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# Motion of Particles Position Vector, Velocity and Acceleration

Introduction
If a particle is moving along a curved path rather than a straight path, then it is said to be in curvilinear motion. It can be in 2D or 3D. These are more complex than the rectilinear motion of particles.

• Position Vector, Velocity, and Acceleration
Position vector, also known as radius vector or location vector, is a Euclidean vector which represents the location or a position of any point P in space with respect to an arbitrary reference origin O. At any time t, for specifying position of particle in x-y reference system, position vector is used as in fig 1.1 a. Since is characterized by its magnitude r and direction with respect to reference axes, it completely defines the position of a particle with respect to those axes. Let us consider, at another time $(t+\Delta t)$, particle occupies new position P’ along with position vector $\vec{r^{'}}$. Now $\Delta \vec{r}$ represents a change in both magnitude and direction of the position vector. The average velocity of particle at time interval $(\Delta t)$ = $\frac{\Delta F}{\Delta t}$,which has direction of $\Delta \vec{r}$ magnitude equal to that of $\Delta \vec{r}$ divided by Δt. (fig 1.1 b) So, instantaneous velocity becomes, $\vec{v}$ = $\lim_{\Delta t\to 0}\frac{\Delta \vec{r}}{\Delta t}$ = $\frac{d\vec{r}}{dt}$ Now P and P’ gets closer as $\Delta \vec{r}$ and Δt becomes shorter and is tangent to the path of the particle. (fig 1.1 c) Here, the length of the segment (PP’) tends to the length of arc Δs, as Δt decreases. $\therefore V$ = $\lim_{\Delta r\to 0}\frac{PP^{'}}{\Delta t}$ = $\lim_{\Delta r\to 0}\frac{\Delta S}{\Delta t}$ = $\frac{ds}{dt}$ here,$\Delta \vec{r}$ can be resolved to and parallel to x and y-axes as in fig 1.1 d.
$\Delta \vec{r}$ = $\vec{PP^{"}} + \vec{P^{"}P^{'}}$ Dividing above equation by Δt and taking the limit as Δt → 0 .
$\vec{v}$ = $\lim_{\Delta t\to 0}\frac{\Delta \vec{r}}{\Delta t}$ = $\lim_{\Delta t\to 0}\frac{\vec{PP^{"}}}{\Delta t} + \lim_{\Delta t\to 0}\frac{\vec{P^{"}P^{'}}}{\Delta t}$
$\therefore \vec{v}$ = $V_x\hat{i} + V_y\hat{j}$
where, $V_x$ = $\lim_{\Delta t\to 0}\frac{\Delta x}{\Delta t}$ = $\frac{dx}{dt}$ = $\dot{x}$
$V_y$ = $\lim_{\Delta t\to 0}\frac{\Delta y}{\Delta t}$ = $\frac{dy}{dt}$ = $\dot{y}$
A positive value of vx gives rightward direction and $V_y$ gives upward direction.
Then, V = $\sqrt{v_x^2 + v_y^2}, tan\alpha$ = $\frac{V_v}{V_x}$ (see fig 1.1e)
In fig 1.2 a, $\vec{v}$ and $\vec{v^{'}}$ be velocities at time t and (t+Δt), that is, tangents at P and P’.
Let, vectors $\vec{v}$ and $\vec{v^{'}}$ be passed through common origin O’ as in fig 1.2 b. Now, the average acceleration of particle over time Δt is = $\frac{\Delta \vec{v}}{\Delta t}$
Also, instantaneous acceleration,
$\vec{a}$ = $\lim_{\Delta t\to 0}\frac{\Delta \vec{v}}{\Delta t}$ = $\frac{d\vec{v}}{dt}$
Again, $\Delta \vec{v}$can be resolved to$\vec{QQ^{"}}\;and\;\vec{Q^{"}Q^{'}}$ parallel to x and y-axes, then
$\Delta \vec{v}$ = $\vec{QQ^{"}} + \vec{Q^{"}Q^{'}}$.
