<p class="MsoNormal"><span style="font-size: 1rem;">In the design of high-speed trains (HSTs), fuel costs are minimized by reducing friction: friction with the air is kept low, while friction with the track is set at a sufficiently high level for safety. The photograph is of the front of a high-speed train. The train is shaped like this to reduce the frictional force (drag) caused by moving through a fluid (air). The resulting low ‘drag coefficient’ reduces the energy loss at high speed and so increases the train’s maximum speed. The rear of the train, too, needs to be the right shape, to minimize air turbulence, which also wastes energy.</span></p><p class="MsoNormal" style="text-align: center; "><img src="https://res.cloudinary.com/drrz8xekm/image/upload/v1571415906/w2gosdzfewscmf9bimye.jpg" data-filename="w2gosdzfewscmf9bimye" style="width: 325.5px;"><span style="font-size: 1rem;"><br></span></p><p class="MsoNormal" style="text-align: center; "><a href="https://pixabay.com/photos/chinese-high-speed-train-model-2825951/" target="_blank"><sup>Chinese High-Speed Train. Pixabay</sup></a><br></p><p class="MsoNormal"><span lang="">The mass of the train depends on the number of passengers, so the applied forces needed to accelerate and brake have to adjust to this. In emergencies, deceleration may have to be quite rapid but still safe, so great care is taken in designing the braking systems. The geography of the track is important: a longer route may save energy if it avoids inefficient changes between kinetic and potential energy, such as when a loaded train goes up and downhills.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">A modern vehicle with low drag travelling at high speed experiences the same kind of force that allows aircraft to fly – aerodynamic lift. Trains and cars are unlikely to take off – but the effect is to reduce the ‘supporting’ force between the ground and the vehicle’s wheels. There must be this supporting force to produce the frictional forces that a train needs in order to accelerate or decelerate. If friction is reduced too much, traction is lost and the driving wheels will simply skid.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">When a train travels round bends at speed it needs a centripetal force to keep it moving in the arc of the circle. This is provided by the contact force of the track against the wheels. It would be better if the centripetal force acted through the centre of mass of the train, which is higher than the track contact. As it doesn’t, there is a tendency for the train to topple over as a result of the turning effect of the weight and the contact force acting at a lower point. The faster the train, the worse the effect – the centripetal force varies as the square of the speed. It helps stability if the centre of mass is low, and this is a major feature for all HSTs. In the French TGV the centre of mass is so low that even when it is derailed the chance of the compartments turning over is very small. No modern trains have ever fallen sideways because they round a bend too quickly.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">It also helps if the track is tilted (banked) at the bend so that the centripetal force acts closer to the centre of mass, reducing the turning effect. But if the track is also to be used for slower trains, passengers will tend to slide ‘downhill’ and start to get worried. It is equally worrying for passengers when HSTs go around a tilted track at high speed – they are forced ‘uphill’ to the outside of the bend. Also, their drinks and plates slide off the table.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">Actually they don’t – they are just stubbornly obeying Newton’s first law of motion, carrying on in a straight line while the table slides away beneath them!