Torque: Kinetic Energy Essay

WHAT IS TORQUE? Torque is a measure of how much a force acting on an object causes that object to rotate. The object rotates about an axis, which we will call the pivot point, and will label ‘O’. We will call the force ‘F’. The distance from the pivot point to the point where the force acts is called the moment arm, and is denoted by ‘r’. Note that this distance, ‘r’, is also a vector, and points from the axis of rotation to the point where the force acts. (Refer to Figure 1 for a pictoral representation of these definitions. ) | Figure 1 Definitions| Torque is defined as r x F = r F sin(). In other words, torque is the cross product between the distance vector (the distance from the pivot point to the point where force is applied) and the force vector, ‘a’ being the angle between r and F. Using the right hand rule, we can find the direction of the torque vector. If we put our fingers in the direction of r, and curl them to the direction of F, then the thumb points in the direction of the torque vector. Imagine pushing a door to open it. The force of your push (F) causes the door to rotate about its hinges (the pivot point, O).

How hard you need to push depends on the distance you are from the hinges (r) (and several other things, but let’s ignore them now). The closer you are to the hinges (i. e. the smaller r is), the harder it is to push. This is what happens when you try to push open a door on the wrong side. The torque you created on the door is smaller than it would have been had you pushed the correct side (away from its hinges). Note that the force applied, F, and the moment arm, r, are independent of the object. Furthermore, a force applied at the pivot point will cause no torque since the moment arm would be zero (r = 0).

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Another way of expressing the above equation is that torque is the product of the magnitude of the force and the perpendicular distance from the force to the axis of rotation (i. e. the pivot point). Let the force acting on an object be broken up into its tangential (Ftan) and radial (Frad) components (see Figure 2). (Note that the tangential component is perpendicular to the moment arm, while the radial component is parallel to the moment arm. ) The radial component of the force has no contribution to the torque because it passes through the pivot point.

So, it is only the tangential component of the force which affects torque (since it is perpendicular to the line between the point of action of the force and the pivot point). | Figure 2 Tangential and radial components of force F| There may be more than one force acting on an object, and each of these forces may act on different point on the object. Then, each force will cause a torque. The net torque is the sum of the individual torques. Rotational Equilibrium is analogous to translational equilibrium, where the sum of the forces are equal to zero. In rotational equilibrium, the sum of the torques is equal to zero.

In other words, there is no net torque on the object. * Note that the SI units of torque is a Newton-metre, which is also a way of expressing a Joule (the unit for energy). However, torque is not energy. So, to avoid confusion, we will use the units N. m, and not J. The distinction arises because energy is a scalar quanitity, whereas torque is a vector. Torque If a net force is applied to an object’s center of mass, it will not cause the object to rotate. However, if a net force is applied to a point other than the center of mass, it will affect the object’s rotation. Physicists call the effect of force on rotational motion torque.

Torque Defined Consider a lever mounted on a wall so that the lever is free to move around an axis of rotation O. In order to lift the lever, you apply a force F to point P, which is a distance r away from the axis of rotation, as illustrated below. Suppose the lever is very heavy and resists your efforts to lift it. If you want to put all you can into lifting this lever, what should you do? Simple intuition would suggest, first of all, that you should lift with all your strength. Second, you should grab onto the end of the lever, and not a point near its axis of rotation.

Third, you should lift in a direction that is perpendicular to the lever: if you pull very hard away from the wall or push very hard toward the wall, the lever won’t rotate at all. Let’s summarize. In order to maximize torque, you need to: 1. Maximize the magnitude of the force, F, that you apply to the lever. 2. Maximize the distance, r, from the axis of rotation of the point on the lever to which you apply the force. 3. Apply the force in a direction perpendicular to the lever. We can apply these three requirements to an equation for torque, :

In this equation, is the angle made between the vector for the applied force and the lever. Torque Defined in Terms of Perpendicular Components There’s another way of thinking about torque that may be a bit more intuitive than the definition provided above. Torque is the product of the distance of the applied force from the axis of rotation and the component of the applied force that is perpendicular to the lever arm. Or, alternatively, torque is the product of the applied force and the component of the length of the lever arm that runs perpendicular to the applied force.

We can express these relations mathematically as follows: where and are defined below. Torque Defined as a Vector Quantity Torque, like angular velocity and angular acceleration, is a vector quantity. Most precisely, it is the cross product of the displacement vector, r, from the axis of rotation to the point where the force is applied, and the vector for the applied force, F. To determine the direction of the torque vector, use the right-hand rule, curling your fingers around from the r vector over to the F vector.

