$\large v= v_{0}+at$

$\large x= x_{0}+v_{0}t+\frac{1}{2}at^2$

$\large v^{2}=v_{0}^2+2a(x-x_{0})$

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}]

$v_{1x}=v_{1}\cos\theta_{1}$ | $v_{2x}=v_{2}\cos\theta_{2}$ |

$v_{1y}=v_{1}\sin\theta_{1}$ | $v_{2y}=v_{2}\sin\theta_{2}$ |

$v_{Rx}=v_{1x}+v_{2x}$ | $\tan\theta_{R}=\frac{v_{Ry}}{v_{Rx}}$ |

$v_{Ry}=v_{1y}+v_{2y}$ | $v_{R}=\sqrt{v_{Rx}^2+v_{Ry}^2}$ |

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

$\vec{r}\,\mathrm{m}=x\,\mathrm{m}\,\hat{i}+y\,\mathrm{m}\,\hat{j}+z\,\mathrm{m}\,\hat{k}$

$\Large a_{x}=0$ | $\Large a_{y}=-g$ |

$\Large v_{x}=v_{x0} $ | $\Large v_{y}=v_{y0}-gt$ |

$\Large x=x_0+v_{x0} t$ | $\Large y=y_{0}+v_{y0} t -\frac{1}{2}gt^{2}$ |

$\Large a_{x}=0$ | $\Large a_{y}=-g$ |

$\Large v_{x}=v_{0}\cos\theta$ | $\Large v_{y}=v_{0}\sin\theta-gt$ |

$\Large x=x_0+v_{0}\cos\theta t$ | $\Large y=y_{0}+v_{0}\sin\theta t -\frac{1}{2}gt^{2}$ |

Dealing with relative velocity is a particularly important application of addition and subtraction of vectors.

We can adopt a notation which can be helpful.

$\vec{v}_{AB}$: velocity of A relative to B

If we want to know the velocity of A relative to C if we know the velocity of A relative to B and the velocity of B relative to C

$\vec{v}_{AC}=\vec{v}_{AB}+\vec{v}_{BC}$

Whereas if we want to find the velocity of B relative to C

$\vec{v}_{BC}=\vec{v}_{AC}-\vec{v}_{AB}$

Is motion of an object fundamentally different to it being at rest? Our experience with reference frames should tell us otherwise. An object with a *constant* velocity in one reference frame would be at rest in a reference frame which had the same *constant* velocity as the object. It follows then that motion with a *constant* velocity is not intrinsically different to an object being at rest. This idea, which begins with Galileo, is the basis for Newton's first law of motion.

*Every object continues in its state of rest, or of uniform velocity in a straight line, as long as no net force acts on it.*

This law essentially describes the property of an object that we call **inertia**.

How much inertia does an object have? This is determined by it's **mass**, for which the SI unit is kg. The more mass an object has the harder it is to get it to change it's motion.

For an object to change it's velocity and thus have an acceleration, it needs to be subjected to a net force. The degree of acceleration produced by the sum of the forces is determined by the size of the force and the mass of the object according to the formula.

$\large\Sigma\vec{F}=m\vec{a}$

Dimensional analysis shows us that force must have units $\mathrm{kgms^{-2}}$. These units are normally expressed as Newtons and denoted by the symbol $\mathrm{N}$.

*Whenever one object exerts a force on a second object, the second exerts an equal force in the opposite direction on the first.*

Another way of saying this: *Forces come in action-reaction pairs and the sum of the forces in a pair is zero.*

Be careful with the third law, the equal and opposite forces are acting on **different** objects. The motion of an object is determined only by the sum of the forces acting on **that** object.