Dividing above equation by Δt and taking limit as Δt→ 0, we get
$\vec{a}$ = $\lim_{\Delta t\to 0}\frac{\Delta \vec{v}}{\Delta t}$ = $\lim_{\Delta t\to 0}\frac{\vec{QQ^{"}}}{\Delta t} + \lim_{\Delta t\to 0}\frac{\vec{Q^{"}Q^{'}}}{\Delta t}$
$\therefore \vec{a}$ = $a_x\hat{i} + a_y\hat{j}$
where $a_x$ = $\lim_{\Delta t\to 0}\frac{\Delta {v_x}}{\Delta t}$ = $\frac{dv_x}{dt}$ = $\dot{v}x$ = $\ddot{x}$
$a_y$ = $\lim_{\Delta t\to 0}\frac{\Delta \vec{v_y}}{\Delta t}$ = $\frac{dv_y}{dt}$ = $\dot{v}y$ = $\ddot{y}$
a = $\sqrt{(a_x)^2 + (a_y)^2}, tan\beta$ = $\frac{a_y}{a_x}$ (see fig 3.2C)
• Derivatives of Vector Functions
A vector function is a function which has one or more variables whose range is a set of infinite-dimensional vectors or multidimensional vectors. Let us consider $\vec{P}(u)$ as a vector function of a scalar variable u. If a vector $\vec{P}$ is drawn from a fixed origin O and scalar u is allowed to vary then the tip of $\vec{P}$ will tell about a given curve in space. Considering vectors $\vec{P}$ corresponding, respectively, to the values u and $(u+\Delta u)$ of the scalar variables as shown in fig 1.3 a. Then, $\Delta \vec{P}$ = $\vec{P} (u+\Delta u)- \vec{P}(u)$
Dividing above equation by Δu and let Δu → 0,
$ie, \frac{\vec{P}}{du}$ = $\lim_{\Delta u\to 0}\frac{\Delta \vec{P}}{\Delta u}$ = $\lim_{\Delta u\to 0}{\frac{\vec{P}(u+\Delta u)-\vec{P}(u)}{\Delta u}}$-----1{i}
As Δu → 0, the line of action of  $\Delta \vec{P}$ becomes tangent to the curve of fig 1.3 a.
Thus $\frac{d\vec{P}}{du}$of the vector function P(u) is tangent to the curve as shown in fig 1.3 b.
Considering the sum of two vector functions $\vec{P}(u)\;and\;\vec{Q}(u)$ of the same scalar variable u. Then the derivative of the vector $(\vec{P}+\vec{Q})$ is given by,
$\frac{d(\vec{P}+\vec{Q})}{du}$ = $\lim_{\Delta u\to 0}\frac{\Delta (\vec{P}+\vec{Q})}{\Delta u}$ = $\lim_{\Delta u\to 0}(\frac{\Delta \vec{P}}{\Delta u} + \frac{\Delta \vec{Q}}{\Delta u})$ = $\lim_{\Delta u\to 0}\frac{\Delta \vec{P}}{\Delta u} + \lim_{\Delta u\to 0}\frac{\Delta \vec{Q}}{\Delta u}$
$\frac{d(\vec{P}+\vec{Q})}{du}$ = $\frac{d\vec{P}}{du}+\frac{d\vec{Q}}{du}$ ----------(ii)
Again, considering the product of a scalar function f(u) and of a vector function$\vec{P}(u)$ of the same scalar variable u. Then derivative of the vector f$\vec{P}$is given by,
$\frac{d(f\vec{P})}{du}$ = $\lim_{\Delta u\to 0}\frac{(f+\Delta f)(\vec{P}+\Delta \vec{P}) - f\vec{P}}{\Delta u}$ = $\lim_{\Delta u\to 0}(\frac{\Delta f}{\Delta u }\vec{P }+f \frac{\Delta \vec{P}}{\Delta u})$
$\therefore$ $\frac{d(f\vec{P})}{du}$ = $\frac{df}{du }\vec{P }+f \frac{d\vec{P}}{du}$ ----------- (iii)
Similarly, scalar product and vector product of two vector functions $\bar{P}(u)\;and\;\bar{Q}(u)$ may be obtained as:
$\frac{d(\vec{P}.\vec{Q})}{du}$ = $\frac{d\vec{P}}{du}.\vec{Q}+\vec{P}.\frac{d\vec{Q}}{du}$ ----------(iv)
$\frac{d(\vec{P}\times \vec{Q})}{du}$ = $\frac{d\vec{P}}{du}\times \vec{Q}+\vec{P}\times \frac{d\vec{Q}}{du}$ ---------- (v)
Again,
$\vec{P}$ = $P_x\hat{i} + P_y\hat{j}+P_z\hat{K}$ --------- (vi)
where Px, Py, Pz are the rectangular scalar components of the vector
$\vec{P}\;and\;\hat{i},{j},{k}$ be thr units vectors along x,y &z-axis respectively be the unit vectors along x, y & z-axes respectively.
