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SPACE Section 2. Projectile Motion. A projectile is an object that is projected (that means thrown, dropped or launched) into the air, but not propelled as is a rocket.
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Projectile Motion • A projectile is an object that is projected (that means thrown, dropped or launched) into the air, but not propelled as is a rocket. • This includes a ball being thrown, a football being kicked, a golf ball being struck, a bullet being fired, or cargo being dropped from a plane. • Basically Projectile = Unpowered
Galileo & Projectiles • Aristotle: • Everyobject has a certain amount of impetus and thenwhenthisruns out, the objectfalls to Earth.
Galileo & Projectiles • Galileo: • Studied projectile motion and came to realiseit has a parabolic nature. What is the nature of this curved path? How to test this out? Things fall way too fast!
Galileo & Projectiles • Galileo’sinclined plane: Problem 1: Things fall too fast. Soln: Inclined planes make things fall a lot slower. Problem 2: How to accurately time? Soln: Being a musician GG realised that the ear is very good at keeping track of time. Thus he designed his inclined plane to have movable bells that rung everytime an object went down the plane. He adjusted these backwards and forwards until he got a perfectly even beat as the object travelled down. This allowed him to work out the distance of the object in equal time intervals.
Galileo & Projectiles 1 4 9 16 Galileo's genius: Distance proportional to time squared in a sense: y = x2 which is...A PARABOLA! Galileo was able to use MATHS to prove that the nature of any projectile is a parabola.
Galileo & Projectiles Galileo showed that horizontal motion is completely independent of vertical motion. Looking at the horizontal motion we can see the ux = vx However vertical motion has a square relationship
Galileo & Projectiles Galileo showed that horizontal motion is completely independent of vertical motion. Looking at the horizontal motion we can see the ux = vx However vertical motion has a square relationship
Galileo & Projectiles Galileo's analysis of Projectile Motion Summary · Inclined plane · Distance ∝to time2 · Horizontal motion independent of vertical motion · Projectiles follow a parabolic path
Projectile Motion A launched projectile, in the absence of any air resistance, will follow a parabolic path until it strikes the Earth. This path is called the trajectory of a projectile. • To analyse the trajectory of a projectile separate it into horizontal and vertical motion (NOTE: We ignore air resistance). • The horizontal motion, is not subject to any forces, and therefore experiences no acceleration. • The vertical motion, is subject to the vertical weight force, and therefore experiences the downwards acceleration due to gravity. (Recall that there is only one force acting – the vertical weight.)
Questions 1. A rifle with a muzzle velocity (the speed the bullet comes out of the barrel) of 450 ms-1 is fired level at the horizon. Determine: how fast the bullet is travelling 0.3 seconds after firing how far it has travelled horizontally in that time. ANS a) Our first formula tells us that vx = ux, that is, the final velocity equals the initial velocity over any time period. In other words, the horizontal velocity is the same all the way through the motion. Therefore, the velocity after 0.3 s is still 450 ms-1 (horizontally). ANS b) x = uxt = 450 x 0.3 = 135 m That is, after 0.3 s the bullet has travelled 135 m.
Questions 2. An air hockey puck is pushed so that it glides along its table at 0.15 ms-1. If the table is 1.2 m long, determine: how long the puck takes to travel the length of the table its velocity when it gets there ANS a) ux= 0.15 ms-1, x = 1.2 m, t = ? x = u x t 1.2 = 0.15 t t = 8.0 s ANS b) vx= ux = 0.15 ms-1
Questions 3. A stone is thrown horizontally at 8.0 ms-1. If it takes 0.5 s to fall to the ground, how far horizontally will it have travelled in this time? ANS ux= 8.0 ms-1, t = 0.5 s, x = ? x = uxt = 8.0 x 0.5 = 4.0 ms-1 That is, the trajectory of the stone will have a range of 4.0 m.
Vertical Motion • In this regard it is much like any object thrown straight up. If thrown up from ground level, an object will rise up to a peak height, stop momentarily in the air, then return to the ground, speeding up as it does so. • The second half of the motion is symmetrical with the first part, so that the time taken for the fall will equal the time taken for the rise. Also, the speed with which the object strikes the ground will equal the speed with which it was launched (only the direction will be down instead of up).
