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## Kinematics

We introduce some basic kinematics examples of motion.

### Free Fall

Finding the trajectory of a free falling body has puzzled scientist for a long time. Only in the 17th century Galilei discovered the law of motion of free fall. A free fall is a motion in a vacuum with only the force of gravity working on the object.

Galilei discovered that the distance travelled in a serie of equal units of time is proportional to the odd numbers and is independent of the mass of the body. Let c be the distance travelled in the first interval then the distance in the subsequent intervals of time develops as: 3c,5c,7c,9c,.... From this serie follows that the total distance travelled is a serie of squares: c,4c,9c,16c,25c,..... or in a mathematical formula, where x is the total distance travelled: $x(t)=ct^2$ From mathematics we know that the average velocity of a very small interval of time around a point approaches the instantenous velocity at this point. We have therefore for the velocity the following relation: $v(t)=2ct$ The speed increase with 2c. The increase of speed is called the acceleration $a(t)=2c$ The constant change in velocity is called the gravitational constant, denoted with $g$, and is about $g=2c=9.8 ms^{-2}=32 fts^{-2}$ around the surface of the earth.

We learned from Galilei: The gravitational force causes an object to accelerate uniformly with g. As a result the change in distance in the vertical direction is a quadratic function in relation to time and independent of the mass of the body. Although the calculations are mathematical in nature, the law itself is an observational result, and does not follow from mathematics. Mathematics is only the tool to describe this law.

We can use differential calculus for the case of uniform accelaration to establish the relationships of distance, velocity and acceleration as function of time as follows: \begin{align*} a(t)&=g \to \\ a(t)dt&=gdt \to \\ \int{a(t)dt}&=\int{gdt} \to \\ v(t)&=gt+v_0 \end{align*} \begin{align*} v(t)dt&=(gt+v_0)dt \to \\ \int{v(t)dt}&=\int{(gt+v_0)dt} \to \\ x(t)&=\tfrac{1}{2}gt^2 + v_0t+x_0 \end{align*}

Motion takes place in a three dimensionsal space: length, breadth and height. The mathematical object of a 3 dimensional vector is usefull in physics to represent position, velocity and acceleration in these 3 dimensions. One of the remarkable things of mechanics is the insight that Galilea brought us too: Motion in each of the three dimensions of space are mutually independent. The gravitional force acts only in one dimension, height, and does not impact the other dimensions. If no other forces acts on the body the velocity in these other dimensions remain unchanged during the free fall motion.

We will now turn to a variant of free falling motion, where there is an initial velocity in the other dimensions.

## Projectile motion

A stone dropped from an airplane will accelerate in the vertical direction due to gravity and move with constant speed in the horizontal direction. A passenger in the plane will see a straight line motion, whereas an observer on the ground will see a curved motion. This type of curved motion (motion in more than one dimension), with a constant acceleration, is called projectile motion. It is the idealized motion of the flight of a ball. The motion of a projectile is a two dimensional motion. We choose a coordinate system with the positive y-axis vertically upward and the origin at the point where the projectile begins its flight. Furhter we assume that the projectiles height is limited, so the same gravitational force can be applied during its flight, and also that the reach of the projectile is limited so the earth can be taken as flat for this motion. We can use a constant force vector for this type of motion. The projectile motion starts with a force that launches the projectile with an initial velocity $\vec{v_0}$. After launch only the gravitational force acts on the projectile. The gravitational force is given by: $\vec{F}_g=-mg\hat{\vec{ y}}$ If we apply Newtons second law $\vec{F}=m\vec{a}$ we get: \begin{align*} -mg\hat{\vec{y}}=m\vec{a} \\ \vec{a}=-g\hat{\vec{y}} \end{align*} This shows clearly that the mass of the projectile plays no role in the projectile motion. We can write down the equations for the vector components for acceleration and velocity: \begin{align*} a_x&=0 \\ a_y&=-g \\ v_{x0}&=v_0\cos\alpha \\ v_{y0}&=v_0\sin\alpha \end{align*} From these equations we see that there is no change in the velocity in the horizontal direction. The only force is the gravittional force and this force works only in the vertical direction. Notice also that a body in a projectile motion and free fall fall the same distance in the same time. We can use this information to write down the equations of motion with constant acceleration for this special case: \begin{align*} v_x&=v_0\cos\alpha \\ v_y&=v_0\sin\alpha - gt \\ x &= (v_0\cos\alpha)t \\ y &= (v_0\sin\alpha)t -\tfrac{1}{2}gt^2 \end{align*} We can derive also the equation which gives the relation between the x and y coordinates by combining the last two equations and eliminating $t$. We get: \begin{align*} y&=\tan\alpha - \frac{g}{2(v_0\cos\alpha)^2}x^2 \end{align*} This shows us that the projectile motion is a parabolic motion.