$\therefore\frac{d\vec{P}}{du}$ = $\frac{dP_x}{du}\hat{i} + \frac{dP_y}{du}\hat{j} + \frac{dP_z}{du}\hat{k}$-------(vii)
when the vector  $\vec{P}$ is a function of the time t, its derivative $\frac{d\vec{P}}{dt}$ represents the rate of change of vector (i.e  $\vec{P}$ ) with respect to the frame oxyz, Then,
$\frac{d\vec{P}}{dt}$ = $\frac{dP_x}{dt}\hat{i} + \frac{dP_y}{dt}\hat{j} + \frac{dP_z}{dt}\hat{k}$ --------- (viii)
• Rectangular Components of Velocity and Acceleration
When the position of a particle P is defined at any instant by its rectangular coordinates x,y and z then position vector can be written as,
$\vec{r}$ = $x\hat{i} + y\hat{j}+ z\hat{K}$ ---------- (i)
Differentiating with respect to time, we get,
$\vec(v)$ = $\frac{d\vec{r}}{dt}$ = $\dot{x}\hat{i} + \dot{y}\hat{j}+ \dot{z}\hat{K}$ ----------- (ii)
Again differentiating with respect to time,
$\vec{a}$ = $\frac{d\vec{v}}{dt}$ = $\ddot{x}\hat{i} + \ddot{y}\hat{j}+ \ddot{z}\hat{K}$ ----------- (iii)
where,
$\dot{x}$ = $V_x$, $\dot y$ = $v_y$ and $\dot{z}$ = $v_z$,    $\ddot{x}$ = $a_x$, $\ddot{y}$ = $a_y$ and $\ddot{z}$ = $a_z$ are the components of velocity and acceleration along x, y and z-axes as shown in fig 1.4 a & b. If $a_x$ is independent of y and $v_y$ and $a_y$ is independent of x and $v_x$ then use of rectangular components to describe the position, velocity and acceleration of a particle are effective. It means that the motion in x and y-direction are considered separately.
For example, in projectile motion, if air resistance is neglected, the components of accelerations are $a_x$ = 0 and $a_y$ = -g .
i.e, $\dot{v_x}$ = 0 and $\dot{v_y}$ = $-g$ --------- (iv)
Integrating above equation with respect to time we get
$v_x$ = $C_1$ and $v_y$ = gt + $C_2$ [$C_1$ and $C_2$ are integration constants]
at time t =  0 sec, $(v_x)$ = $(v_x)_0$ and $v_y$ = $(v_y))_0$, then
$C_1$ = $(v_x)_0$ & $C_2$ = $(v_y)_0 -gt$-----(v)
Again integrating equation (v) with respect to time we get,
x = $(v_x)_0t + C_3$ and y = $(v_y)_0 t-\frac{gt^2}{2}+C_4$
where, $C_3$ and $C_4$ are integration constant.
At time t = 0 sec, x = 0 & y = 0, then $C_3$ = 0, $C_4$ = 0
Hence, $x$ = $(v_x)_0 t$ (uniform motion) & $y$ = $(v_y)_0 t-\frac{gt^2}{2}$ --------- (vi)
The motion of a projectile may thus be replaced by two independent rectilinear motions. The path followed by the projectile is parabolic as shown in fig. 1.5.
• Motion Relative to a Frame in Translation
Considering two particles A and B moving in the same plane as shown in fig 1.6 .
$\vec{r_A}\;and\;\vec{r_B}$ be their position vector with respect to X — Y axes centered at O.
Again considering a new system of axes (X' - Y') centered at A and parallel to the original axes, X — Y. Here orientation of the particle remains same and hence axes are said to be in translation . The vector $\vec{r}_\frac{B}{A}$ is the position of particle B with respect to X'Y' axes. It is the position vector of B relative to the reference system centred at $A_1$ or position vector of B relative to A.