<o:p></o:p></span></p><p class="MsoNormal"><span lang="">The current solution to passenger comfort is to use tilting trains. The carriage is made to swing as sensors detect that the track is tilting and an accelerometer measures the inward acceleration. This data is used by a computer to tilt the carriage just enough to keep passengers and coffee cups stable. It works because the contact force between cup and table (and passenger and seat) has a component towards the centre of the curve and so provides, with the help of friction, enough force to keep things stable.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">Tilting trains have been in use in Italy, Spain and Scandinavia for several years. In Britain, Virgin Rail has upgraded the West Coast line between London and Glasgow to take tilting trains with a maximum speed of 225 km/h (140 mph).<o:p></o:p></span></p><p class="MsoNormal" style="text-align: center; "><img src="https://res.cloudinary.com/drrz8xekm/image/upload/v1571417914/bkm9ieoc2p7qedvbspqe.jpg" data-filename="bkm9ieoc2p7qedvbspqe" style="width: 325.5px;"><span style="font-size: 1rem;"><br></span></p><p class="MsoNormal" style="text-align: center; "><a href="https://commons.wikimedia.org/wiki/File:JRS_2000_series_shimanto_2119.JPG" target="_blank"><sup>The JR Shikoku 2000 series DMU negotiating a tight curve on Shikoku's mountainous railway network. Mitsuki-2368, CC BY-SA 3.0</sup></a><span style="font-size: 1rem;"><br></span></p><p class="MsoNormal"><b><span style="font-size: 1rem;">The ideas in this article…..</span><br></b></p><p class="MsoNormal"><span lang="">Newton’s laws of motion are particularly important for an understanding of the topic of transport – and for mechanical engineering in general. The modern world relies on an ability to move materials and people quickly, efficiently and safely from place to place. We shall apply fundamental ideas about mass, force, velocity, acceleration, momentum, impulse and energy not only to the motion of vehicles. We shall also look at the behaviour of fluids (liquids and gases) to understand how ships float and aeroplanes fly. Transport systems have to take account of the fact that vehicles move through a fluid (usually air) and that the fluid is an important restraint on motion.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">But a moving fluid carries kinetic energy and so can be used to do work. Moving water is used to drive turbines in hydroelectric power stations. The energy of moving air is used in wind turbines, a cheap source of energy but with an impact on the visual environment which can be controversial.<o:p></o:p></span></p><h2><span lang="">CONSERVATION OF MOMENTUM AND ENERGY IN TRANSPORT SYSTEMS<o:p></o:p></span></h2><p class="MsoNormal"><span lang="">This chapter will deal with the key ideas of momentum and energy in relation to moving objects. Both total energy and momentum are conserved when engines are used to move vehicles such as cars, trains and rockets – but kinetic energy is not conserved.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">Think about a rocket: the exhaust gases and the rocket itself gain kinetic energy from burning fuel, but a significant fraction of the energy in the fuel mixture is transferred to the random motion of gas particles and does not appear as useful kinetic energy of the rocket vehicle itself. The same applies to road and rail transport vehicles, which gain their energy from burning fuels. Even electric traction engines ultimately depend on the fuels used in power stations to generate electricity.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">Not all the energy transferred by a transport system appears as kinetic energy. Some energy is usually transferred to or gained from gravitational potential energy, for example when aircraft climb and land, and ground vehicles go up and downhills. But what goes up must come down and there is no net change in gravitational potential energy for a vehicle that always returns to its starting level. There are usually some friction losses: for example, energy is lost to the surroundings when vehicles brake (the brake discs get hotter).