PHSY 1110 Topic 3 ebooklet PDF

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UNSW 2021 - 2022 term 2 PHSY 1110 week 3' s lecture notes...

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PHYS1110 Everyday Physics Topic 3 How does a hot air balloon work

1. Introducti on Have you ever wondered about how a hot air balloon works? Well in this topic you are going to learn all about it. We will start by considering gravity, since the hot air balloon must overcome gravity in order to lift off. We will also have a quick look at air resistance which happens in the atmosphere. If we dropped an object over the side of a hot air balloon it would experience some air resistance on the way down, this is why sky divers reach what is known as a terminal velocity. For a good understanding of hot air balloons, we will need to understand a little about forces. I will introduce you to Newton’s first and second law. These help us relate the forces acting on an object to the movement of the object. Another important force in a hot air balloon is the buoyancy force, this is the upwards force that is greater than the weight force as the hot air balloon takes off. The buoyancy force is due to hot air inside the balloon having a lower density than the cold air outside. We will also examine Archimedes Principle which was a very exciting finding and according to legend was the first eureka moment in science. In order to understand how the density of air inside the balloon changes we will be learning about the ideal gas law. The ideal gas law tells us what happens to gas as its temperature changes and provides an explanation for why hot air has a lower density than cold air. I hope you enjoy this topic as much as I do! Supplementary resources You may want to watch this very interesting Todd Sampson documentary where he makes use of buoyancy to take a trip using Helium balloons. https://ap01a.alma.exlibrisgroup.com/leganto/readinglist/citation/20784815430001731?institute=61UNSW _INST&auth=LOCAL The ideas in this topic are covered in sections 2.6, 11.6, 14.1 and 14.2 of Cutnell and Johnson.

2. Gravity, weight and Ai r resistance In this section, we're going to be looking at gravity, weight, and air resistance. All objects on Earth experience gravity. Gravity is what makes objects fall when we let go of them. Gravity is a force. Later, we will more closely examine what a force is exactly, but for now, a force is a push or a pull. Gravity pulls all objects down towards the centre of the Earth. We have a formula to describe the gravity felt by objects on the surface of the Earth. Because gravity is a force, we write this equation as F is equal to the mass of the object (m), multiplied by the acceleration it feels due to gravity (g). Acceleration describes how quickly the speed or velocity of an object changes with time. When I drop an object, not only does it start to fall towards the Earth, but it falls at an ever-increasing rate. It falls faster and faster as it gets closer to the floor. The longer the time it has been falling for, the faster it is going.

𝐹 = 𝑚𝑔 Newton came up with the Law of Universal Gravitation. Newton said that it is not just objects that are attracted to the Earth. Any two objects in the universe are attracted to each other, and this is caused by the mass of these objects. Newton wrote down the Law of Universal Gravitation which can be written as F is equal to the gravitational constant (G) which is equal to 6.673 x 10-11 Nm2kg-2, multiplied by the mass of the first object (m1), multiplied by the mass of the second object (m2), divided by the square of the distance between the objects (r 2).

𝐹=

𝐺𝑚& 𝑚'  , where 𝐺 = 6.673 × 106&&777 Nm' kg 6' 𝑟'

Let's have a look now at how we can use Newton's Law of Universal Gravitation, along with our first simple equation, F = mg to calculate the acceleration of objects on the surface of the Earth. The acceleration of objects on the surface of the Earth is given the symbol g. We know that 𝐹 = 𝑚𝑔 which describes the force felt by an object dropped on the surface of the Earth, and F =

=>? >@ A@

is the force felt due to gravitation of any two objects which are attracting

each other. When we drop an object on the surface of the Earth, mass one ( 𝑚& ) is the Earth and mass two (𝑚') is the object. We can equate these two forces. They are both describing the same thing, i.e. the force felt by an object when it is dropped on the surface of the Earth. The mass of the object ( 𝑚B ) is equal to the gravitational constant (G) multiplied by the mass of the Earth (𝑚C ), multiplied by the mass of the object ( 𝑚B ), divided by 𝑟 ' :

𝑚D g =

𝐺𝑚C 7𝑚D 7 𝑟'

