Particle Kinetics

Introduction

Particle kinetics relates the forces acting on a particle to its motion. Unlike kinematics, which describes motion without considering forces, kinetics uses Newton's Second Law to determine how forces cause acceleration.

Newton’s Second Law, Weight, and Friction

Newton’s Second Law relates the net force acting on a particle to its acceleration:

Weight ($W = mg$) acts vertically down. $W$ is in units of $N$ or $lb$, and $m$ is in units of $kg$ or $slug$

Normal forces ($N$) act perpendicular to contact surfaces.

Static friction: $f_{s} = \mu_{s}N$

Kinetic friction: $f_{k} = \mu_{k}N$

Normal-Tangential (Curvilinear) Components

For motion along a curved path:

Tangential (along direction of motion):

$\Sigma F_{t} = ma_{t}$

$a_{t} = \frac{dv}{dt}$

Normal (toward center of curvature):

$\Sigma F_{n} = ma_{n}$

$a_{n} = \frac{v^{2}}{\rho}$

Radial-Transverse (Cylindrical) Components

For particles moving in polar or cylindrical coordinates:

Radial Direction:

$\Sigma F_{r} = ma_{r}$

$a_{r} = \ddot{r} - r\dot{\theta}\ ^{2}$

Transverse Direction:

$\Sigma F_{\theta} = ma_{\theta}$

$a_{\theta} = r\ddot{\theta} + 2\dot{r}\dot{\theta}$

For 3-D Motion:

$\Sigma F_{z} = ma_{z}$

$a_{z} = \ddot{z}$

Common Procedure

Draw a free-body diagram, choose the appropriate coordinate system ($x,\ y\ ;\ t,\ n\ ;\ r,\ \theta,\ z)$, determine corresponding acceleration components ($a_{x},\ a_{y},\ a_{t},\ a_{n},\ a_{r},\ a_{\theta},\ a_{z})$, and apply $\Sigma F = ma$ in each coordinate direction to solve for unknowns.

Example 1:
Find the velocity of 10-kg block after 5 seconds (block starts from rest).

1) $\Sigma F_{y} = 0$ since the y axis is perpendicular to the slope.

$F_{N} - Wcos(30) = 0\ \Rightarrow \ F_{N} = 98.1cos(30) = 84.96\ N$

2) $f_{k} = \mu_{k}F_{N} = (0.2)(84.96\ N) = 17\ N$

3) $\Sigma F_{x} = ma_{x}\ \Rightarrow \$$- 17\ N + Wsin(30) = (10\ kg)(a_{x})$

$- 17\ N + 98.1sin(30) = 10a_{x} \Rightarrow a_{x} = 3.205\ m/s^{2}$

4) After 5 seconds, ${v = at = (3.205\ m/s^{2})(5\ s) = 16.025\ m/s}_{}$

Example 2:
Find the max constant speed that a 70-kg pilot can travel so that he experiences a maximum acceleration of 78.5 m/s² and the normal force he exerts on the seat when traveling at this speed and is at the lowest point.

Since speed is constant, $a_{t} = 0$. Therefore, $a_{n} = 78.5\ m/s^{2}$.

$a_n = \frac{v^2}{\rho} \Rightarrow v = \sqrt{\rho a_n} = \sqrt{(800\,m)(78.5\,m/s^2)} = 250.6\,m/s$

$\Sigma F_{N} = ma_{n}$

$F_{N} - W = ma_{n}$

$F_{N} = ma_{n} + W = (70\ kg)(78.5\ m/s^{2}) + (70\ kg)(9.81\ m/s^{2}) = 6182\ N$

Example 3:
Find the magnitude of resultant force acting on a 5-kg particle at t = 2s if the particle is moving along a horizontal path defined by the equations $r = (2t + 10)\ m$ and $\theta = (1.5t^{2} - 6t)\ rad$.

1) Position and derivatives at $t = 2\ s$: $r = 14\ m$ ; $\dot{r} = 2\ m/s$ ; $\ddot{r} = 0$

2) Angle and derivatives at $t = 2\ s$: $\theta = - 6\ rad$ ; $\dot{\theta} = 0$ ; $\ddot{\theta} = 3\ rad/s^{2}$

3) Radial acceleration: $a_{r} = \ddot{r} - r\dot{\theta}\ ^{2} = 0 - (14\ m)(0)^{2} = 0$

4) Transverse acceleration: $a_{\theta} = r\ddot{\theta} + 2\dot{r}\dot{\theta} = (14\ m)(3\ rad/s^{2}) + 2(2\ m/s)(0) = 42\ m/s^{2}$

5) Resultant acceleration: $a = \sqrt{a_r^2 + a_\theta^2} = \sqrt{0^2 + 42^2} = 42\,m/s^2$

6) Resultant force: $F = ma = (5\ kg)(42\ m/s^{2}) = 210\ N$