<o:p></o:p></span></p><p class="MsoNormal"><span lang="">Nevertheless, in all these changes both total energy and momentum are conserved although it is more difficult to keep track of the energy transfers than the momentum changes.<o:p></o:p></span></p><h2><span lang="">TRACTION AND BRAKING<o:p></o:p></span></h2><p class="MsoNormal"><span lang="">The control of a car – in accelerating, turning and braking – relies on its contact with the road surface. The key factor is friction between the tyres and the road surface. The area of contact is quite small as is shown in the diagram below.</span></p><p class="MsoNormal" style="text-align: center; "><img src="https://res.cloudinary.com/drrz8xekm/image/upload/v1571419784/t0r0vizbtzj0thuezbkc.jpg" data-filename="t0r0vizbtzj0thuezbkc" style="width: 325.5px;"><span lang=""><b><o:p><br></o:p></b></span></p><p class="MsoNormal" style="text-align: center; "><a href="https://commons.wikimedia.org/wiki/File:Studless_tire_2.jpg" target="_blank"><sup>Patterned surface of a tyre showing the area in contact with the ground. Lombroso, Public Domain</sup></a><span lang=""><b><o:p><br></o:p></b></span></p><h3><span lang="">EXAMPLE<o:p></o:p></span></h3><p class="MsoNormal"><span lang="">The mass of a given car is 1000 kg. Its tyre pressure is rated as 2.4 </span><span lang="" style="font-family: Candara, sans-serif;">×</span><span lang=""> 10<sup style="font-size: 10.5px;">5</sup> Pa above normal atmospheric pressure: that is, about 3.4 </span><span lang="" style="font-family: Candara, sans-serif;">×</span><span lang=""> 10<sup style="font-size: 10.5px;">5</sup> Pa. What area of the tyre is in contact with the ground?<o:p></o:p></span></p><p class="MsoNormal"><span lang="">Using the definition, pressure = force / area, we get:<o:p></o:p></span></p><p class="MsoNormal"><span lang="">Area of contact = weight of car / Tyre pressure = 1000 x 9.8 N / 3.4 </span><span lang="" style="font-family: Candara, sans-serif;">×</span><span lang=""> 10<sup style="font-size: 10.5px;">5</sup> N m<sup style="font-size: 10.5px;">-2</sup> = 0.029 m<sup style="font-size: 10.5px;">2</sup><o:p></o:p></span></p><p class="MsoNormal"><span lang="">This is a total area of 290 cm<sup style="font-size: 10.5px;">2</sup>. Thus the area of contact per tyre is 72 cm<sup style="font-size: 10.5px;">2</sup> – roughly the area of the palm of your hand.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">As the car wheel turns, the part of the tyre in contact with the ground is instantaneously at rest with reference to the ground. If the car is accelerating or braking, the force that ultimately acts on the car as a whole must be due to the friction between the tyres and the ground. If there were zero friction the wheels would spin freely and the car would skid out of control – as tends to happen on icy roads. In a figure showing the forces acting on an accelerating car. The maximum value of the frictional force F is decided by the nature of the two surfaces in contact and the normal force N between them. Here the word ‘normal’ has the special mathematical meaning of ‘at right angles to the surfaces at the point of contact’. This force is often called the normal reaction. The relationship is simple:<o:p></o:p></span></p><h4 align="center" style="text-align: center;"><span lang="">F = </span><span lang="">μ</span><span lang="">N<o:p></o:p></span></h4><p class="MsoNormal"><span lang="">Where </span><span lang="">μ</span><span lang=""> is a dimensionless number called the coefficient of friction. Its value depends on the nature of the surfaces, and </span><span lang="">μ</span><span lang=""> is usually lower for surfaces sliding against each other than it is for the same surfaces when they are at rest relative to each other. We, therefore, distinguish between the coefficient of static friction between two particular surfaces and their coefficient of sliding friction. For rubber against a paved road, the static coefficient varies between 0.7 and 0.9. The coefficient for sliding friction is between 0.5 and 0.8; on a wet road, this reduces to between 0.25 and 0.7.