Lecture 3 - 2 and 3 dimensional kinematics
In this lecture we look at vectors in 2 and 3 dimensional motion and study a specific example of 2 dimensional motion - projectile motion.
If you need a pdf version of these notes you can get it here
Video of lecture
Vectors and scalars
Vector quantities with number, direction and units:
- Displacement $\vec{r}$ [m]
- Velocity $\vec{v}$ [ms-1]
- Acceleration $\vec{a}$ [ms-2]
Scalar quantities number and units only
- Distance traveled [m]
- Speed [ms-1]
Graphical representation of vectors and components
It is frequently useful to draw two dimensional vectors as arrows, and to split them in to components that lie along the coordinate axes. The choice of coordinate axes is up to you..but choosing the right ones will make the problem easier or harder.
We can take a look at the acceleration due to gravity as vector using a phone accelerometer, using this tool. (This will work on a idevice with an accelerometer running iOs 4.2 or higher. It also works on Android devices using the default browser and Firefox, but not Chrome). As it turns out even my Mac has an accelerometer in it somewhere!
Adding and subtracting vectors
Vectors - Components
| $v_{1x}=v_{1}\cos\theta_{1}$ | $v_{2x}=v_{2}\cos\theta_{2}$ |
| $v_{1y}=v_{1}\sin\theta_{1}$ | $v_{2y}=v_{2}\sin\theta_{2}$ |
| $v_{Rx}=v_{1x}+v_{2x}$ | $\tan\theta_{R}=\frac{v_{Ry}}{v_{Rx}}$ |
| $v_{Ry}=v_{1y}+v_{2y}$ | $v_{R}=\sqrt{v_{Rx}^2+v_{Ry}^2}$ |
Unit Vectors
It can be useful to express vector quantities in terms of unit vectors. These are dimensionless vectors of length = 1 that point along the coordinate axes. They are usually denoted with carets (hats), i.e. $(\hat{i},\hat{j},\hat{k})$
For example:
$\vec{v}\,\mathrm{ms^{-1}}=v_{x}\,\mathrm{ms^{-1}}\,\hat{i}+v_{y}\,\mathrm{ms^{-1}}\,\hat{j}+v_{z}\,\mathrm{ms^{-1}}\,\hat{k}$
or
$\vec{r}\,\mathrm{m}=x\,\mathrm{m}\,\hat{i}+y\,\mathrm{m}\,\hat{j}+z\,\mathrm{m}\,\hat{k}$
Vectors and motion
$\vec{r_{1}}=x_{1}\,\hat{i}+y_{1}\,\hat{j}+z_{1}\,\hat{k}$
$\vec{r_{2}}=x_{2}\,\hat{i}+y_{2}\,\hat{j}+z_{2}\,\hat{k}$
$\Delta\vec{r}=\vec{r_{2}}-\vec{r_{1}}=(x_{2}-x_{1})\,\hat{i}+(y_{2}-y_{1})\,\hat{j}+(z_{2}-z_{1})\,\hat{k}$
Average velocity: $\vec{v_{ave}}=\frac{\Delta\vec{r}}{\Delta t}$
Instantaneous velocity: $\vec{v}=\frac{d\vec{r}}{dt}=\frac{dx}{dt}\,\hat{i}+\frac{dy}{dt}\,\hat{j}+\frac{dz}{dt}\,\hat{k}=v_{x}\,\hat{i}+v_{y}\,\hat{j}+v_{z}\,\hat{k}$
Average acceleration: $\vec{a_{ave}}=\frac{\Delta\vec{v}}{\Delta t}$
Instantaneous acceleration: $\vec{a}=\frac{d\vec{v}}{dt}=\frac{dv_x}{dt}\,\hat{i}+\frac{dv_y}{dt}\,\hat{j}+\frac{dv_z}{dt}\,\hat{k}=\frac{d^{2}x}{dt^2}\,\hat{i}+\frac{d^{2}y}{dt^2}\,\hat{j}+\frac{d^{2}z}{dt^2}\,\hat{k}$
Projectile motion
If $g=9.8ms^{-2}$
| $\Large a_{x}=0$ | $\Large a_{y}=-g$ |
| $\Large v_{x}=v_{x0} $ | $\Large v_{y}=v_{y0}-gt$ |
| $\Large x=x_0+v_{x0} t$ | $\Large y=y_{0}+v_{y0} t -\frac{1}{2}gt^{2}$ |
Projectile motion for a "cannon"
| $\Large a_{x}=0$ | $\Large a_{y}=-g$ |
| $\Large v_{x}=v_{0}\cos\theta$ | $\Large v_{y}=v_{0}\sin\theta-gt$ |
| $\Large x=x_0+v_{0}\cos\theta t$ | $\Large y=y_{0}+v_{0}\sin\theta t -\frac{1}{2}gt^{2}$ |
Motion path
Taking $x_{0}$,$y_{0}$ as $(0,0)$
$t=\frac{x}{v_{0}\cos\theta}$
$y=\frac{\sin\theta}{\cos\theta}x-\frac{g}{(2v_{0}^{2}\cos^2\theta)} x^{2}$
Horizontal Range → Solve for $y=0$
$0=x(\frac{\sin\theta}{\cos\theta}-\frac{g}{(2v_{0}^{2}\cos^2\theta)}x)$
$x=0$ and $x=\frac{2v_{0}^2\sin\theta\cos\theta}{g}$ → $R=\frac{2v_{0}^2\sin\theta\cos\theta}{g}$
Projectile motion video
Note that in terms of the equations shown previously $\theta=(180^o-angle)$ as the motion is from right to left.
Measurement of initial velocity
0.04/.0095=4.2m/s
Maple Worksheet to plot trajectories can be downloaded here. (Right click and “Save As”. You need to have Maple installed on your computer to use this.)
Angle for which maximum range is acheived
$R=\frac{2v_{0}^2\sin\theta\cos\theta}{g}=\frac{v_{0}^{2}}{g}\sin 2\theta$
$\frac{dR}{d\theta}=\frac{2v_{0}^2}{g}(\frac{d(\sin\theta)}{d\theta}\cos\theta)+\sin\theta\frac{d(\cos\theta)}{d\theta})$
$=\frac{2v_{0}^2}{g}(\cos^2\theta-\sin^2\theta)=\frac{2v_{0}^2}{g}\cos2\theta$
or
$\frac{dR}{d\theta}=\frac{v_{0}^2}{g}\frac{d(\sin{2\theta})}{d\theta}\frac{d(2\theta)}{d\theta}=\frac{2v_{0}^2}{g}\cos2\theta$
$\frac{dR}{d\theta}=0$ at $\theta=45^o,135^o$
$45^o$ and $135^o$ correspond to maximum range.
Predict range for straight shot
$\Large y=\frac{\sin\theta}{\cos\theta}x-\frac{g}{(2v_{0}^{2}\cos^2\theta)} x^{2}$
Relative velocity
As in 1 dimensional motion, to find the velocity of an object relative to a moving reference frame, subtract the reference frame velocity from that of the object. Of course you have to do this correctly observing vector rules! Usually you are going to have to break things down in to appropriate components. Several of the homework problems are relative velocity problems and we'll talk more about this in the next lecture.