In the example of lifting the lever, the torque would be represented by a vector at O pointing out of the page. Example | | | A student exerts a force of 50 N on a lever at a distance 0. 4 m from its axis of rotation. The student pulls at an angle that is 60A? above the lever arm. What is the torque experienced by the lever arm? | | Let’s plug these values into the first equation we saw for torque: This vector has its tail at the axis of rotation, and, according to the right-hand rule, points out of the page. Energy Basics: | Issac Says: “Let’s Learn about Potential and Kinetic Energy! | Potential Energy: Potential energy exists whenever an object which has mass has a position within a force field. The most everyday example of this is the position of objects in the earth’s gravitational field. The potential energy of an object in this case is given by the relation: PE = mgh where * PE = Energy (in Joules) * m = mass (in kilograms) * g = gravitational acceleration of the earth (9. 8 m/sec2) * h = height above earth’s surface (in meters) Kinetic Energy: Kinetic Energy exists whenever an object which has mass is in motion with some velocity.

Everything you see moving about has kinetic energy. The kinetic energy of an object in this case is given by the relation: KE = (1/2)mv2 where * KE = Energy (in Joules) * m = mass (in kilograms) * v = velocity (in meters/sec) Conservation of Energy This principle asserts that in a closed system energy is conserved. This principle will be tested by you, using the experimental apparatus below. In the case of an object in free fall. When the object is at rest at some height, h, then all of its energy is PE. As the object falls and accelerates due to the earth’s gravity, PE is converted into KE.

When the object strikes the ground, h=0 so that PE=0, the all of the energy has to be in the form of KE and the object is moving it at its maximum velocity. (In this case we are ignoring air resistance). These concepts are merely an outgrowth of Newton’s second law as discussed in an earlier unit. Newton’s second law (Fnet = m • a) stated that the acceleration of an object is directly proportional to the net force acting upon the object and inversely proportional to the mass of the object. When combined with the definition of acceleration (a = change in velocity / time), the following equalities result.

If both sides of the above equation are multiplied by the quantity t, a new equation results. This equation represents one of two primary principles to be used in the analysis of collisions during this unit. To truly understand the equation, it is important to understand its meaning in words. In words, it could be said that the force times the time equals the mass times the change in velocity. In physics, the quantity Force • time is known as impulse. And since the quantity m•v is the momentum, the quantity m•v must be the change in momentum. The equation really says that the Impulse = Change in momentum

One focus of this unit is to understand the physics of collisions. The physics of collisions are governed by the laws of momentum; and the first law which we discuss in this unit is expressed in the above equation. The equation is known as the impulse-momentum change equation. The law can be expressed this way: In a collision, an object experiences a force for a specific amount of time which results in a change in momentum. The result of the force acting for the given amount of time is that the object’s mass either speeds up or slows down (or changes direction).

The impulse experienced by the object equals the change in momentum of the object. In equation form, F • t = m • v. What Is Linear Momentum? Linear momentum is a vector quantity defined as the product of an object’s mass, m, and its velocity, v. Linear momentum is denoted by the letter p and is called “momentum” for short: Note that a body’s momentum is always in the same direction as its velocity vector. The units of momentum are kg · m/s. Fortunately, the way that we use the word momentum in everyday life is consistent with the definition of momentum in physics.

For example, we say that a BMW driving 20 miles per hour has less momentum than the same car speeding on the highway at 80 miles per hour. Additionally, we know that if a large truck and a BMW travel at the same speed on a highway, the truck has a greater momentum than the BMW, because the truck has greater mass. Our everyday usage reflects the definition given above, that momentum is proportional to mass and velocity. Linear Momentum and Newton’s Second Law In Chapter 3, we introduced Newton’s Second Law as F = ma. However, since acceleration can be expressed as , we could equally well express Newton’s Second Law as F = .

Substituting p for mv, we find an expression of Newton’s Second Law in terms of momentum: In fact, this is the form in which Newton first expressed his Second Law. It is more flexible than F = ma because it can be used to analyze systems where not just the velocity, but also the mass of a body changes, as in the case of a rocket burning fuel. Impulse The above version of Newton’s Second Law can be rearranged to define the impulse, J, delivered by a constant force, F. Impulse is a vector quantity defined as the product of the force acting on a body and the time interval during which the force is exerted.