Let’s look at this diagram of Earth (below). We can imagine that all that mass of the Earth is concentrated at the very centre of the Earth, and our object is being dropped just above the surface of the Earth. The distance between the centre of the Earth and where we're dropping the object from is the radius of the Earth. When looking at our two equations, you can see that the mass of the object appears on both sides, so we can cancel that out. Now we have an expression for g which is the acceleration due to gravity on the surface of the Earth. g is equal to G multiplied by 𝑚C divided by 𝑟 ' . G (universal gravitational constant) = 6.673 x 10-11; 𝑚C (mass of the Earth) = 5.98 x 1024 kg; 𝑟C ' (radius of the Earth) = (6.37 x 106)2 m. Solving on the calculator, we get 9.83 ms-2. When we measure the acceleration due to gravity on the surface of the Earth, we get 9.8, so this is in agreement.

This formula lets us calculate the acceleration of any object (g) on the surface of the Earth, but we can use it on any planet or on the Moon. If this was Mars, we could put the mass of the Mars and the radius of the Mars. To get the weight force of the object, we can simply use this formula with the g for the specific planet that we are considering.

Previously we have mentioned that 𝐹 = 𝑚𝑔 and g is fixed on the surface of the Earth, which is saying that all objects should accelerate at the same rate on the surface of the Earth. This is true for relatively heavy objects. For example, if I drop these two balls which are not identical (one is a rubber ball and one is a Happy Sack) from the same height, they should hit the floor at about the same time. However, what happens when I drop a heavy object, and something that is really light, like a leaf? They will not hit the floor at the same time. So, what's going wrong with the physics in this case? The answer is air resistance. The air around us is made up of little particles which try to stop the motion of the objects through them. Air resistance provides another force, and this force works to oppose the pull of the gravity. Air resistance is more significant the lighter an object is. Air resistance affects the leaf a lot because the leaf has a large surface area and is very light. On the other hand, the ball has a slightly smaller surface area and more mass. The gravitational force is larger for the ball, because the gravitational force is given by mg. The gravitational force is smaller for the leaf, and so the air resistance is more significant for the leaf. This is because it has more surface area, and it has less gravitational force to overcome. For this reason, these objects do not hit the floor together. But what would happen if we could get rid of the air? What would happen if we dropped two objects on the Moon, where there is no atmosphere? On the Moon, the leaf and the ball would drop at the same rate, and we will prove this in the next demonstration. In the demonstration we will use a Perspex tube from which we can remove all the air. At the bottom of the tube, we have a cupcake patty case and a ceramic ball. The cupcake patty case is very light and has a large surface area. The ceramic ball is heavy and has a small surface area. The air is removed by attaching a vacuum to the top of the tube. Let's look at what happens to this cupcake patty case and the ball in a vacuum when we turn the tube the other way up. The cupcake patty case and the ceramic ball hit the bottom of the tube at almost the same time. Therefore, in a vacuum, the two objects fall at approximately the same rate. After letting air in the tube, the cupcake patty case falls a lot slower than the ball due to

air resistance. Around us, we have air resistance, thus 𝐹 = 𝑚𝑔 is correct, but it is only an approximation. Objects with a large surface area that do not have much mass are affected by air resistance and a hot air balloon needs to overcome this gravitational force to rise into the sky. One interesting thing which happens with air resistance is that everything feels a weight force downwards. The size of the air resistance force depends on the velocity of the object which is falling. For example, a cat that is falling from a high-rise apartment (it accidently fell, it wasn't pushed) feels an air resistance force upwards. When it starts going fast enough, this air resistance force is equal to the weight force. At that point when the air resistance is equal to the weight force, it stops accelerating, which means that this cat starts to fall at a constant velocity. This velocity is called the terminal velocity of the cat, and no matter how long the cat falls for, its velocity will not increase. A study of all the cats that had fallen out of their apartments in New York found that the highest mortality rate was for the cats which had fallen from the 7th floor. The belief was that cats will reach their terminal velocity when they have fallen seven floors. If a cat falls from higher than that, then it falls for a longer time, and more time to relax in the air and get into a suitable position to land. Cats have survived falls from the 45th floor, but the 7th floor is the most dangerous since they reach their terminal velocity and have less time to prepare to hit the ground. Velocity is another word for speed, except that velocity has a direction. What we can draw is a speed vs time graph for the falling cat. Initially, it has a lot of acceleration, but the air resistance force increases as the speed increases, and so the speed of the cat levels off at the terminal velocity:

Equations Since the acceleration due to gravity is represented by 𝑔, for an object dropped on Earth,

𝐹 = 𝑚𝑔! 𝐹=

𝐺𝑚& 𝑚'  , where 𝐺 = 6.673 × 106&&777 Nm' kg 6' 𝑟'

𝑚D g =

𝐺𝑚C 7𝑚 D 7 𝑟'

Supplementary resources If you would like to find out more about gravity, I recommend looking at Joe Wolfe's PHYSCLIPS: Weight and Contact forces: http://www.animations.physics.unsw.edu.au/mechanics/chapter6_weightandcontactforces. html Here is a Mythbusters episode where they look at terminal velocity. They use American units (pounds and feet) rather than SI ones but it is worth watching if you have time. The whole Mythbusters team tackle one massive myth tonight, the "Seesaw Saga". Could a skydiver whose parachute failed to open hit a playground seesaw and send a small girl flying seven stories high? And would she survive? http://er.library.unsw.edu.au/er/cgibin/eraccess.cgi?url=http://search.informit.com.au/documentSummary;dn=TEX20093504182; res=TVNEWS Section 2.6 and 4.7-4.8 of Cutnell and Johnson covers freely falling bodies and the gravitational force. 3. Newtow n’s first and Second Laws Newton’ s fi rst law In this section we're going to be looking at forces and Newton's laws of motion. Previously, we stated that a force is a push or a pull. Here we will look at forces in a lot more detail. You will be introduced to Newton's laws, which describe forces, and you will see that it is vital that we have a force in order to start a car moving. Newton's First Law of motion states that an object continues in a state of rest or in a state of motion at a constant velocity unless compelled to change that state by a net force. Another way of stating this is that an object at rest remains at rest, and an object in motion remains in exactly that state of motion unless acted upon by a net force. A net force is the sum of all the forces acting on the body. Forces are vectors. When we sum vectors, we sum them vectorially, which means summing them up head to tail. Let's have a look at an example now of how we could calculate the net force acting on an object. Below is a picture of a force board. On this force board, there is a central brass plate with a mass of about 100 g. A 150 g weight is attached to the central brass plate, so these two combined have a weight force going downwards. Around the edge, each string has 250 g of mass attached to it. We can represent these weight forces by drawing lines parallel to our strings on the picture (white arrows). To remember in which direction those forces were acting, I have drawn arrowheads. We have drawn all the forces which are acting in this picture as vectors.

If we take these arrows as vectors and connect them, it would form a perfect pentagon. This would indicate that there is no resultant force. As the central mass is stationary, and therefore at rest, there should be no net force acting on it. This body is what we call in equilibrium i.e. there is no net force acting on it. What we see is that an object at rest remains at rest and an object in motion will remain in motion unless acted upon by an outside force. What happens when a book is pushed? If I push this book (see below), I'm applying a force on it, so that will start it moving. But why does it come to a stop? If you said because there are frictional forces acting on it to slow it down, then you're correct. If there was no friction, the book would keep moving forever. In space, if you push an object, it keeps travelling as there is no friction or air resistance. Newton's First Law leads us to the concept of inertia -- a body at rest wants to remain at rest. Inertia means that a body wants to stay where it is. To start moving a body, you need to apply a force. The inertia of a body is related to its mass. The larger a body, the larger the force needed to move it. We will see why that is with Newton's Second Law, but for now, let's have a look at a couple of demonstrations showing the concept of inertia.

In the first demonstration of inertia, I have a table and a tablecloth. On top is a glass bottle of water and a glass of water. I will try to remove the tablecloth from below the glass and the bottle. If I do this fast enough, the inertia of the glass and the bottle should keep them stationary where they are.

In the second demonstration of inertia, I have a glass with a plate on top, and an egg in the plate. If I remove the plate quickly, the inertia of the egg should keep it where it is. Therefore, once the plate is gone, the egg will experience a gravitational force and should drop straight into the glass. Let’s give it a go. After quickly hitting the plate away, the egg landed in the glass. As soon as the plate disappeared, the egg was left where it was, and it dropped down directly into the glass.