Angle of sight
A plane is flying with a constant horizontal velocity of 500km/h at a height of 5km toward a target point on the ground. At what angle of sight $\theta$ should a projectile be released to hit the target on the ground ?
Solution

Motion of a thrown ball

A ball is thrown at an angle $\theta$ and initial velocity $v_0$. Calculate:

• The time when the ball reaches the highest point
• The maximum height
• The time of flight
• The range of the ball
• The velocity of the ball when it hits the ground

Solution
Projectile trajectory

Graph the trajectory of a ball launched at an angle of $37^\circ$ and with an inital speed of $15m/s$.

Solution
Gun fired at monkey
A famous physics lecture demonstration is a gun sighted at an elavated target which is released in free fall as the buller leaves the muzzle. The question is does the bullet hit the target ? Solution

## Uniform circular motion

Another type of motion is uniform circular motion. In this type of motion the magnitude of the velocity remains te same during a revolutuin, but changes only in direction. The direction is always tangent to the circle. Because the velocity changes, there is an acceleration. This acceleration is directed towards the center of the circle and the following relation holds between the magnitudes of the acceleration $a$,velocity $v$ and radius $r$ of the circle: $a=\frac{v^2}{r}$

Orbit satellite around the earth
Calculate the speed of an earth satellite assume that it is travelling at an altitude of 225km above the surface of the earth where $g=9.1m/s^2$. What is the reason that the satellite does not fall down to the earth ? Calculate the path of the satellite.
Solution
Orbit moon
The moon revolves around the earth in 27.3 days. Assume that the orbit is circular with a radius of 385000 km, with the center at the middle of the earth. Find the acceleration of the moon towards the earth.
Solution

## Planetary motion

In this section we explore the planetary motions. Analyzing motion starts with finding the acceleration. Finding the acceleration starts with finding the applicable forces with their force law. For planetary motion this force law is Newtons universal gravitation law. This law implies a force between two particles along the line of joining them. For the celestial bodies we assum they are spherical and therefore we can treat these bodies as pointlike particles.

Kepler's law of periods

The third law of Kepler states that for any object orbitting the sun: $\frac{T^2}{S^3}=C$ where the $T$ the period of one revolution about the sun and $S$ the average distance to the sum is, C is a constant.

Calculate $C$.

Show based on the universal law of Newton that this ratio is a constant.

Solution
Orbit earth
Calculate the orbit of the earth around the sun.
Solution

## Newton's Laws

Newton's laws are demonstrated by a serie of examples.

## Trajectory

Newton's second law states in essence that each motion is determined by a second order differential equation with two initial conditions for each degree of freedom: $\ddot{x}(t_0)=\frac{1}{m}F(x(t_0),\dot{x}(t_0),t_0)$

We describe a numerical method to find the trajectory $x(t)$. We know that a smooth functioncan be extrapolated with a Taylor expansion if we use small timesteps $\Delta t$: $x(t_0+\Delta t) = x(t_0)+\Delta t \dot{x}(t_0)+...$ and of course also $\dot{x}(t_0+\Delta t)=\dot{x}(t_0)+\Delta t \ddot{x}(t_0)+...$ but we have: $\ddot{x}(t_0)=\frac{1}{m}F(x(t_0),\dot{x}(t_0),t_0)$ Repeat this process $n$ times to compute $x(t+n\Delta t)$.

