r$), translational speed ($v$), and translational acceleration ($a$). The angular parameters differ dimensionally from the translational parameters only by a factor having the unit of length.

The equations of motion for uniformally circular accelaration are similar to their translational counterparts:

### Equations of motion

#### Circular motion

$v=v_0+at$ $\omega=\omega_0+\alpha t$
$s=s_0+v_0t+1/2at^2$ $\theta=\theta_0+\omega_0t+1/2\alpha t^2$
$v^2=v_0^2+2a(s-s_0)$ $\omega^2=\omega_0^2+2\alpha(\theta-\theta_0)$
$s_n=s_0+1/2(v_n+v_0)t$ $\theta=\theta_0+1/2(\omega+\omega_0)t$

### Angular vector quantities

Before the angular quantities were treated as scalars. Are these angular quantities vector quantities ? The answer depends on whether the angular displacement is a vector quantity. In general $\theta$ is not a vector quantity, as can be seen from the fact that $\theta_1+\theta_2 \neq \theta_2+\theta_1$.

However if angular displacements are made infinitesimal this differences disappears. So we can conclude that $\mathrm{d}\theta$ is a vector quantity. From this follows that $\omega = \mathrm{d}\theta / \mathrm{dt}$, is also a vector quantity. The direction of this vector is defined as normal to the plane of motion and in the direction of the right-hand screw rule. The angular acceleration $\alpha = \mathrm{d}\omega / \mathrm{dt}$ is also a vector normal to the plane of motion and in the direction of $\omega$ if $\alpha>0$ and in opposite direction if $\alpha <0$.

### Circular motion

Circular motion is motion in the plane along a circle. A circle is defined as the locus of points with a fixed distance, the radius $r$, to the center of the circle and is described by the following equation: $x^2+y^2=r^2$. Describing circular motion is most convenient with polar coordinates, $(r,\theta)$, instead of Cartesian coordinates $(x,y)$. If the radius $r$ is not constant with time the motion described becomes a curved path. Circular motion is equivalent to rotational motion around a fixed axis in space.

For circular motion the velocity vector is normal to the position vector or in symbols: $r(t).\dot{\vec{r}}(t)=0$ PROOF. This follows directly from the fact $\norm{r(t)}=c$. \begin{align*} c^2&=r(t).r(t) \\ 0 &= \frac{d}{dt}r(t).r(t) \\ &=\dot{\vec{r}}(t).r(t)+r(t).\dot{\vec{r}}(t)\\ &=2\dot{\vec{r}}(t).r(t) \end{align*} From which follows $\dot{\vec{r}}(t).r(t)=0$

#### UNIFORM CIRCULAR MOTION

Uniform motion with constant acceleration is called uniform circular motion. The constant acceleration is called centripetal acceleration and is directed toward the center of the circle. The centripetal acceleration does not influence the speed, but only changes the direction of the motion along the circle.

The forces causing the centripetal acceleration are one of the familiar sources of force, it is not a new kind of force. The force is the pull or push on the particle to keep its path curving along the circle. For the motion of the Earth around the Sun, the centripetal force is gravity. For an object sitting on a rotating turntable, the centripetal force is friction. For a rock whirled horizontally on the end of a string, the magnitude of the centripetal force is the tension in the string. Once the force is removed the particle continues its path along a straight a line (or projectile motion in a gravitational field).

There is often confusion about the forces acting on a particle in uniform circular motion. If we analyse the motion of the particle from an inertial reference frame, then there is only one net force acting on the particle, causing a centripetal acceleration. The particle itself exerts a reaction force on the body that maintains the circular motion (such as a string, surrounding or gravity). If we analyze the motion from the rotating, non-inertial frame, the situation is different. In that case we have two forces acting on the particle, the force causing the centripetal acceleration and an inertial force, called centrifugal force. In the rotating frame the particle is at rest and so these forces cancel each other. That the centrifugal force is not a Newtonian force follows from the fact that it has not a reaction force on another body. Once the force causing the centripetal acceleration is removed the particle moves away in a straight line for the observer in the rotational frame. This acceleration is then attributed to this centrifugal force.