<o:p></o:p></span></p><h3><span lang="">Sliding down a slope<o:p></o:p></span></h3><p class="MsoNormal">The figure below shows<span lang=""> the forces acting on an object that is sliding steadily down a slope. It is not accelerating, so the component of the object’s weight down the slope must equal the frictional force acting up the slope. </span></p><p class="MsoNormal" style="text-align: center; "><img src="https://res.cloudinary.com/drrz8xekm/image/upload/v1571420337/ctmvbh7zi7acwphxbfl5.png" data-filename="ctmvbh7zi7acwphxbfl5" style="width: 325.5px;"><span lang=""><br></span></p><p class="MsoNormal" style="text-align: center; "><a href="https://commons.wikimedia.org/wiki/File:Free_body1.3.svg" target="_blank"><sup>A body on an inclined plane. Jer S, CC BY-SA 4.0</sup></a><span lang=""><br></span></p><p class="MsoNormal"><span lang="">The frictional force is </span><span lang="">μ</span><span lang="">N where N is the supporting force between the object and the surface of the slope. Thus:<o:p></o:p></span></p><h4 align="center" style="text-align: center;"><span lang="">mg sin </span><span lang="">θ</span><span lang=""> = </span><span lang="">μ</span><span lang="">N<o:p></o:p></span></h4><p class="MsoNormal"><span lang="">But N is also due to the weight of the object. In fact, N = mg cos </span><span lang="">θ</span><span lang="">, so we can write:<o:p></o:p></span></p><h4 align="center" style="text-align: center;"><span lang="">mg sin </span><span lang="">θ </span><span lang="">= </span><span lang="">μ</span><span lang="">mg cos </span><span lang="">θ</span><span lang=""><o:p></o:p></span></h4><p class="MsoNormal"><span lang="">which simplifies to:<o:p></o:p></span></p><h4 align="center" style="text-align: center;"><span lang="">μ = sin θ / cos θ = tan θ<o:p></o:p></span></h4><p class="MsoNormal"><span lang="">This relationship can be used as a simple way of measuring the coefficient of friction between two surfaces. Note that the frictional force F does not depend on the area of the surfaces in contact.<o:p></o:p></span></p><h2><span lang="">FRICTION AND ACCELERATION<o:p></o:p></span></h2><p class="MsoNormal"><span lang="">Over-keen drivers ‘burn rubber’ at traffic lights when they try to accelerate with forces greater than can be provided by static friction between tyres and road. The tyre moves relative to the road surface and work done against sliding friction produces local heating. Similarly you may see skid marks when vehicles brake so strongly that wheels ‘lock’, so that again there is relative movement between road and tyre as the tyres slide along the ground. A car may skid if it is made to turn too sharply, that is, in too small a circle. A central force is needed to make a car move in a circle, and again this must be provided by the force of friction between car and ground. Look back to the start of this post to see why a turning vehicle might topple over.<o:p></o:p></span></p><h3><span lang="">EXAMPLE<o:p></o:p></span></h3><p class="MsoNormal"><span lang="">A typical value for the coefficient of static friction between a particular tyre and a road is 0.7. Estimate the maximum acceleration that a car of mass 1000 kg can have.<o:p></o:p></span></p><h3><span lang="">ANSWER<o:p></o:p></span></h3><h4 align="center" style="text-align: center;"><span lang="">Maximum frictional force F = μN = 0.7 × 1000 × 9.8 = 6.8 kN<o:p></o:p></span></h4><p class="MsoNormal"><span lang="">The acceleration produced by this force is:<o:p></o:p></span></p><p class="MsoNormal"><span lang="">a = F / m = 6.8 × 10<sup style="font-size: 10.5px;">3</sup> / 1000 = 6.8 ms<sup style="font-size: 10.5px;">-2</sup><o:p></o:p></span></p><p class="MsoNormal"><span lang="">But this is a very rough estimate – the two drive wheels share the weight of the car with the other pair of wheels, so the normal reaction N is less than the weight of the car. This means that the maximum possible acceleration is less than that calculated. Motor engineers still find the simple formula F = </span><span lang="">μ</span><span lang="">W useful, however, where W is now the load on the driving axle, which may be measured (with difficulty at speed) or calculated from theory.