If we want to have an accurate result we cannot just plug in the $\dot{x}(t_0+\Delta t)$ to calculate $x(t_0+\Delta t)$ because during the time interval the velocity has not been constant. Therefore we assume that during the small step of time, where the acceleration is more or less constant, we calculate the velocity and then the change in distance as follows: $\Delta x_n = v_{avg}\Delta t$ with: $v_{avg}=\tfrac{1}{2}((a_n\Delta t+v_{n-1}) + v_{n-1})=\tfrac{1}{2}a_n\Delta t + v_{n-1}$ So $\Delta x_n = \tfrac{1}{2}a_n(\Delta t)^2 + v_{n-1}\Delta t$ To have the cumulative numbers we just add up these small intervals: $x_n=x_0+\sum_{i=1}^{n}{\Delta x_i}=x_0+\sum_{i=1}^{n}\tfrac{1}{2}a_n(\Delta t)^2 + v_{n-1}\Delta t$ $v_n=v_0+\sum_{i=1}^{n}{a_i}\Delta t$

To avoid the calculation of the average of the velocities, we can follow another approach to calculate the velocity at the time $t+0.5\Delta t$ instead of at time $t+\Delta t$. To start we calculate $v(0.5\Delta t)=v(0)+0.5\Delta t a(0)$ and then use following equations for each time step: \begin{align*} x(t+\Delta t)&=x(t)+v(t+0.5\Delta t)\Delta t \\ v(t+0.5\Delta t)&=v(t-0.5\Delta t)+a(t)\Delta t \\ a(t)&=F(t)/m \end{align*}

This approach clearly shows that the trajectory is determined by two initial conditions per degree of freedom. A degree of freedom is a coordinate in which an object is free to move. The fundamental law of mechanics is equivalent to a second order differential equation and its solution is $x(t)$ with two integration constants $x(t_0),\dot{x}(t_0)$.

These relationships can be concisely programmed in a language as Haskell. Haskell is a functional programming language and very usefull for mathematic type of programming. We simply defined the vector as a list and defined simple operations for adding and multiplication with a scalar.

## Conservation linear momentum

#### Bullet and gun

Newton's third law states that if two bodies $A$ and $B$ interact then the forces between them are equal and opposite. $\vec{F}_{AB}=-\vec{F}_{BA}$ the notation $\vec{F}_{AB}$ means the force on $A$ caused by $B$. Now its second law states that a force causes an acceleration inversely proportional to the mass of an object. An acceleration is just a change of velocity. So we have: \begin{align*} \Delta \vec{v}_A &= \frac{\vec{F}_{AB}}{m_A} \\ \Delta \vec{v}_B &= \frac{\vec{F}_{BA}}{m_A} \\ \end{align*} but form this follows: $m_A\Delta v_A=-m_B\Delta v_B$ or $m_A\Delta v_A+m_B\Delta v_B=0$ we clearly see that total of the quantity $mv$, which is called momentum, is not changed by the interaction of bodies, but remains constant (of course not neccessarily zero). The change in velocity of one body,resulting from the force, is compensated by the other body. This ratio of their velocities is inversely proportional to the ratio of their masses: $\frac{\Delta v_A}{\Delta v_B}=\frac{m_B}{m_A}$ Let us assume that $m_A>m_B$ and let us change our notation $M=m_A$ and $m=m_B$ it is immediately clear that $\Delta v_A \less \Delta v_B$ Let us from now on denote $v=v_A$ and $V=v_B$ to remind us of this fact. If we reformulate our earlier result: $\frac{\Delta V}{\Delta v}=\frac{M}{m}$ but as speed determines distance we have the same relationship for the distance travelled by $m$ which we denote by $D$ and $M$ which we denote by $d$: $\frac{\Delta D}{\Delta d}=\frac{M}{m}$ the distance $\Delta D$ is $\frac{M}{m}$ times bigger than the distance $\Delta d$. The ratios of their masses determines therefore the ratio of their changed velocities and distances. We could define a point $C$ on the line between $M$ and $m$ such that: $\frac{Cm}{CM}=\frac{D}{d}=\frac{M}{m}$ during uniform motion this point would be at rest. This point is called the center of gravity between two masses. Suppose a bullet has a mass of 1 ounce (= 1/16 pound) and a muzzle velocity of 1600 feet per second. Its forward momentum is then 1 / 16 x 1600 = 100 pounds feet per second. The gun must therefore have a momentum of -100 pounds feet per second. If the gun’s mass is 12.5 pounds, its recoil velocity would be -100 pounds feet per second divided by 12.5 pounds = - 8 feet per second. In other words, the gun’s recoil velocity is 8 feet per second in the direction opposite that of the bullet.