If the force changes during the time interval, F is the average net force over that time interval. The impulse caused by a force during a specific time interval is equal to the body’s change of momentum during that time interval: impulse, effectively, is a measure of change in momentum. The unit of impulse is the same as the unit of momentum, kg · m/s. Example | | | A soccer player kicks a 0. 1 kg ball that is initially at rest so that it moves with a velocity of 20 m/s. What is the impulse the player imparts to the ball? If the player’s foot was in contact with the ball for 0. 1 s, what was the force exerted by the player’s foot on the ball? | | What is the impulse the player imparts to the ball? Since impulse is simply the change in momentum, we need to calculate the difference between the ball’s initial momentum and its final momentum. Since the ball begins at rest, its initial velocity, and hence its initial momentum, is zero. Its final momentum is: Because the initial momentum is zero, the ball’s change in momentum, and hence its impulse, is 2 kg · m/s. What was the force exerted by the player’s foot on the ball?

Impulse is the product of the force exerted and the time interval over which it was exerted. It follows, then, that . Since we have already calculated the impulse and have been given the time interval, this is an easy calculation: Impulse and Graphs SAT II Physics may also present you with a force vs. time graph, and ask you to calculate the impulse. There is a single, simple rule to bear in mind for calculating the impulse in force vs. time graphs: The impulse caused by a force during a specific time interval is equal to the area underneath the force vs. ime graph during the same interval. If you recall, whenever you are asked to calculate the quantity that comes from multiplying the units measured by the y-axis with the units measured by the x-axis, you do so by calculating the area under the graph for the relevant interval. Example | | | What is the impulse delivered by the force graphed in the figure above between t = 0 and t = 5? | | The impulse over this time period equals the area of a triangle of height 4 and base 4 plus the area of a rectangle of height 4 and width 1. A quick calculation shows us that the impulse is:

Work When we are told that a person pushes on an object with a certain force, we only know how hard the person pushes: we don’t know what the pushing accomplishes. Work, W, a scalar quantity that measures the product of the force exerted on an object and the resulting displacement of that object, is a measure of what an applied force accomplishes. The harder you push an object, and the farther that object travels, the more work you have done. In general, we say that work is done by a force, or by the object or person exerting the force, on the object on which the force is acting.

Most simply, work is the product of force times displacement. However, as you may have remarked, both force and displacement are vector quantities, and so the direction of these vectors comes into play when calculating the work done by a given force. Work is measured in units of joules (J), where 1 J = 1 N · m = 1 kg · m2/s2. Energy Energy is one of the central concepts of physics, and one of the most difficult to define. One of the reasons we have such a hard time defining it is because it appears in so many different forms.

There is the kinetic and potential energy of kinematic motion, the thermal energy of heat reactions, the chemical energy of your discman batteries, the mechanical energy of a machine, the elastic energy that helps you launch rubber bands, the electrical energy that keeps most appliances on this planet running, and even mass energy, the strange phenomenon that Einstein discovered and that has been put to such devastating effect in the atomic bomb. This is only a cursory list: energy takes on an even wider variety of forms. Energy, like work, is measured in joules (J).

In fact, work is a measure of the transfer of energy. However, there are forms of energy that do not involve work. For instance, a box suspended from a string is doing no work, but it has gravitational potential energy that will turn into work as soon as the string is cut. We will look at some of the many forms of energy shortly. First, let’s examine the important law of conservation of energy. Conservation of Energy As the name suggests, the law of conservation of energy tells us that the energy in the universe is constant. Energy cannot be made or destroyed, only changed from one form to another form.

Energy can also be transferred via a force, or as heat. For instance, let’s return to the example mentioned earlier of the box hanging by a string. As it hangs motionless, it has gravitational potential energy, a kind of latent energy. When we cut the string, that energy is converted into kinetic energy, or work, as the force of gravity acts to pull the box downward. When the box hits the ground, that kinetic energy does not simply disappear. Rather, it is converted into sound and heat energy: the box makes a loud thud and the impact between the ground and the box generates a bit of heat. Forms of Energy

Though energy is always measured in joules, and though it can always be defined as a capacity to do work, energy manifests itself in a variety of different forms. These various forms pop up all over SAT II Physics, and we will look at some additional forms of energy when we discuss electromagnetism, relativity, and a number of other specialized topics. For now, we will focus on the kinds of energy you’ll find in mechanics problems. Kinetic Energy Kinetic energy is the energy a body in motion has by virtue of its motion. We define energy as the capacity to do work, and a body in motion is able to use its motion to do work.