Questions An arrow is fired directly upwards with a velocity of 55 ms-1. Assume that it is fired from ground level and that there is no air resistance. a) How fast is the arrow moving when it returns to the ground? b) What is the time of flight of the arrow? Solution: • By the symmetry of the motion, you can say that the arrow will have a velocity of 55 ms-1 down. • A useful strategy to solve this problem is to focus on the arrow’s rise up to its peak height. Assume that ‘up’ is the positive direction. You can now say that: uy = 55 ms-1, vy = 0 ms-1, ay = -9.8 ms-2, t = ? Theequationto use is: v = u + at 0 = 55 + (–9.8)t t = 5.6 s That is, the arrow will take 5.6 s to rise to its peak height. By symmetry, it must take just as long to fall back, so: trip time = 2 x5.6 = 11.2 s.
Components The mathematical expressions you need to use therefore are: • the horizontal component, ux = u cos • the vertical, uy= u sin , where is measured from horizontal.
Putting it all together • In order to find the velocity of the projectile at other times during its flight you need to separately calculate the velocity in the x direction and the y direction, and then add them together using two-dimensional vector addition. • Resolve initial velocity u into components ux and uy. • Note that the horizontal velocity is uniform and doesn’t change (vx = ux). • Consider the vertical motion and calculate the velocity vy after the specified time. • You must now add vx and vy together as shown in the following diagram:
Questions • A tennis ball is struck, this time at 5.0 ms-1, 55° above horizontal. What is the velocity of the tennis ball 1.2 s after being struck? a) The components of the initial velocity are: The horizontal component, ux = u cos = 5.0 cos 55 = 2.87 ms-1 Theverticalcomponent, uy = u sin = 5.0 sin 55 = 4.10 ms-1 b) vx = ux = 2.87 ms-1 c) In analysing the vertical motion weshalltake ‘up’ to be the positive direction. vy = uy + at=4.10 + (-9.8 x 1.2) = -7.66 ms-1= 7.66 ms-1down d) The final step is toaddvxandvytogether as shown in thediagram. That is, after 1.2 s the velocity of the tennis ball is 8.2 ms-1 at 69° below horizontal.
Maximum height and trip time • The strategy for working out these quantities is to consider just the vertical motion up to the peak. At the peak, the projectile has stopped moving in the vertical direction so that you can say that vy = 0 in this portion of the motion. The method then follows these steps. a) Resolve the vertical component uy of the initial velocity u. b) Consider the vertical motion up to the peak. c) Note that vy = 0 for this portion of the motion. d) Select an acceleration equation that enables calculation of unknowns from the data. then either calculate y, which is the maximum height. or calculate t, which is the time for the projectile to rise up to the peak. e) Double this time to find the trip time. (Making use of the symmetry of the motion here, because it takes as long to fall as it does to rise up to the peak.)
Questions • A tennis player plays a half volley off the ground, so that the ball leaves the racquet with a velocity of 7.2 ms-1 at 36° above horizontal. Calculate the maximum height achieved by the ball, and the time it takes to bounce for the first time (that is, the trip time). The first step is to calculate uy: uy = 7.2 sin 36° = 4.2 ms-1 uy = 4.2 ms-1, vy = 0 ms-1, ay = -9.8 ms-2, y = ? Thecalculationneededtofindthemaximumheightis: vy2 = uy2 + 2ayy 0 = 4.22 + (-9.8)y y = 0.9 m Tocalculatethetrip time youneedtofindthe time toreachthepeak. uy = 4.2 ms-1, vy = 0 ms-1, ay = -9.8 ms-1, y = 0.9 m, t = ? A suitablecalculationtofindthe time tothepeakis: v = u + at 0 = 4.2 + (-9.8)t t = 0.43s and hence, trip time = 2t = 2 0.43 = 0.86 s
Range • Your final strategy is concerned with calculating the maximum horizontal displacement of a projectile, that is, the range of the trajectory. In order to make this calculation you will need to follow these steps: a) Resolve initial velocity u into components uy and ux. b) Analyse the vertical motion to find the trip time as shown above. c) Consider only the horizontal motion and calculate the range using, x = uxt
Questions • Back to the tennis shot played in the earlier sample problem. The ball was struck from ground level at 7.2 ms-1 at 36° above horizontal. What will be the range of its trajectory? The first step, as usual, is to resolve the initial velocity into components. ux = 7.2 cos 36 = 5.8 ms-1 uy = 7.2 sin 36= 4.2 ms-1 Normallyyouwouldhavetocalculatethetrip time byanalysingthe vertical motion, butweknowthatthetrip time is 0.86s. The final stepthenistoanalysethe horizontal motiontofindthemaximumdisplacement. ux = 5.8 ms-1, trip time t = 0.86s, x = ? The required calculation is: x = uxt = 5.8 X 0.86 = 5.0m
Newton’s concept of escape velocity Isaac Newton was the first to write about the possibility of an artificial satellite of the Earth. He imagined a cannon ball fired from a very tall mountain. Such a projectile could follow a trajectory as shown, eventually hitting the Earth. !! boom trajectory
Newton’s concept of escape velocity • Fired at a greater velocity, the projectile travels further… • The projectile increases its speed as it falls due to the conversion of gravitational potential energy to kinetic energy • The ball reaches its greatest speed as it hits the ground.
