Kinematics

By Prof. Seungchul Lee
http://iai.postech.ac.kr/
Industrial AI Lab at POSTECH

Table of Contents

1. Two Trains and Fly¶

Two trains are on the same track a distance 100 mi apart heading towards one another, each at a speed of 10 mi/h.

A fly starting out at the front of one train, flies towards the other at a speed of 20 mi/h.

Upon reaching the other train, the fly turns around and continues towards the first train.

How many kilometers does the fly travel before getting squashed in the collision of the two trains?

http://mathworld.wolfram.com/TwoTrainsPuzzle.html

2. Kinematics¶

way to describe the motion
- translation
- rotation (will be covered later)

3. Translation motion¶

Position

The motion of an object is described by the position, $x(t)$, as a function of time $t$, measured in a given reference system.

Velocity

Displacement

The displacement $\Delta x(t_1)$ over the time interval from $t=t_1$ to $t=t_1+\Delta t$ is defined as:

$$ \Delta x(t_1) = x(t_1+\Delta t) - x(t_1)$$

Average Velocity

The average velocity from $\Delta x(t_1)$ over the time interval from $t=t_1$ to $t=t_1+\Delta t$ is defined as:

$$ \bar v(t_1) = \frac{x(t_1+\Delta t) - x(t_1)}{\Delta t} = \frac{\Delta x(t_1)}{\Delta t}$$

Instantaneous Velocity

The instantaneous velocity is defined as the time derivative of the position:

$$ v(t) = \lim_{\Delta t \to 0}\frac{x(t+\Delta t) - x(t)}{\Delta t} = \frac{d x(t)}{d t} = \dot x (t)$$

Acceleration

average acceleration

The average acceleration over a time interval $\Delta t$ from $t$ to $t + \Delta t$ is:

$$ \bar a(t) = \frac{v(t+\Delta t) - v(t)}{\Delta t} = \frac{\Delta v(t)}{\Delta t}$$

instantaneous acceleration

The instantaneous acceleration is defined as:

$$ a(t) = \lim_{\Delta t \to 0}\frac{v(t+\Delta t) - v(t)}{\Delta t} = \frac{d v(t)}{d t} = \frac{d}{dt}\frac{dx}{dt} = \frac{d^2x}{dt^2} = \dot v (t) = \ddot x(t)$$

3.1. Constant acceleration¶

$$ \begin{align*} & a = \text{const.} \\ & v = v_0 + at \\ & x = x_0 + v_0 t + \frac{1}{2} a t^2 \end{align*}$$

zero acceleration

$$ \begin{align*} & a = 0 \\ & v = v_0 \\ & x = x_0 + v_0 t \end{align*}$$

free fall

$$ \begin{align*} & a = -g \quad \left(+ \uparrow \right) \\ & v = v_0 -gt \\ & x = x_0 + v_0 t - \frac{1}{2}gt^2 \end{align*}$$

for the motion of a falling tennis ball when there is no air resistance

% 1D

g = -9.8;
y0 = 1.6;
v0 = 3;
t = 0:0.05:1;
y = y0 + v0*t + 1/2*g*t.^2;

plot(t,y)
xlabel('time in sec')
ylabel('y')

3.2. Kinematics in 2-D¶

$$ \begin{align*} \vec{a} & = \text{const.} = a_x \hat{x} + a_y \hat{y} \\ \\ \vec{v} & = \left( v_{0x} + a_x t \right) \hat{x} + \left( v_{0y} + a_y t \right) \hat{y} \\ \\ \vec{r} & = \left( x_{0x} + v_{0x} t + \frac{1}{2} a_x t^2 \right) \hat{x} + \left( y_{0x} + v_{0y} t + \frac{1}{2} a_y t^2 \right) \hat{y} \\ \\ & \implies \; \text{Two independent 1-D motions} \end{align*}$$

% 2D

g = -9.8;

x0 = 0;
y0 = 1.6;

v0x = 3;
v0y = 3;

t = 0:0.05:1;
x = x0 + v0x*t;
y = y0 + v0y*t + 1/2*g*t.^2;

plot(x,y), axis equal
xlabel('x')
ylabel('y')

4. Data fitting from experimental measurements¶

4.1. Zero acceleration¶

experiment video
regression (data fitting or approximation)

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</iframe>

t = [0 0.033 0.067 0.1 0.133 0.167 0.2 0.233 0.267 0.3 0.334 0.367 0.4 0.434 0.467]';
x = [0.001 -0.028 -0.059 -0.087 -0.118 -0.149 -0.175 -0.21 -0.237 -0.262 -0.285 -0.312 -0.344 -0.378 -0.405]';

plot(t,x,'.')
xlabel('t')
ylabel('measured x')