For instance, a cue ball on a pool table can use its motion to do work on the eight ball. When the cue ball strikes the eight ball, the cue ball comes to a stop and the eight ball starts moving. This occurs because the cue ball’s kinetic energy has been transferred to the eight ball. There are many types of kinetic energy, including vibrational, translational, and rotational. Translational kinetic energy, the main type, is the energy of a particle moving in space and is defined in terms of the particle’s mass, m, and velocity, v: For instance, a cue ball of mass 0. kg moving at a velocity of 2 m/s has a kinetic energy of 1/2 (0. 5 kg)(2 m/s)2 = 1 J. The Work-Energy Theorem If you recall, work is a measure of the transfer of energy. An object that has a certain amount of work done on it has that amount of energy transferred to it. This energy moves the object over a certain distance with a certain force; in other words, it is kinetic energy. This handy little fact is expressed in the work-energy theorem, which states that the net work done on an object is equal to the object’s change in kinetic energy: For example, say you apply a force to a particle, causing it to accelerate.

This force does positive work on the particle and increases its kinetic energy. Conversely, say you apply a force to decelerate a particle. This force does negative work on the particle and decreases its kinetic energy. If you know the forces acting on an object, the work-energy theorem provides a convenient way to calculate the velocity of a particle. Example | | | A hockey puck of mass 1 kg slides across the ice with an initial velocity of 10 m/s. There is a 1 N force of friction acting against the puck. What is the puck’s velocity after it has glided 32 m along the ice? | |

If we know the puck’s kinetic energy after it has glided 32 m, we can calculate its velocity. To determine its kinetic energy at that point, we need to know its initial kinetic energy, and how much that kinetic energy changes as the puck glides across the ice. First, let’s determine the initial kinetic energy of the puck. We know the puck’s initial mass and initial velocity, so we just need to plug these numbers into the equation for kinetic energy: The friction between the puck and the ice decelerates the puck. The amount of work the ice does on the puck, which is the product of the force of friction and the puck’s displacement, is negative.

The work done on the puck decreases its kinetic energy, so after it has glided 32 m, the kinetic energy of the puck is 50 – 32 = 18 J. Now that we know the final kinetic energy of the puck, we can calculate its final velocity by once more plugging numbers into the formula for kinetic energy: We could also have solved this problem using Newton’s Second Law and some kinematics, but the work-energy theorem gives us a quicker route to the same answer. Potential Energy As we said before, work is the process of energy transfer. In the example above, the kinetic energy of the puck was transferred into the heat and sound caused by friction.

There are a great number of objects, though, that spend most of their time neither doing work nor having work done on them. This book in your hand, for instance, is not doing any work right now, but the second you drop it—whoops! —the force of gravity does some work on it, generating kinetic energy. Now pick up the book and let’s continue. Potential energy, U, is a measure of an object’s unrealized potential to have work done on it, and is associated with that object’s position in space, or its configuration in relation to other objects.

Any work done on an object converts its potential energy into kinetic energy, so the net work done on a given object is equal to the negative change in its potential energy: Be very respectful of the minus sign in this equation. It may be tempting to think that the work done on an object increases its potential energy, but the opposite is true. Work converts potential energy into other forms of energy, usually kinetic energy. Remove the minus sign from the equation above, and you are in direct violation of the law of conservation of energy! There are many forms of potential energy, each of which is associated with a different type of force.

SAT II Physics usually confines itself to gravitational potential energy and the potential energy of a compressed spring. We will review gravitational potential energy in this section, and the potential energy of a spring in the next chapter. Gravitational Potential Energy Gravitational potential energy registers the potential for work done on an object by the force of gravity. For example, say that you lift a water balloon to height h above the ground. The work done by the force of gravity as you lift the water balloon is the force of gravity, –mg, times the water balloon’s displacement, h.

So the work done by the force of gravity is W = –mgh. Note that there is a negative amount of work done, since the water balloon is being lifted upward, in the opposite direction of the force of gravity. By doing –mgh joules of work on the water balloon, you have increased its gravitational potential energy by mgh joules (recall the equation ). In other words, you have increased its potential to accelerate downward and cause a huge splash. Because the force of gravity has the potential to do mgh joules of work on the water balloon at height h, we say that the water balloon has mgh joules of gravitational potential energy.

For instance, a 50 kg mass held at a height of 4 m from the ground has a gravitational potential energy of: The most important thing to remember is that the higher an object is off the ground, the greater its gravitational potential energy. Mechanical Energy We now have equations relating work to both kinetic and potential energy: Combining these two equations gives us this important result: Or, alternatively, As the kinetic energy of a system increases, its potential energy decreases by the same amount, and vice versa. As a result, the sum of the kinetic energy and the potential energy in a system is constant.