<o:p></o:p></span></p><h2><span lang="">ROTATION: WHEELS AND ROTATIONAL INERTIA<o:p></o:p></span></h2><p class="MsoNormal"><span lang="">A spinning wheel has both kinetic energy and momentum due to its rotation – even when the wheel is not moving forward. A turning force (or torque) is needed to get a wheel to rotate, and the force does work that appears as kinetic energy. Torque is measured, like the moment of a force, in newton metres (N m). Tests show that it is harder to spin a wheel of large diameter than a wheel of smaller diameter even when both wheels have the same mass. Just as we call the resistance of a mass to being accelerated its inertia (m), so we can also define the resistance of a wheel to being rotated by a turning force as <b>rotational</b> <b>inertia</b> I (also called moment of inertia). The units of rotational inertia are of mass × (distance)<sup style="font-size: 10.5px;">2</sup>, that is kg m<sup style="font-size: 10.5px;">2</sup>.</span></p><p class="MsoNormal"><img src="https://res.cloudinary.com/drrz8xekm/image/upload/v1571422229/vxsjhvjipia5jjbbzup5.jpg" data-filename="vxsjhvjipia5jjbbzup5" style="width: 325.5px; float: left;" class="note-float-left"><div style="text-align: center;"><a href="https://commons.wikimedia.org/wiki/File:%D0%9C%D0%B0%D1%85%D0%BE%D0%B2%D0%B8%D0%BA.jpg" target="_blank"><a href="https://commons.wikimedia.org/wiki/File:%D0%9C%D0%B0%D1%85%D0%BE%D0%B2%D0%B8%D0%BA.jpg" target="_blank" style="background-color: rgb(255, 255, 255); font-size: 1rem;"><sup>Flywheel of a factory steam engine. Витольд Муратов - Self-photographed, CC BY-SA 3.</sup>0</a><br></a></div></p><p class="MsoNormal"><span lang="" style="font-size: 1rem;">Rate of spin is measured by the angle through which a radius line turns per second, that is, its angular velocity </span><span style="font-size: 1rem;">ω</span><span lang="" style="font-size: 1rem;">. Angular velocity is measured in radians per second. Just as velocity is a vector, so is angular velocity. With </span><span lang="" style="font-size: 1rem;">α</span><span lang="" style="font-size: 1rem;"> as angular acceleration, comparing linear and rotational motion:</span><br></p><h4 align="center" style="text-align: center;"><span lang="">Force = mass × acceleration<o:p></o:p></span></h4><h4 align="center" style="text-align: center;"><span lang="">F = ma<o:p></o:p></span></h4><h4 align="center" style="text-align: center;"><span lang="">torque = rotational inertia × angular acceleration<o:p></o:p></span></h4><h4 align="center" style="text-align: center;"><span lang="">T = I</span><span lang="">α</span><span lang=""><o:p></o:p></span></h4><h3><span lang="EL">EXAMPLE<o:p></o:p></span></h3><p class="MsoNormal"><span lang="EL">A torque of 20 Nm acts on a wheel that is initially still and has </span>r<span lang="EL">otational inertia of 600 kg m<sup style="font-size: 10.5px;">2</sup>.<o:p></o:p></span></p><p class="MsoNormal"><span lang="EL">a) What is the angular acceleration produced?<o:p></o:p></span></p><p class="MsoNormal"><span lang="EL">b) At what angular velocity is the wheel spinning after 2 minutes?<o:p></o:p></span></p><p class="MsoNormal"><span lang="EL">c) The wheel is 30 cm in radius. How fast (in ms</span><sup style="font-size: 10.5px;">-1</sup><span lang="EL">) is a point on the rim</span><span lang="EL"> </span><span lang="EL">moving at this time?