Newton’s concept of escape velocity • Fired at a still greater velocity, the projectile follows a trajectory closely matching the curvature of the Earth, so that it never hits the ground. • If the orbit is elliptical, the projectile increases its speed as it gets closer to the Earth. • The force of gravity results in the the projectile being in a continuous state of free fall… • There is a net force, gravity, towards the centre of the Earth. This force is the centripetal force
Newton’s concept of escape velocity Escape velocity Fired with a sufficiently large velocity, the projectile would never return to Earth. The minimum velocity required to achieve this result is called the escape velocity. The escape velocity from the Earth’s surface is . . .~ 11.2 km s-1
Newton’s concept of escape velocity Escape velocity A satellite given a velocity equal to or greater than the escape velocity never returns because its initial kinetic energy exceeds the change in potential energy as it increases its altitude (distance from the central body) Escape velocity is the minimum velocity that must be imparted to an object at a specific location to cause it to completely escape from the gravitational pull of the planet A projectile with a large velocity may travel so fast that it escapes the Earth’s gravitational influence.
Newton’s concept of escape velocity Not on sheet Learn it!
Forces on an Astronaut During Launch For a rocket at rest • The downward force due to gravity… is equal to • the upward reaction force of the Earth against the rocket. • The two forces are in equilibrium. • The net force on the rocket is zero.
Forces on an Astronaut During Launch • As the rocket lifts off the weight of the rocket (W) is less than the force produced by the thrust of the engines (FT) • The two forces are not in equilibrium. • Because the net force on the rocket is upward… the rocket accelerates in upward direction (Newton’s second law in action) aR W = mg FT> mg
Fweight = mg Fthrust Forces on an Astronaut During Launch • As a rocket takes off, the thrust produced by the rocket engines produces a force that exceeds the weight of the vehicle • The resulting net upward force causes the rocket to accelerate away from the Earth’s surface • The acceleration increases as the mass decreases as fuel is burned [F=ma] because the mass of the rocket is rapidly decreasing. Atlantis blasts off - September 10th 2006
Forces on an Astronaut During Launch During Launch Freaction • The astronaut experiences two forces • A gravitational force downward • A reaction force upward • These are equal in magnitude and opposite in direction - there is zero net force on the astronaut • The astronaut is said to be experiencing a force of “1G” Rocket at Rest • The astronaut experiences two forces • A downward gravitational force • - which remains constant • An upward reaction force • - which exceeds that of gravity • The sum of these two forces (the resultant) produces a net upward force. • If the rocket is accelerating upward at 9.8 m s–2, the astronaut experiences a reaction force of “2G” Freaction W=mg W=mg
Forces on an Astronaut During Launch • If the thrust produced by the engines remains constant… • As the mass of the rocket decreases due to the fuel being expelled… • - the acceleration of the rocket, and hence the astronaut in the rocket, increases • Hence the upward reaction force on the astronaut increases… • - reaching a typical maximum during a launch of 3G (e.g. the space shuttle) • As the rocket mass decreases, the engines may be throttled back to avoid excessive accelerations which could damage the rocket • The reaction force that the astronaut experiences is often called a “g-force”. • Once the spacecraft is orbit, there is no reaction force - gravity is the only force acting on the astronaut - a condition sometimes called “zero g” Freaction As the rocket accelerates upwards… W=mg
‘g’ Forces • The term ‘g force’ is used to express a person’s apparent weight as a multiple of their normal true weight (that is, weight when standing on the surface of the Earth).