For uniform circular motion the acceleration vector is normal to the velocity vector or in symbols: $\dot{\vec{r}}(t).\ddot{\vec{r}}(t)=0$ This follows from $\norm{\dot{\vec{r}}(t)}=c$.The proof follows the same approach as the proof for the velocity vector.

The speed of uniform circular motion is given by:

#### Speed

translational variable: $$v=\frac{2\pi R}{T} \; (m.s^{-1})$$ angular variable: $$\omega=\frac{2\pi }{T} \; (rad.s^{-1})$$ with $R$ the radius of the circle and $T$ the time in seconds for one complete revolution. The inverse of $T$ is called the frequency $f$, with unit $1\text{Hz}=1s^{-1}$.

For the centripetal acceleration we have:

#### Centripetal acceleration

$$$a_c=\frac{v^2}{R} =\frac{4\pi^2R}{T^2}$$$

PROOF. By definition we have for the average acceleration: $$$\bar{a}_c=\frac{\Delta v}{\Delta t} \;\; (1)$$$ Notice that the angle $\theta$ between the position vectors is equal to the angle between the velocity vectors because the velocity vectors are always normal to the position vectors. The figure below shows that the two triangles formed by the position vectors $\vec{r_i},\vec{r_f},\Delta\vec{r}$, and the velocity vectors $\vec{v_i},\vec{v_f},\Delta\vec{v}$ are similar. This follows from the fact that both are isosceles triangles with the same vertex angle, so all angles in the two triangles are the same.

This implies following: $\frac{\norm{\Delta \vec{r}}}{r}=\frac{\norm{\Delta \vec{v}}}{v}$ which implies: $\Delta v=\frac{v}{r}\Delta r$ if we substitute this in (1) we have: $\bar{a}_c=\frac{v}{r}\frac{\Delta r}{\Delta t}$ now if we take limits $\Delta t \rightarrow 0$ then $\frac{\Delta r}{\Delta t}\rightarrow v$ and we get: $a_c=\frac{v^2}{r}$

#### RELATION BETWEEN translational AND ANGULAR SCALARS

For circular motion the radius, $r$, is a constant and we start with the following relation between translational and angular scalar position: $s=\theta r$ with $\theta$ in radians.

Differentiating both sides of this equation with respect to the time we obtain: $\frac{ds}{dt}=\frac{d\theta}{dt} r$ Now $ds/dt$ is the translational speed and $d\theta/dt$ is the angular speed so that: $v=\omega r$

Differentiating this equation with respect to time, we have $\frac{dv}{dt}=\frac{d\omega}{dt} r$ But $dv/dt$ is the scalar tangential component of acceleration of the particle and $d\omega/dt$ is the scalar angular acceleration, so that $a_t=\alpha r$

We have seen that the scalar radial component of acceleration is $v^2/r$ this can be expressed in terms of angular speed by using $v=\omega r$. We have $a_r=\frac{v^2}{r}=\omega^2 r$

#### VECTOR REPRESENTATION OF CIRCULAR MOTION

Describing circular motion is most convenient with polar coordinates, $(r,\theta)$, instead of Cartesian coordinates $(x,y)$.

We have the following relationships between Cartesian and polar coordinates: $\begin{gather} r=\sqrt{x^2+y^2} \\ \theta = \tan^{-1} \left(\frac{y}{x}\right) \end{gather}$ and $\begin{gather} x=r\cos \theta \\ y=r\sin \theta \end{gather}$

For polar coordinates we define two new unit vectors: $\vec{e_r}$ the radial unit vector at any point which is in the direction of $\vec{r}$, radially outward from the origin, and $\vec{e_\theta}$ the angular unit vector at any point which is in the direction of increasing $\theta$, tangent to a circle through the point in a counterclockwise direction. These new unit vectors change from point to point and are thus not constant vectors like their Cartesian counterparts $\vec{e_x},\vec{e_y}$ but their direction depend on $\theta$. These unit vectors are orthogonal to each other at each point: $\vec{e_\theta} \perp \vec{e_r}$.