<o:p></o:p></span></p><h3><span lang="EL">A</span>NSWER<o:p></o:p></h3><p class="MsoNormal">a. <span lang="EL">Angular acceleration </span>= <span lang="EL">torque</span><span lang="EL"> </span><span lang="EL">/</span><span lang="EL"> </span><span lang="EL">rotational inertia:<o:p></o:p></span></p><p class="MsoNormal"><span lang="">α =</span><span lang="EL"> 20/600 </span>= <span lang="EL">0.033 rad s<sup style="font-size: 10.5px;">-2</sup><o:p></o:p></span></p><p class="MsoNormal">b. <span lang="EL">Angular velocity </span>= a<span lang="EL">ngular acceleration </span><span lang="">×</span><span lang="EL"> time<o:p></o:p></span></p><p class="MsoNormal"><span lang="EL">=</span><span lang="EL"> </span><span lang="EL">0.033 </span><span lang="">×</span><span lang="EL"> 120 </span>= <span lang="EL">4 rad s<sup style="font-size: 10.5px;">-1</sup><o:p></o:p></span></p><p class="MsoNormal"><span lang="EL">S</span>in<span lang="EL">ce </span>b<span lang="EL">y the definition of a radian:</span> d<span lang="EL">istance covered per second by a point on the rim<o:p></o:p></span></p><p class="MsoNormal"><span lang="EL">= angle turned through per second </span><span lang="">×</span><span lang="EL"> radius<o:p></o:p></span></p><p class="MsoNormal"><span lang="EL">So the speed of a point on the rim is 4 </span><span lang="">×</span><span lang="EL"> 0.30</span> = <span lang="EL">1.2 ms<sup style="font-size: 10.5px;">-1</sup><o:p></o:p></span></p><h2><span lang="EL">ANGULAR MOMENTUM<o:p></o:p></span></h2><p class="MsoNormal"><span lang="EL">Just as lin</span>e<span lang="EL">ar momentum is defined as mass </span><span lang="">×</span><span lang="EL"> velocity, so we define angular</span> m<span lang="EL">omentumn as:<o:p></o:p></span></p><p class="MsoNormal"><b><span lang="EL">Angular momentum</span></b><b> = </b><b><span lang="EL">Rotational inertia x angular velocity</span></b><b> = I</b><b>ω</b><b><o:p></o:p></b></p><p class="MsoNormal"><span lang="EL">As with linear motion, Newton’s second law applies:<o:p></o:p></span></p><p class="MsoNormal"><span lang="EL">Linear force</span> = <span lang="EL">Rate of change of momentum<o:p></o:p></span></p><h4 align="center" style="text-align: center;">F = <span lang="">Δ (mv) / Δt = Δp / Δt</span><o:p></o:p></h4><p class="MsoNormal" align="center" style="text-align: center;"><b>r</b><b><span lang="EL">otational torque </span></b><b>= </b><b><span lang="EL">rate of change of angular momentum<o:p></o:p></span></b></p><h4 align="center" style="text-align: center;">T = <span lang="">Δ (I</span>ω) / <span lang="">Δt</span><o:p></o:p></h4><h2><span lang="EL">ROTATIONAL INERTIA OF EVERYDAY OBJECTS<o:p></o:p></span></h2><p class="MsoNormal"><span lang="EL">The structure of a wheel may be simplified as shown in </span>the f<span lang="EL">igure below</span><span lang="EL">. The spokes</span> a<span lang="EL">re assumed to have negligible mass compared with the rim. All the mass (</span>m<span lang="EL">)</span> i<span lang="EL">s then at the same distance r from the centre of rotation and the rotational</span> i<span lang="EL">nertia is simply mr</span><sup style="font-size: 10.5px;">2</sup>.<o:p></o:p></p><p class="MsoNormal"><img src="https://res.cloudinary.com/drrz8xekm/image/upload/v1571422679/twgxyxaeu5amzybfs1k9.png" data-filename="twgxyxaeu5amzybfs1k9" style="width: 325.5px; float: left;" class="note-float-left"><div style="text-align: center;"><a href="https://pixabay.com/vectors/wheel-tire-bicycle-round-bike-307316/" target="_blank"><sup><a href="https://pixabay.com/vectors/wheel-tire-bicycle-round-bike-307316/" target="_blank" style="background-color: rgb(255, 255, 255); font-size: 1rem;">A wheel. Pixabay</a><br></sup></a></div></p><p class="MsoNormal"><span lang="EL" style="font-size: 1rem;">In many cases, however, the mass of a spinning wheel is not evenly</span><span lang="EL" style="font-size: 1rem;"> </span><span style="font-size: 1rem;">d</span><span lang="EL" style="font-size: 1rem;">istributed on the rim. The rotational inertia can be calculated if the</span><span style="font-size: 1rem;"> </span><span style="font-size: 1rem;">d</span><span lang="EL" style="font-size: 1rem;">istribution is fairly simple, but may in practice have to be measured</span><span style="font-size: 1rem;"> </span><span style="font-size: 1rem;">e</span><span lang="EL" style="font-size: 1rem;">xperimentally. The value of the rotational inertia of a car wheel