Another example is a balloon from which air escapes from the nozzle. The shrinking balloon will careen crazily until it becomes deflated. The balloon is propelled forward by pushing its contents rearward through the nozzle. This same principle is the basis for rocket propulsion, which expels high-velocity exhaust gases from the rocket engine. The rocket motor is thus a self-sufficient means of propulsion. It can therefore operate in “outer space".

## Forces on mass

### Force on a mass on a horizontal surface The most simple situation is a force on a single mass. In analyzing mechanics problems we isolate the different masses and draw all the forces on an individual mass to find the net resultant force acting on this mass. We draw these forces in a free body diagram. In this case there are 3 forces at work on the mass. The gravitational force $F_g$, the normal force $F_N$ and the applied force $F$. The gravitational force and the normal force act in the vertical direction and cancel each other. The acceleration of the mass is in the direction of the applied net force $F$ and is proportional to the applied force and inversely proportional to the mass: $a=\frac{F}{m}$. ### Force on a mass on a horizontal surface at an angle When the force acts at an angle the force is resolved in a vertical and horizontal component. The vertical component is cancelled out by the normal force that the support surface exerts on the mass. ## Strings

Consider a block of mass $m$ suspended from a fixed beam by means of a string. We suppose the string has neglectible mass compared to the block, and the string is inextensible. The beam and the block pull on the string. As reaction the string pulls the beam and block together. This force is called the tension force $\vec{F}_T$ on the beam and the block. The tension force at each end of the string is equal in magnitude and opposite in direction. These forces act so as to oppose the stretching of the string: i.e., the beam experiences a downward force, whereas the block experiences an upward force. When the block is in equilibrium, no acceleration, we have $T-mg=0$

Consider now a block mass m is suspended by three strings. Determine the tension forces $F_T$, $F_{T_1}$ and $F_{T_2}$.

It is easily seen that the two forces on the block must cancel so we have again $F_T=mg$. To find the other tension forces we draw a free body diagram. The net resultant force in the horizontal and vertical direction are zero. So we resolve the three forces in these directions. We take the direction to the right and up as positive. For the horizontal directions we have: \begin{align*} F_{T_1}\cos 60^\circ - F_{T_2}\cos 30^\circ=&0 \\ \frac{F_{T_1}}{2}-\frac{\sqrt{3}F_{T_2}}{2}=0 \\ F_{T_1}=\sqrt{3}F_{T_2} \end{align*} For the vertical direction we have: \begin{align*} F_{T_1}\sin 60^\circ + F_{T_2}\sin 30^\circ -mg=&0 \\ \frac{\sqrt{3}F_{T_1}}{2}+\frac{F_{T_2}}{2}-mg&=0 \\ \end{align*}

Substitution of $F_{T_1}=\sqrt{3}F_{T_2}$ into the last equation we find: \begin{align} F_{T_2}&=\frac{mg}{2} \\ F_{T_1}&=\frac{\sqrt{3}mg}{2} \end{align}

### Inclined motion

With an incline we can demonstrate different forces of motion in action. As well as the usage of vector components in the different directions of motion.

Suppose we have the following situation of a block with mass $m$ on a wedge. We analyze different types of problems.

Motion of a block on a fixed wedge without friction
We assume the incline is fixed to the earth, so it does not move. Determine the acceleration of the block in the horizontal and vertical dimension.
Solution
Motion of a block on a non-fixed wedge without friction
Now suppose that the wedge can move along the ground and has mass $M$. Determine the acceleration of the wedge,$a_w$, and the acceleration of the block relative to the wedge, $a_{wb}$.
Solution

Motion of a block on a non-fixed wedge without friction
Suppose now we apply a force to the wedge such that the block remains stationary on the wedge. Determine the force $F$ and the acceleration $a$ of the wedge.
Solution
Motion along a wedge with friction
Find the statical friction coefficient, $\mu_s$.
Solution
Universal law of gravitation
Derive Newton's universal law of gravitation for circular motion.
Solution