‘g’ Forces A person with mass, m, who is located at or near the surface of the Earth will always have some weight W=mg. When a person stands on a scale, the reading (the number of kilograms or newtons) on the scale is actually the Normal Force that the scale exerts back towards the person to support the person's mass/weight. This Normal Force is what we call Apparent Weight. However when the person and the scale experience acceleration, things get complicated.
‘g’ Forces Consider a person travelling in an elevator whilst standing on measuring scales. If the acceleration of the elevator is zero, then there are two possible scenarios; the elevator can be at rest (stationary, zero velocity) or moving with a constant speed (no acceleration if velocity does not change).
‘g’ Forces In this case, the action and reaction force pair between the person and the scale is just the weight. The person pushes down on the scale with a force of -W=-mg (negative direction) and the scale pushes back up against the man with a Normal Force in the positive direction. Because the reading on the scale is the magnitude of the normal force, the scale will read the true weight when the elevator is NOT accelerating.
‘g’ Forces If the elevator is going up, there is an acceleration on the person and the scales. The inertia of the person would prefer to stay stationary, so the elevator floor and scale must push up on the person to accelerate him upward along with the elevator. (The person doesn't sink into the floor when the elevator accelerates up. The elevator and the scale and the person all move together.) The scale therefore has to push upward with extra force on the person to accelerate the person's mass upward. This results in a greater contact force between the scale and the person. Therefore the Normal Force is larger, so the reading on the scale is a number that is GREATER than the true weight.
‘g’ Forces The inertia of the person would prefer to stay at rest, so the elevator floor and scale effectively drop out a little bit from underneath the person as the elevator accelerates down. The person doesn't float upward here also, because again the elevator and the person move together, but the contact force between the person and the scale is reduced. The scale therefore has to push upward with less force on the person to support the person's weight. Therefore the Normal Force is smaller, so the reading on the scale is a number that is LESS than the true weight. If the elevator is going down, there is a deceleration on the person and the scales.
‘g’ Forces If the elevator cable were to break, the whole elevator-scale-person system would all begin to accelerate downward due to the force of gravity. All objects in freefall accelerate downward with the same magnitude (acceleration due to gravity, g). The scale and the person are freefalling together, so there is NO contact force (Normal Force) between the scale and the person. (When they are both falling together, there is no way that the scale can support any of the person's weight.)
Forces on an Astronaut During Launch As a rocket ascends from Earth’s surface FT The G-force experienced by the astronaut when the rocket is accelerating vertically is… W=mg a = Fnet/m Fnet = FT – W The Earth’s gravitational acceleration is almost constant over the distances involved in LEO. LEO orbits range up to about 500 km altitude.
Question about g forces • Identify why the term ‘g forces’ is used to explain the forces acting on an astronaut during launch. (4M) Answer continued. . . The force applied to the astronaut by the seat is the sum of the reaction force against gravity plus the additional force needed to produce the acceleration. To the astronaut this force feels just like the reaction force of the seat when the rocket is stationary - only greater. The force experienced feels like that of gravity - the “g-force” - only larger, so the force is called the “g-force”. The astronaut feels him/herself being “pushed into the seat” however, rather than being aware of the net upward force. [Compare this to the effect felt by a person in a car when the car accelerates rapidly.] Therefore, “g-force” is used because it quantitatively relates the effect of acceleration on an astronaut during a rocket launch to the familiar effect, called 1-g, that is produced by the force of gravity when the person is at rest on Earth’s surface. Answer The seat supporting the astronaut provides an upward force on the astronaut that counteracts the force of gravity, even when the astronaut is at rest on the launch pad. During launch a net upward force must act on the astronaut to increase the astronaut’s upward speed. This force is also applied through the seat.
Variation in g forces during launch • G-forces encountered during the launch of a Saturn V rocket were significantly greater than those experienced during a space shuttle launch. • When the rocket accelerates vertically upward at 9.8 m s–2, the astronaut experiences a reaction force of “2G”.