#### Uniform circular motion in polar coordinates

For uniform circular motion we have following velocity vector in polar coordinates:

#### Velocity vector

$\vec{v}=\vec{e_\theta} v \tag{1}$
We derive the acceleration vector as follows (taken into account that $v$ is a constant): \begin{align} \vec{a}=\frac{\mathrm{d}\vec{v}}{\mathrm{dt}}&=\vec{e_\theta} \frac{\mathrm{d}v}{dt}+\frac{\mathrm{d}\vec{e_\theta}}{\mathrm{dt}}v \\ &=\frac{\mathrm{d}\vec{e_\theta}}{\mathrm{dt}}v \tag{2} \end{align} Now $\Delta \vec{e_\theta}$ points radially inward toward the origin in the limiting case as $\Delta t \to 0$, so $\mathrm{d} \vec{e_\theta}$ at any point has the direction of $-\vec{e_r}$. This gives: $\frac{\mathrm{d}\vec{e_\theta}}{\mathrm{dt}}=-\vec{e_r}\frac{\mathrm{d}\theta}{\mathrm{dt}} = -\vec{e_r} \omega \tag{3}$ Combining (2 and (3 gives: $\vec{a}=-\vec{e_r}\frac{\mathrm{d}\theta}{\mathrm{dt}}v=-\vec{e_r} \omega v \tag{4}$ For uniform circular motion we have $\omega = \dfrac{v}{r}$, so finally we have the vector equation:

#### Tangential acceleration vector

$\vec{a}=-\vec{e_r} \frac{v^2}{r} \tag{5}$

#### GENERAL CIRCULAR MOTION

We now will describe curved motion in the plane where the radius $r$ and velocity $v$ are not constants. We will again use polar unit vectors. The relation between polar unit vectors and Cartesian unit vectors will assist us in deriving some important relationships: \begin{align} \vec{e_r}&=\cos \theta \vec{e_x} + \sin \theta \vec{e_y} \\ \vec{e_\theta} &= -\sin \theta \vec{e_x} + \cos \theta \vec{e_y} \end{align}

We see from this easily the following relationships: \begin{align} \frac{\mathrm{d}\vec{e_r}}{\mathrm{d}\theta}&=\vec{e_\theta} \tag{6} \\ \frac{\mathrm{d}\vec{e_\theta}}{\mathrm{d}\theta}&=-\vec{e_r} \tag{7} \\ \end{align} and also \begin{align*} \vec{e_r} \times \vec{e_\theta} &= \left( \begin{array}{ccc} \vec{e_x} & \vec{e_y} & \vec{e_z} \\ \cos \theta & \sin \theta & 0\\ -\sin \theta & \cos \theta & 0 \\ \end{array} \right) \\ &=(\cos^2 \theta + \sin^2 \theta) \vec{e_z} \\ &= \vec{e_z} \tag{8} \end{align*} Furthermore following relations are also of interest: $\frac{\mathrm{d}\vec{e_r}}{\mathrm{dt}}=\frac{\mathrm{d}\vec{e_r}}{\mathrm{d}\theta}\frac{\mathrm{d}\theta}{\mathrm{dt}}=\omega\vec{e_\theta} \tag{9}$ $\frac{\mathrm{d}\vec{e_\theta}}{\mathrm{dt}}=\frac{\mathrm{d}\vec{e_\theta}}{\mathrm{d}\theta}\frac{\mathrm{d}\theta}{\mathrm{dt}}=-\omega\vec{e_r} \tag{10}$