T = [ones(size(t)) t];
b = regress(x,T)

b =

   -0.0019
   -0.8621

T = [ones(size(t)) t];
b = regress(x,T);

tp = 0:0.01:0.5;
plot(t,x,'o'), hold on
plot(tp,b(1)+b(2)*tp), hold off

xlabel('t')
ylabel('height')

4.2. Constant acceleration¶

experiment video
regression (data fitting or approximation)

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Experiment result

t = 0:0.1:0.5;
t = t(:);
y = [1.6 1.55 1.4 1.16 0.82 0.39]';

plot(t,y,'o')

xlabel('t')
ylabel('height')

t = 0:0.1:0.5;
t = t(:);
y = [1.6 1.55 1.4 1.16 0.82 0.39]';

T = [ones(size(t)) t.^2];
b = regress(y,T)

plot(t,y,'o'), hold on

tp = 0:0.01:0.5;
plot(tp,b(1)+b(2)*tp.^2), hold off

xlabel('t')
ylabel('height')

b =

    1.5968
   -4.8382

We can learn about the magnitude of the gravity

$$ 4.8383 \approx \frac{9.8}{2}$$

5. Data analysis from acceleration¶

Usually, we do not know the position as a function of time. Instead, we want to determine the motion based on measurements of the acceleration (or velocity)

Integral

$$ \begin{align*} v(t) &= v(t_i) + \int_{t_i}^{t} a(s)\; ds\\ x(t) &= x(t_i) + \int_{t_i}^{t} v(s)\; ds \end{align*} $$

discrete integration (Euler's method)

$$ \begin{align*} \upsilon(t_0) &= \upsilon_0\\ x(t_0) &= x_0\\ \\ \upsilon(t_1) &= \upsilon(t_0) + a(t_0)\Delta t\\ x(t_1) &= x(t_0) + \upsilon(t_0)\Delta t\\ \\ & \vdots\\ \\ \upsilon(t_i + \Delta t) &= \upsilon(t_i) + a(t_i)\Delta t\\ x(t_i + \Delta t) &= x(t_i) + \upsilon(t_i)\Delta t \end{align*} $$

path = [pwd, '\files'];
temp = load([path,'\therocket.dat']);

t = temp(:,1);
a = temp(:,2);

plot(t,a)
xlabel('time in sec')
ylabel('a [m/s^2]')

%plot -s 600,800

t = temp(:,1);
a = temp(:,2);

dt = t(2) - t(1);

n = length(t);
v = zeros(n,1);
x = zeros(n,1);

v(1) = 0;
x(1) = 0;

for i = 1:n-1
    v(i+1) = v(i) + a(i)*dt;
    x(i+1) = x(i) + v(i)*dt;
end

subplot(3,1,1), plot(t,x), ylabel('x [m]')
subplot(3,1,2), plot(t,v), ylabel('v [m/s]')
subplot(3,1,3), plot(t,a), ylabel('a [m/s^2]'), xlabel('t [s]')

6. Data analysis from position¶

Motion of a falling tennis ball

motion (position, or displacement) $\rightarrow$ velocity and acceleration
derivative

$$ \begin{align*} v(t_i) &\approx \frac{r(t_{i}+\Delta t)-r(t_i)}{\Delta t}\\ a(t_i) &\approx \frac{v(t_{i}+\Delta t)-v(t_i)}{\Delta t} \end{align*} $$

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<iframe 
width="560" height="315" src="https://www.youtube.com/embed/HBsRfaix2SA" frameborder="0" allowfullscreen>
</iframe>

%plot -s 500,400

path = [pwd, '\files'];
temp = load([path,'\fallingtennisball02.dat']);

t = temp(:,1);
y = temp(:,2);
plot(t,y)
xlabel('t [s]');
ylabel('y [m]');

$$ \begin{align*} v(t_i) &\approx \frac{r(t_{i}+\Delta t)-r(t_i)}{\Delta t}\\ a(t_i) &\approx \frac{v(t_{i}+\Delta t)-v(t_i)}{\Delta t} \end{align*} $$

%plot -s 600,800
n = length(t);
dt = t(2) - t(1);

v = zeros(n-1,1);
for i = 1:n-1
    v(i) = (y(i+1) - y(i))/dt;
end

a = zeros(n-2,1);
for i = 1:n-2
    a(i) = (v(i+1) - v(i))/dt;
end

subplot(3,1,1), plot(t,y), ylabel('x [m]')
subplot(3,1,2), plot(t(2:n),v), ylabel('v [m/s]')
subplot(3,1,3), plot(t(3:n),a), ylabel('a [m/s^2]'), xlabel('t [s]')

%plot -s 500,400

imax = max(find(t<=0.5));

plot(t(3:imax),a(3:imax));
xlabel('t [s]')
ylabel('a [m/sˆ2]')

The acceleration is clearly not a constant in this case. It starts at $−9.8m/s^2$, but its magnitude becomes smaller with time. (This is due to air resistance).