From vector triangle,
$\vec{r}_B$ = $\vec{r}_B + \vec{r}_\frac{B}{A}$ ---------- (i)
In terms of x and y components, this relation becomes,
$\left.\begin{matrix} \vec{x}_B=\vec{x}_A + \vec{x}_\frac{B}{A} \\ \vec{y}_B = \vec{y}_A + \vec{y}_\frac{B}{A} \end{matrix}\right\}$ -------------- (ii)
where,
$X_B,Y_B$ are coordinates of B with respect to X-Y axes.
$X_A,Y_A$ are coordinates of A with respect to X-Y axes.
$X_\frac{B}{A},Y_\frac{B}{A}$ are coordinates of B with respect to $X^{'}-Y^{'}$ axes.
Differentiating equation (i) with respect to time we get,
$\vec{V}_B$ = $\vec{V}_B + \vec{V}_\frac{B}{A}$ ---------- (iii)
In scalar form,
$\left.\begin{matrix} \dot{x}_B=\dot{x}_A + \dot{x}_\frac{B}{A} \\ \dot{y}_B = \dot{y}_A + \dot{y}_\frac{B}{A} \end{matrix}\right\}$----------(iv)
where,
$\dot{X_A},\dot{Y_A}$ are  x & y components of $\vec{V_A}$ of particles A,
$\dot{X_B},\dot{Y_B}$ are x & y components of $\vec{V_B}$ of particles B,
$\dot{X_\frac{B}{A}},\dot{Y_\frac{B}{A}}$ are x & y components of $\vec{V_\frac{B}{A}}$
Again differentiating equation (iii) with respect to time we get
$\vec{a}_B = \vec{a}_A + \vec{a}_\frac{B}{A}$ ----------- (v)
In scalar form,
$\left.\begin{matrix} \ddot{x}_B=\ddot{x}_A + \ddot{x}_\frac{B}{A} \\ \ddot{y}_B = \ddot{y}_A + \ddot{y}_\frac{B}{A} \end{matrix}\right\}$ ---------- (iv)
• Tangential and Normal components
If there is a vector at a point on a curve then the vector can be decomposed as the sum of two vectors uniquely.The one which is tangent to a curve is called tangential vector and next which is normal to a curve is called normal components of a vector.
The velocity of a particle is a vector tangent to the path of the particle but that the acceleration vector is in general, not tangent to the path.The acceleration vector may be resolved into components directed respectively, along the tangent and normal to the path of the particle. $\hat{e_t}\;and\;\hat{e_n}$ be the unit vectors, directed along the tangent and along the normal or towards the center of curvature of the path respectively. Unit vector $\hat{e_t}\;and\;\hat{e_n}$  will generally change in their directions as the particle moves from one point to another along the curve path. From Fig 3.7a, $\Delta \vec{r}$ = $\vec{r^{'}} -\vec{r}$, $\lim_{\Delta s\to 0}|\Delta \vec{r}|$ = $\Delta s$
then, $\hat{e_t}$ = $\lim_{\Delta s\to 0}\frac{\Delta \vec{r}}{\Delta s}$ = $\frac{d\vec{r}}{ds}$
$\therefore \hat{e_t}$ = $\frac{d\vec{r}}{ds}$ ------------ (i)
Again, $|\hat{e_t}|$ = $\lim_{\Delta s\to 0}\frac{|\Delta \vec{r}|}{|\Delta s|}$ = $c{\Delta s}$ = 1
Thus, $\hat{e_t}$ is a unit vector and is tangent to the path of the particle at the point P. Let ρbe the radius of curvature of the path at the point P as shown in fig 1.8 a. $\hat{e_t}\;and\;\hat{e_t^{'}}$ be the tangent unit vectors at P and P'. $\Delta \hat{e_t}$ be the change in the unit vector while the particle moves from P to P' as shown if fig 1.8 b. $\Delta s$ = PP' = $\rho \Delta \theta$
$\Delta\hat{e_t}$ = $\hat{e_t}^{'}- \hat{e_t}$ $\cong$ $\Delta \theta \hat{e_n}$
$\frac{d\theta}{ds}$ = $\lim_{\Delta s\to 0}\frac{\Delta \theta}{\Delta s}$ = $\lim_{\Delta s\to 0}\frac{\Delta \theta}{\rho\Delta\theta}$ = $\frac{1}{\rho}$