is quite</span><span style="font-size: 1rem;"> </span><span style="font-size: 1rem;">d</span><span lang="EL" style="font-size: 1rem;">ifficult to calculate mathematically, for example. As a general rule the</span><span style="font-size: 1rem;"> </span><span style="font-size: 1rem;">m</span><span lang="EL" style="font-size: 1rem;">athematics involves adding up the contributions of all the small masses,</span><span style="font-size: 1rem;"> </span><span style="font-size: 1rem;">t</span><span lang="EL" style="font-size: 1rem;">aking into account the square of their distance from the centre of rotation:</span><br></p><h4 align="center" style="text-align: center;"><span lang="EL">I =</span><span lang="EL"> </span>Σ mr<sup style="font-size: 11.2455px;">2</sup><o:p></o:p></h4><p class="MsoNormal"><span lang="EL">Note that the</span> value of <span lang="EL">rotational inertias of some simple shapes</span><span lang="EL"> </span><span lang="EL">also depends on how the object is spun round an axis.</span><o:p></o:p></p><h2><span lang="EL">ROTATIONAL ENERGY<o:p></o:p></span></h2><p class="MsoNormal"><span lang="EL">The kinetic energy of a rotating body is </span>½Iω<sup style="font-size: 10.5px;"><span lang="EL">2</span></sup><span lang="EL">. This is the equivalent of the</span> l<span lang="EL">inear formula</span><span lang="EL"> </span>½mv<sup style="font-size: 10.5px;"><span lang="EL">2</span></sup>. <span lang="EL">If the wheel has mass m and is also rolling forwards with speed v it will</span> a<span lang="EL">lso have translational kinetic energy </span>½<span lang="EL">mv<sup style="font-size: 10.5px;">2</sup>. The total kinetic energy E</span><sub style="font-size: 10.5px;">k</sub><span lang="EL"> of a</span> r<span lang="EL">olling wheel is therefore:<o:p></o:p></span></p><p class="MsoNormal"><span lang="EL">E</span><sub style="font-size: 10.5px;">k</sub><span lang="EL">= </span>½Iω<sup style="font-size: 10.5px;"><span lang="EL">2</span></sup> + ½mv<sup style="font-size: 10.5px;"><span lang="EL">2</span></sup><o:p></o:p></p><p class="MsoNormal"><span lang="EL">Large vehicles (like some electric locomotives) can reduce the waste of</span> e<span lang="EL">nergy in braking by storing linear kinetic energy as rotational energy in</span> m<span lang="EL">assive rotating wheels (flywheels).</span><span lang="EL"> </span><span lang="EL">Track wheels are linked through a</span> g<span lang="EL">earing system to the flywheel instead of to the usual friction brakes</span>. <span lang="EL">As the track wheels slow down, the flywheel speeds up, and vice versa.<o:p></o:p></span></p><h2><span lang="EL">CONSERVATION OF ANGULAR MOMENTUM<o:p></o:p></span></h2><p class="MsoNormal"><span lang="EL">Hold the axis of a bicycle wheel in your hands and get someone to spin it in</span> t<span lang="EL">he vertical plane. Then try to tip the wheel through a small angle away from</span> t<span lang="EL">he vertical. You will experience a mysterious force that seems to oppose the</span> c<span lang="EL">hange in direction of the axis. In fact, this is a consequence of the</span> c<span lang="EL">onservation of angular momentum </span><span lang="">– </span>a<span lang="EL">nd Newton’s third law of motion.</span><span lang="EL"> </span><span lang="EL">Newton’s third law states that,</span> w<span lang="EL">hen two bodies interact, the</span> f<span lang="EL">orce exerted by one body</span><span lang="EL"> </span><span lang="EL">(here, the holder tipping the wheel)</span> b<span lang="EL">rings about an equal and</span> o<span lang="EL">pposite force exerted by the</span> o<span lang="EL">ther body (the wheel).</span> So, c<span lang="EL">hanging the direction of the axis of spin means altering the angular</span> m<span lang="EL">omentum, not in size but in direction. This requires a force, and as forces</span> o<span lang="EL">ccur in pairs there is an equal and opposite force exerted on you as the</span> w<span lang="EL">heel changes direction. This ‘resistance’ of a spinning wheel to changing its</span> d<span lang="EL">irection of action is what makes a bicycle so stable – as long as the wheels</span> a<span lang="EL">re spinning! </span>Conservation <span lang="EL">of angular momentum </span>is also of<span lang="EL"> importance to</span> t<span lang="EL">he origin and nature of the Solar System</span>.