$\large \Sigma \vec{F}_{\parallel} = \vec{F}_{G}\sin\theta=m\vec{g}\sin\theta$

$\large \Sigma \vec{F}_{\perp} = \vec{F}_{G}\cos\theta+\vec{F}_{N}=0 $

For this system (Atwood's Machine) need to consider free-body diagrams for two objects. $\vec{a}$ is the same in magnitude for each weight, the sign relative to gravity must be opposite on one side from another. The weights are both connected by the same rope and thus the force due to tension is the same for each object. The concept behind this device is used in elevators and funiculars |

$\large m_{1}\vec{a}=\vec{T}-m_{1}\vec{g}$ $\large m_{2}\vec{a}=m_{2}\vec{g}-\vec{T}$

$\large \vec{T}=m_{1}\vec{a}+m_{1}\vec{g}$ $\large \vec{T}=m_{2}\vec{g}-m_{2}\vec{a}$

$\large m_{1}\vec{g}+m_{1}\vec{a}=m_{2}\vec{g}-m_{2}\vec{a}$

$\large (m_{1}+m_{2})\vec{a}=(m_{2}-m_{1})\vec{g}$

$\large \vec{a}=\vec{g}\frac{m_{2}-m_{1}}{m_{2}+m_{1}}$

$\large \vec{T}=\vec{g}\frac{2m_{2}m_{1}}{m_{2}+m_{1}}$

To answer this question we need to compare the applied force to the **maximum** force that static friction can provide, which is $F_{fr}=\mu_{s} N$. We should note that this force only is present when a force is applied and up to the point where the component of the applied force in the direction of motion exceeds the maximum possible static friction force the static friction force will be exactly equal and opposite to the component of the applied force in the direction of motion.

Newton's second law tells us we can only have a centripetal acceleration towards the center of a circular motion path if the sum of the forces in the radial direction

$\Sigma F_{R}=ma_{R}=m\frac{v^{2}}{r}$

This **centripetal** force must be provided in some way for circular motion to occur.

Do not be misled in to thinking that such a thing as **centrifugal** force exists. As we saw earlier if the centripetal force is removed an object continues in a tangential, not perpendicular path to the motion. Of course in situations where you are in circular motion, it can feel as if you are being pushed out from the center, but this is in fact you applying the equal and opposite reaction force required by Newton's 3rd Law. The force on **you** is directed inwards.

If we now try to move the mass on spring in a vertical circle we can see it doesn't happen. The force required is different at the top and bottom because gravity alternatively assists or hinders us at the top and bottom of the path and as the spring changes it's length according to the tension on it the resulting motion is not circular.

A ball on a string works, providing the velocity is greater than $v=\sqrt{rg}$

Roads designed for high speed traffic will often used banked turns to increase the maximum speed for which slipping does not occur. A well designed banked turn means the car should not rely on friction. (It also makes the problem easier..)

The relationship between maximum speed, radius and the bank angle can be found from considering the forces

$F_{N}\sin\theta>\frac{mv^{2}}{r}$

$F_{N}\cos\theta=mg$

$mg\frac{\sin{\theta}}{\cos{\theta}}=\frac{mv^{2}}{r}$

$\tan{\theta}=\frac{v^2}{rg}$

or with friction

$F_{N}\sin\theta+\mu F_{N}\cos\theta=\frac{mv^{2}}{r}$

$F_{N}\cos\theta-\mu F_{N}\sin\theta=mg$

$\frac{\sin\theta+\mu\cos\theta}{\cos\theta-\mu\sin\theta}=\frac{v^2}{rg}$

We can define a Gravitational Constant $G$ such that the force between two masses is

$F=G\frac{m_{1}m_{2}}{r^{2}}$

The value of $G$ is $6.67\times10^{-11}\mathrm{Nm^{2}kg^{-2}}$.

For a circular orbit a satellite of mass $m$ must have a velocity such that the gravitational acceleration is the same as the centripetal acceleration.

$G\frac{mM_{E}}{r^{2}}=\frac{mv^{2}}{r}$

$v_{circular}=\sqrt{\frac{GM_{E}}{r}}$

GPS satellites are essentially extremely accurate clocks that continually broadcast their position and the time, which an Earth based receiver can use to calculate it's own position, so long as it can receive at least 4 signals. It is not desirable for GPS satellites to be in geostationary orbits.

A GPS satellite orbits about 20,200 km above the surface of the Earth (or at a radius of 26,600 km from the center of the Earth.)