We can now derive the position, velocity and acceleration vectors in polar coordinates for curved motion in the plane. We write the position vector $\vec{r}$ of circular motion as follows: $\vec{r}=r\vec{e_r}$

We now proceed with finding the two other kinematic quantities in polar coordinates: $\begin{gather} \vec{v}=\frac{\mathrm{d}\vec{r}}{\mathrm{dt}}\\ \vec{a}=\frac{\mathrm{d}^2\vec{r}}{\mathrm{dt}} \end{gather}$

We start with the velocity: \begin{align} v&=\frac{\mathrm{d}\vec{r}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(r \vec{e_r} \right) \\ &=\frac{\mathrm{d\vec{r}}}{\mathrm{dt}}\vec{e_r}+r\frac{\mathrm{d}\vec{e_r}}{\mathrm{dt}} \\ &=\dot{\vec{r}}\vec{e_r}+r\frac{\mathrm{d}\vec{e_r}}{\mathrm{d}\theta}\frac{\mathrm{d}\theta}{\mathrm{dt}} \\ &=\dot{\vec{r}}\vec{e_r}+r\omega\vec{e_\theta} \end{align} The scalars $\dot{\vec{r}}$ and $r\omega$ are called respectively the radial and tangential components of velocity. The speed of $v$ is given by: $\norm{v}=\sqrt{v \cdot v}=\sqrt{\dot{\vec{r}}^2+r^2\omega^2}$

We continue with the acceleration \begin{align} \vec{a} &=\frac{\mathrm{d}}{\mathrm{dt}}\left(\dot{\vec{r}}\vec{e_r}+r\omega\vec{e_\theta}\right) \\ &=(\ddot{\vec{r}}\vec{e_r}+\dot{\vec{r}}\frac{\mathrm{d}\vec{e_r}}{\mathrm{d}\theta})+(\dot{\vec{r}}\omega\vec{e_\theta}+r\alpha\vec{e_\theta}+r\omega\frac{\mathrm{d}\vec{e_\theta}}{\mathrm{dt}}) \\ &=(\ddot{\vec{r}}\vec{e_r}+\dot{\vec{r}}\omega\vec{e_\theta})+(\dot{\vec{r}}\omega\vec{e_\theta}+r\alpha\vec{e_\theta}-r\omega^2 \vec{e_r}) \\ &=(\ddot{\vec{r}}-r\omega^2)\vec{e_r}+(r\alpha+2\dot{\vec{r}}\omega)\vec{e_\theta} \end{align} The first two terms are the radial acceleration components: the translational acceleration $\ddot{\vec{r}}$, and centripetal acceleration,$-r\omega^2$. The second to terms are the tangential acceleration components: the translational tangential acceleration $r\alpha$ and the Coriolis acceleration $2\dot{\vec{r}}\omega$.

#### RELATION BETWEEN translational AND ANGULAR VECTORS

We write the vector relations between translational and angular quantities as follows (based on the equation of curved motion with constant $r$): \begin{align} \vec{v}&=\vec{\omega} \times \vec{r} \\ \vec{a}&= \vec{\omega} \times \vec{v} +\vec{\alpha} \times \vec{r} \end{align}

NOTE 1) $\vec{\omega}$ and $\vec{r}$, $\vec{\omega}$ and $\vec{v}$, $\vec{\alpha}$ and $\vec{r}$ each mutually perpendicular, thus the angle $\phi$ for each of these three pair of vectors is $90^\circ$, then in the cross product between these vectors $\sin \phi=1$. (2) the direction of $c=a \times b$ is given by the right-hand screw rule, using this rule gives the correct directions: $v \perp r$, $a_t \parallel v$ and $a_r$ opposite to $r$.

### Combined motion

The general motion of a system of particles, e.g. a rigid body, can always be separated into translational motion of its center of mass and rotational motion about its center of mass. The axis of rotation passes through the center of mass and has a fixed direction.