In this section we have introduced Newton's First Law which tells us that an object at rest will remain at rest and an object in motion will remain in a state of constant motion unless acted upon by a net force. In terms of what makes a car go, this tells us that in order to get the car moving, the engine needs to provide enough force on the car. You would need to calculate the net force and consider any other forces acting on the car. In order to start the car moving, the engine would need to provide a force forwards which was larger than any forces pushing the car backwards. Newton’ s secon d Law Newton’s second law is a very important law in physics: “The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to the mass”. In equation form, the acceleration of an object (a) is equal to the sum of the forces acting on it (net force) (ΣF), divided by the mass (m):

𝑎=

Σ𝐹 𝑚

Often, Newton's second law is written as the net force acting on an object, so the sum of all the forces is equal to the mass times the acceleration of the object:

Σ𝐹 = 𝑚𝑎 In the example with the masses hanging on the board, all the forces acting on the central mass added up to give zero. Thus, there was no net force. Newton's second law tells us that if there's no net force acting on a mass, then the acceleration of that mass is zero. Which is indeed what we saw. Let's have a look at a couple of demonstrations of Newton's second law in action. I have a heavy mass and a hammer (below). If I was to whack my hand with the hammer, it would hurt a lot because the force from the hammer would all be imparted into my hand. However, if I put the heavy mass on my hand and whack it with the hammer, my hand wouldn’t hurt because the force of the hammer is being distributed over a much larger mass. The acceleration of the large mass acting on my hand is a lot smaller, thus the total force applied on my hand is smaller.

Here I have a trolley (image below). It has a mechanism so that when I push down the lever, the component at the end shoots out. This will be up against the wall and will cause a force pushing the trolley along the ground. Let’s try this with a different amount of mass placed on the trolley (each mass is 1 kg). The one with the most mass accelerates much less than the one with less mass.

Scenario: A stalled car has a mass of 1,100 kg. Two people push the car, one with a force of 3,050 N and the other with a force of 4,200 N. Both forces are directed in the same direction. A frictional force of 6,100 N opposes the motion of the car. Part A) Calculate the net force on the car.

Part B) Calculate the acceleration of the car. Part A) As usual, let’s start by drawing a diagram (below). To calculate the net force, we need to sum all the forces together. We can draw them head to tail (as a vector). We have 3,050 N and a slightly longer one for 4,200 N, and then one going in the opposite direction of 6,100 N. To work out the size of this resultant force, we have 3,050 + 4,200 - 6,100 = 1,150 N. Remember that the frictional force of 6,100 N is in the opposite direction and so it subtracts. We need a direction for the force, and this is in the same direction as the people are pushing. Part B) To calculate the acceleration of the car we will use Newton's second law. The acceleration is given by the net force divided by the mass: 𝑎 =

GH >

We've calculated the net force as 1,150 N, and we’re told that the mass of the car is 1,100 kg. Therefore, the acceleration is 1.05 ms-2. Remember that since acceleration is a vector, we need to provide a direction. The direction of acceleration is the same as the direction of the net force, so it is the same direction as the people pushing the car.

That was a relatively straight forward example, where we only had to worry about the horizontal components of the forces. Let's have a look at a more complicated example in which we will be introducing the normal reaction force. A rope is used to pull along a block as shown in the diagram below. A force of 30 N is applied at an angle of 30" to the horizontal, and the block has a mass of 2 kg. Part A) Will the block be lifted off the surface in this case? Part B) What is the net force acting on the block? Part C) What is the acceleration of the block? Part A) We need to consider all the forces that are acting in the vertical direction. For a block like this, the weight force is acting downwards. The normal reaction force from the table is

acting upwards. In this case we've got an additional force (we'll call it force y) which is the vertical component of this applied force. We can use SOH-CAH-TOA to work out the size of force y. Force y is opposite the 30° and we have the hypotenuse so we will be using sin 30 = opposite (force y) divided by the hypotenuse (30). Therefore, force y = 30 x sin 30 =15 N. We have an upwards force of 15 N and a downwards force equal to mass multiplied by gravitational force. The mass is 2 kg and g is 9.8, so this equals 19.6 N. Our weight force is larger than our upwards force, so that tells us that it is going to remain on the table. The normal reaction force will be the size of the d...