$\frac{d\hat{e_t}}{d\theta}$ = $\lim_{\Delta s\to 0}\frac{\Delta \hat{e_t}}{\Delta \theta}$ = $\lim_{\Delta s\to 0}\frac{\Delta \theta\hat{e_n}}{\Delta \theta}$ = $\hat{e_n}$
$\frac{d\theta}{ds}$ = $\frac{1}{\rho}$ & $\frac{d\hat{e_t}}{d\theta}$ = $\hat{e_n}$ --------(ii)
Then,
v=$\frac{ds}{dt}$ = $\lim_{\Delta s\to 0}\frac{\Delta s}{\Delta t}$ = $\lim_{\Delta s\to 0}\frac{\Delta \theta}{\Delta t}$ = $\rho\frac{d\theta}{dt}$
$\therefore$ V = $\rho \dot{\theta}$------------(iii)
$\vec{v}$ = $\frac{d\vec{r}}{dt}$ = $\frac{d\vec{r}}{ds} \frac{ds}{dt}$ = $\hat{e_t}v$ = $v\hat{e_t}$
$\therefore$$\vec{v}$ = $v\hat{e_t}$----------(iv)
$\vec{a}$ = $\frac{d\vec{v}}{dt}$ = $\frac{d}{dt}(v\hat{e_t})$ = $\frac{dv}{dt}(\hat{e_t}) + v\frac{d\hat{e_t}}{dt}$ = $\frac{dv}{dt}\hat{e_t}+v\frac{d\hat{e_t}}{d\theta}\frac{d\theta}{ds}\frac{ds}{dt}$ = $\dot{v}\hat{e_t}+v(\hat{e_n})(\frac{1}{\rho})(v)$
$\vec{a}$ = $\dot{v}\hat{e_t} + \frac{v^2}{\rho}\hat{e_n}$------------(v)
Again, $\vec{a}$ = $a_t\hat{e_t}+a_n\hat{e_n}$
Hence, Tangential component of acceleration $(a_t)$ = $v$ = $\frac{dv}{dt}$
Normal components of acceleration
$(a_n)$ = $\frac{v^2}{\rho}$ = $\frac{(\rho\dot{\theta})^2}{\rho}$ = $\rho\dot{\theta}^2$
Further, $\hat{e_b}$ = $\hat{e_t}\times \hat{e_n}$ (where, $\hat{e_b}$ is te binormal unit vector)
$\rho$ is the radius of curvature of the path,
i.e, $\rho$ = $\frac{[1+(\frac{dy}{dx})^2]^\frac{3}{2}}{\frac{d^2y}{dx^2}}$
where, y = f(x) is the equation of the path.
Conclusions:
• If the magnitude of the velocity (i.e speed) is increasing, $a_t$ will be in the same sense as that of velocity.
• If the magnitude of velocity is decreasing, $a_t$ will be in the opposite sense to that of velocity.
• If the speed is constant, at will be equal to zero, hence there is only normal component of acceleration.
• $\vec{a_n}$ always directed towards the centre of curvature, irrespective of the direction of $\vec{a_t}\;and\;\vec{v}$
• $|\vec{a_n}|$ (i.e. magnitude of a normal component of acceleration) depends on speed & radius of curvature of the path.
• Higher the value of the velocity of the particle & lesser the value of ρ⇒more difficult to make it change the direction of the motion (i.e $a_n$ must be higher).
• If $\pi$ = infinite (as when the path is a straight line) ⇒$a_n$ = 0 (i.e. there is only tangential component of acceleration).
In some problems of plane motion of particles, it is desirable to use polar coordinates to describe the motion. The position of the particle at P is defined by the coordinates r and $\theta$ as shown in fig 1.9 a. where r is the length and $\theta$ is the angle in radians. The unit vectors in radial and transverse directions are denoted by $\hat{e_r}\;and\;\hat{e_\theta}$ respectively. In the fig, $\hat{e_r}$ is directed outward along $\vec{OP}$ and $\hat{e_\theta}$ is obtained by rotating through 90° in anticlockwise direction. As the particle moves from P to P', the unit vectors $\hat{e_r}\;and\;\hat{e_\theta}$  change to $\hat{e_r}^{'}$ and $\hat{e_\theta}^{'}$ by amounts $\Delta \hat{e_r}$ and $\Delta \hat{e_\theta}$ respectively as shown if fig 1.9 b & c.