<span lang="EL"><o:p></o:p></span></p><p></p><h2>REFERENCES</h2><p><a href="https://www.explainthatstuff.com/motion.html" target="_blank">https://www.explainthatstuff.com/motion.html</a><br></p><p><a href="https://www.khanacademy.org/computing/computer-programming/programming-natural-simulations/programming-forces/a/newtons-laws-of-motion" target="_blank">https://www.khanacademy.org/computing/computer-programming/programming-natural-simulations/programming-forces/a/newtons-laws-of-motion</a><br></p><p><a href="https://onlinelibrary.wiley.com/doi/abs/10.1002/pssa.2211450215" target="_blank">https://onlinelibrary.wiley.com/doi/abs/10.1002/pssa.2211450215</a><br></p><p><a href="https://en.wikipedia.org/wiki/Momentum" target="_blank">https://en.wikipedia.org/wiki/Momentum</a><br></p><p><a href="https://en.wikipedia.org/wiki/Transport_phenomena" target="_blank">https://en.wikipedia.org/wiki/Transport_phenomena</a><br></p><p><a href="http://www.sjsu.edu/people/burford.furman/docs/me190/Review%20of%20traction%20and%20braking%20control%20for%20electric%20vehicle.pdf" target="_blank">Traction and braking control system</a><br></p><p><a href="https://en.wikipedia.org/wiki/Traction_control_system" target="_blank">https://en.wikipedia.org/wiki/Traction_control_system</a><br></p><p><a href="http://dynref.engr.illinois.edu/ava.html" target="_blank">http://dynref.engr.illinois.edu/ava.html</a><br></p><p><a href="https://physics.stackexchange.com/questions/56472/finding-the-force-of-friction-of-a-moving-object-and-its-change-when-it-accelera" target="_blank">https://physics.stackexchange.com/questions/56472/finding-the-force-of-friction-of-a-moving-object-and-its-change-when-it-accelera</a><br></p><p><a href="https://physics.info/friction/practice.shtml" target="_blank">https://physics.info/friction/practice.shtml</a><br></p><p><a href="https://sciencing.com/calculate-acceleration-friction-6245754.html" target="_blank">https://sciencing.com/calculate-acceleration-friction-6245754.html</a><br></p><p><a href="http://hpwizard.com/rotational-inertia.html" target="_blank">http://hpwizard.com/rotational-inertia.html</a><br></p><p><a href="https://opentextbc.ca/physicstestbook2/chapter/dynamics-of-rotational-motion-rotational-inertia/" target="_blank">https://opentextbc.ca/physicstestbook2/chapter/dynamics-of-rotational-motion-rotational-inertia/</a><br></p><p><a href="http://homepage.physics.uiowa.edu/~rmerlino/6Spring11/29006_L11.pdf" target="_blank">http://homepage.physics.uiowa.edu/~rmerlino/6Spring11/29006_L11.pdf</a><br></p><p><a href="https://en.wikipedia.org/wiki/List_of_moments_of_inertia" target="_blank">https://en.wikipedia.org/wiki/List_of_moments_of_inertia</a><br></p><p><a href="https://www.livephysics.com/physical-constants/mechanics-pc/moment-inertia-uniform-objects/" target="_blank">https://www.livephysics.com/physical-constants/mechanics-pc/moment-inertia-uniform-objects/</a><br></p><p><a href="https://www.thoughtco.com/moment-of-inertia-2699260">https://www.thoughtco.com/moment-of-inertia-2699260</a><a href="https://www.thoughtco.com/moment-of-inertia-2699260" target="_blank"></a></p><p><a href="https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/Map%3A_University_Physics_I_-_Mechanics%2C_Sound%2C_Oscillations%2C_and_Waves_(OpenStax)/10%3A_Fixed-Axis_Rotation__Introduction/10.4%3A_Moment_of_Inertia_and_Rotational_Kinetic_Energy" target="_blank">Moment of Inertia and Rotational Kinetic Energy</a><br></p><p><a href="https://courses.lumenlearning.com/boundless-physics/chapter/rotational-kinetic-energy/" target="_blank">https://courses.lumenlearning.com/boundless-physics/chapter/rotational-kinetic-energy/</a><br></p><p><br></p>