How many times a day do the GPS Satellites go round the Earth?

$v=\sqrt{\frac{GM_{E}}{r}}$

Mass of Earth $M_{E}=5.9742\times10^{24}\,\mathrm{kg}$

Gravitational Constant $G=6.67\times10^{-11}\,\mathrm{Nm^{2}kg^{-2}}$

Number of seconds for a GPS satellite to go round the Earth $\frac{2\pi r}{v}=2\pi\frac{r^{3/2}}{\sqrt{GM_{E}}}=2\pi\frac{(2.66\times10^{7})^{3/2}}{\sqrt{6.67\times10^{-11}\times5.9742\times10^{24}}}=43,200$

Times round the earth in a day = Seconds in one day/Number of seconds for a satellite to go round the Earth

Seconds in one day = $24\times60\times60=86400$

Times round the earth in one day $\approx 2$

$v=3870\mathrm{m/s}$

Work is a measure of what a force achieves, so when we calculate the work done by a force on an object we consider only it's displacement in the direction of the force.

Mathematically, as we consider both force $\vec{F}$ and displacement $\vec{d}$ to be vectors, and the work done on an object to be a scalar $W$, we say that the work is the scalar product, or dot product of force and displacement. We should note that this is not the displacement from an arbitrary origin, but rather the change in an objects position during a motion.

For a constant force

$W= \vec{F} \cdot \vec{d} $

If the two vectors are at an angle $\theta$ to each other then we can say

$W = Fd\cos\theta$

The work-energy theorem says that the net work done on an object $W_{Net}$ is equal to it's change in kinetic energy $\Delta K$

$W_{Net}=\Delta K$

We also saw today that the change in potential energy **if only conservative forces are active** is

$\Delta U = -\int_{1}^{2}\vec{F}\cdot\,d\vec{l} = - W_{Net} $

So we can see that when only conservative forces act that

$\Delta K + \Delta U = 0 $

We call the sum of the Kinetic Energy and Potential Energy the Total Mechanical Energy E

$E=K+U$

and we can see that under the condition of only conservative forces acting this is a conserved quantity.

Non conservative forces remove mechanical energy from the system, but it is not destroyed, it is simply converted to a different form of energy (frequently, but not always, heat).

The total energy conservation law can also be useful, for example when a frictional force $\vec{F}_{fr}$ is acting and an object travels a distance $d$ while it goes from a height $h_{1}$ to $h_{2}$, changing it's velocity from $v_{1}$ to $v_{2}$, conservation of total energy tells us

$\frac{1}{2}mv_{1}^2+mgh_{1}=\frac{1}{2}mv_{2}^2+mgh_{2}+F_{fr}d$

As we saw previously the force due to gravity on an object of mass $m$ due to an object of mass $M$ at distance $r$ is $-\frac{GmM}{r^2}\hat{r}$

If we raise an object from the surface of a planet to a distance $h$ above it the change on potential energy is

$\Delta U = \int_{R_{E}}^{R_{E}+h}\frac{GmM}{r^2}\,dr=-\frac{GmM}{R_{E}+h}+\frac{GmM}{R_{E}}$

We should consider where a suitable zero of potential energy is

Choosing $r=\infty$ as our reference position is attractive as then the potential energy change in bringing an object from infinity to a position $r$

$\Delta U = \int_{\infty}^{r}\frac{GmM}{r^2}\,dr=-\frac{GmM}{r}$

is the same as the potential energy, so generally we say the gravitational potential energy at point $r$ from the center of a mass $M$.

$U = -\frac{GmM}{r}$

Power is defined as the rate at which work is done.

Average power is given by

$\bar{P}=\frac{W}{t}$

Instantaneous power is given by

$P=\frac{dW}{dt}$

Recall that $dW=\vec{F}\cdot d\vec{l}$, which means that

$P=\frac{dW}{dt}=\vec{F}\cdot \frac{d\vec{l}}{dt}=\vec{F}\cdot\vec{v}$

Units of Power are J/s or W (Watts).