We define this constant as E, the mechanical energy of the system: This law, the conservation of mechanical energy, is one form of the more general law of conservation of energy, and it’s a handy tool for solving problems regarding projectiles, pulleys, springs, and inclined planes. However, mechanical energy is not conserved in problems involving frictional forces. When friction is involved, a good deal of the energy in the system is dissipated as heat and sound. The conservation of mechanical energy only applies to closed systems. Example 1 | | | A student drops an object of mass 10 kg from a height of 5 m.

What is the velocity of the object when it hits the ground? Assume, for the purpose of this question, that g = –10 m/s2. | | Before the object is released, it has a certain amount of gravitational potential energy, but no kinetic energy. When it hits the ground, it has no gravitational potential energy, since h = 0, but it has a certain amount of kinetic energy. The mechanical energy, E, of the object remains constant, however. That means that the potential energy of the object before it is released is equal to the kinetic energy of the object when it hits the ground. When the object is dropped, it has a gravitational potential energy of:

By the time it hits the ground, all this potential energy will have been converted to kinetic energy. Now we just need to solve for v: Example 2 | | | Consider the above diagram of the trajectory of a thrown tomato:| | | 1. | At what point is the potential energy greatest? | | | 2. | At what point is the kinetic energy the least? | | | 3. | At what point is the kinetic energy greatest? | | | 4. | At what point is the kinetic energy decreasing and the potential energy increasing? | | | 5. | At what point are the kinetic energy and the potential energy equal to the values at position A? | The answer to question 1 is point B. At the top of the tomato’s trajectory, the tomato is the greatest distance above the ground and hence has the greatest potential energy. The answer to question 2 is point B. At the top of the tomato’s trajectory, the tomato has the smallest velocity, since the y-component of the velocity is zero, and hence the least kinetic energy. Additionally, since mechanical energy is conserved in projectile motion, we know that the point where the potential energy is the greatest corresponds to the point where the kinetic energy is smallest.

The answer to question 3 is point E. At the bottom of its trajectory, the tomato has the greatest velocity and thus the greatest kinetic energy. The answer to question 4 is point A. At this point, the velocity is decreasing in magnitude and the tomato is getting higher in the air. Thus, the kinetic energy is decreasing and the potential energy is increasing. The answer to question 5 is point C. From our study of kinematics, we know that the speed of a projectile is equal at the same height in the projectile’s ascent and descent. Therefore, the tomato has the same kinetic energy at points A and C.

Additionally, since the tomato has the same height at these points, its potential energy is the same at points A and C. Keep this example in mind when you take SAT II Physics, because it is likely that a similar question will appear on the test. Thermal Energy There are many cases where the energy in a system seems simply to have disappeared. Usually, this is because that energy has been turned into sound and heat. For instance, a coin sliding across a table slows down and comes to a halt, but in doing so, it produces the sound energy of the coin scraping along the table and the heat energy of friction.

Rub your hands together briskly and you will feel that friction causes heat. We will discuss thermal energy, or heat, in greater detail in Chapter 9, but it’s worth noting here that it is the most common form of energy produced in energy transformations. It’s hard to think of an energy transformation where no heat is produced. Take these examples: * Friction acts everywhere, and friction produces heat. * Electric energy produces heat: a light bulb produces far more heat than it does light. * When people talk about burning calories, they mean it quite literally: exercise is a way of converting food energy into heat. Sounds fade to silence because the sound energy is gradually converted into the heat of the vibrating air molecules. In other words, if you shout very loudly, you make the air around you warmer! * Power * Power is an important physical quantity that frequently, though not always, appears on SAT II Physics. Mechanical systems, such as engines, are not limited by the amount of work they can do, but rather by the rate at which they can perform the work. Power, P, is defined as the rate at which work is done, or the rate at which energy is transformed.

The formula for average power is: * * Power is measured in units of watts (W), where 1 W = 1 J/s. * Example | | | A piano mover pushes on a piano with a force of 100 N, moving it 9 m in 12 s. With how much power does the piano mover push? | | * Power is a measure of the amount of work done in a given time period. First we need to calculate how much work the piano mover does, and then we divide that quantity by the amount of time the work takes. * * Be careful not to confuse the symbol for watts, W, with the symbol for work, W. * Instantaneous Power Sometimes we may want to know the instantaneous power of an engine or person, the amount of power output by that person at any given instant. In such cases, there is no value for to draw upon. However, when a steady force is applied to an object, the change in the amount of work done on the object is the product of the force and the change in that object’s displacement. Bearing this in mind, we can express power in terms of force and velocity: * Work Work Done by Gravity Kinetic Energy Work-Energy Theorem Potential Energy Gravitational Potential Energy Mechanical Energy Average Power Instantaneous Power


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