Here,
$\dot{\hat{e_r}}$ = $\frac{d\hat{e_r}}{dt}$ = $\frac{d\hat{e_r}}{d\theta}\frac{d\theta}{dt}$ = $\hat{e_\theta}\dot{\theta}$-------(i)
[Since $d\hat{e_r}$ is directed in the direction of $\hat{e_\theta}$]
$\hat{e_\theta}$ = $\frac{d\hat{e_\theta}}{dt}$ = $\frac{d\hat{e_\theta}}{d\theta}\frac{d\theta}{dt}$ = $-\hat{e_r}\dot{\theta}$-------(ii)
[Since $d\hat{e_\theta}$ is directed in the direction of $-\hat{e_r}$]
We know, $\vec{r}$ = $r\hat{e_r}$
Then $\vec{v}$ = $\frac{d\vec{r}}{dt}$ = $\frac{d(r\hat{e_r})}{dt}$
= $\dot{r}\hat{e_r} + r\dot{\hat{e_r}}$
$\vec{v}$ = $\dot{r}\hat{e_r} + r\dot{\theta}\hat{e_\theta}$ ----------(iii) [$\dot{e_r}$ = $\hat{e_\theta}\dot{\theta}$]
again, $\vec{v}$ = $v_r\hat{e_r}+v_\theta\hat{e_\theta}$
Hence, Radial component.of velocity(vr) = $\dot{r}$ & Transverse component of velocity (vθ) = $r\dot{\theta}$
Similarly, $\vec{a}$ = $\frac{d\vec{v}}{dt}$ = $\frac{d}{dt}(\dot{r}\hat{e_r}+r\dot{\theta}\hat{e_\theta})$
= $\ddot{r}{\hat{e_r}} + \dot{r}\dot{\hat{e_r}} + \dot{r}\dot{\theta}\hat{e_\theta} + r\ddot{\theta}\hat{e_\theta} + r\dot{\theta}\dot{\hat{e_\theta}}$
= $\ddot{r}{\hat{e_r}} + \dot{r}\dot{\theta}\hat{e_\theta} + \dot{r}\dot{\theta}\hat{e_\theta} + r\ddot{\theta}\hat{e_\theta} - r\dot{\theta}^2\hat{e_r}$ [ $\dot{\hat{e_r}}$ = $\hat{e_\theta}\dot{\theta}$ & $\dot{\hat{e_\theta}}$ = $-\hat{e_r}\dot{\theta}$]
$\therefore \vec{a}$ = $(\ddot{r} - r\theta^2)\hat{e}r + (r\ddot{\theta} + 2\dot{r}\dot{\theta})\hat{e_\theta}$--------(iv)
Again, $\vec{a}$ = $a_r\hat{e_r}+a_\theta\hat{e_\theta}$
Hence, Radial component of acceleration $(a_r$ =$(\ddot{r}-r\dot{\theta}^2)$ & Transverse component of acceleration ($a_\theta$) = $(r\ddot{\theta} + 2\dot{r}\dot{\theta})$
In the case of a particle moving along a circular path with its center at the origin O,
we have, r = constant, $\Longrightarrow \dot{r}$ = 0 & $\ddot{r}$ = 0
Then,
$\vec{v}$ = $r\dot{\theta}\hat{e_\theta}$-------(v) &
$\vec{a}$ = $-r\dot{\theta}^2\hat{e_\theta} + r\ddot{\theta}\hat{e_\theta}$--------- (vi)

Reference
Beer F.P. and E.R. Johnson "Vector Mechanics for Engineers", 2nd edition, Tata McGraw hill Publishing Co. Ltd., 1998
Egor .P. Popov "Engineering Mechanics of Solids", 2nd edition, New Dehli, Prentice Hall of India, 1996
Shames, I.H "Engineering Mechanics- Statics and Dynamics", 3rd edition, New Dehli, Prentice Hall of India, 1990
Hibbler, R.C. "Engineering Mechanics" (Statics and dynamics)
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