7. Vector¶

Vector in Coordinate¶

$$ \vec{A} = A_x \hat{x} + A_y \hat{y} + A_z \hat{z} = \begin{bmatrix} A_x \\ A_y \\ A_z \end{bmatrix} \quad \text{coordinate with} \; \{\hat{x}, \hat{y}, \hat{z} \}$$

for example,

$$ \begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix} = 3 \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + 2 \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + 1 \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$$

each component can be treated as a scalar

Decomposition in 2-D¶

Decompose complicated motion (problem) to simple motions (subproblems)

Projectile¶

$$ \begin{array}{c} \text{Given} \; \upsilon \; \text{and} \; \theta \\ H = ? \\ S = ? \end{array} \quad \quad \quad \begin{array}{c} \text{Given} \; \upsilon \\ \max\limits_{\theta}{S} \end{array}$$

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Different Frames¶

$$ \begin{align*} & \text{Mass} \; M \; \text{moves with velocity of} \; v_x \\ & \text{Mass} \; M \; \text{shots ball} \; m \; \text{up with velocity of} \; v_y \end{align*}$$

For frame 1

For frame 2

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8. Uniform Circular Motion¶

motion of an object traveling at a constant speed on a circular path

8.1. Velocity¶

$$\text{change of vector}: \underbrace{\text{change of magnitude}}_{\text{translation}} + \underbrace{\text{change of direction}}_{\text{circular motion}}$$

$$\begin{align*}\\ \vec{v}_{avg} & = \frac{\vec r \left(t + \Delta t\right) - \vec r \left( t \right)}{\Delta t}\\ \vec v & = \lim\limits_{\Delta \rightarrow 0}\frac{\vec r \left( t + \Delta t \right) - \vec r \left( t \right)}{\Delta t} \end{align*}$$

$$\vec{\upsilon} \text{ is tangential to the circle}$$

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frameborder="0" allowfullscreen>
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8.2. Acceleration¶

$$\begin{align*}\\ \vec{a}_{avg} & = \frac{\vec v \left(t + \Delta t\right) - \vec v \left( t \right)}{\Delta t}\\ \vec a & = \lim\limits_{\Delta \rightarrow 0}\frac{\vec v \left( t + \Delta t \right) - \vec v \left( t \right)}{\Delta t} \end{align*}$$

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8.3. Angular Velocity¶

$$\begin{array}\\ & \theta &\triangleq \frac{l}{r} \text{ ( radian ) }\\ & l &= r \theta \\ & \dot l &= \dot r \theta + r \dot \theta\\ \Rightarrow & v &= r \dot \theta = r\omega \end{array}$$

Angular Velocity Vector¶

$$\begin{array}\\ \text{line velocity } &:\vec v, \quad \text{ we know }v = r\omega\text{ (only magnitude)}\\ \text{angular velocity} &: \vec {\omega}\\ \\ \lvert \vec{\omega} \rvert = \omega\\ \vec{\omega}\text{ direction?}\\ \end{array}$$

Cross Product¶

$$\vec A \times \vec B$$

$$\lvert \vec A \times \vec B \rvert = \lvert \vec A \rvert \lvert \vec B \rvert \sin \theta$$

Why do we need to know the cross product?

$$ \lvert \vec{\omega} \rvert = \omega$$

$$\vec{\omega} \text{ dirction ?}$$

Angular Velocity and Angular Acceleration Vector¶

How to represent $\vec{v}$ and $\vec{a}$ in a cross product form

$$ \large \begin{array}\\ \vec{v} &= \vec{\omega} \times \vec r\\ \vec{a} &= \vec{\omega} \times \vec v\\ &= \vec{\omega} \times \left( \vec{\omega} \times \vec r\right)\\ \\ \lvert\vec{a}\rvert & = r\omega^2 = \frac{v^2}{r} \end{array} $$

w = [0 0 1]';
r = [1 0 0]';
v = cross(w,r)
a = cross(w,v)

v =

     0
     1
     0


a =

    -1
     0
     0

theta = 45*pi/180;

w = [0 0 1]';
r = [cos(theta) sin(theta) 0]';
v = cross(w,r)
a = cross(w,v)

v =

   -0.7071
    0.7071
         0


a =

   -0.7071
   -0.7071
         0

Circular Motion or Spiral Motion?¶

$$\lvert\vec{a}\rvert > \frac{v^2}{r}\; ?$$

8.4. Free Fall or Circular Motion?¶

what about satellites ?

9. General motion = translation motion + circular motion¶

particle moving along the curved path

projectile

10. Kinematics of Particles (or Rigid Body)¶

So far, we have been considering the kinematics of a single particle

what if a rigid body whose mass is distributed

The Center of Mass
- the point where all of the mass of the object is concentrated.

treat it as a particle in the center of mass
- the center of mass moves as if it were a point whose mass is equal to the total mass of the system

often enable us to simplify physic problems

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Questions:

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