Physics
Rotational Kinetic Energy
Rotational kinetic energy is the energy an object possesses due to its rotation around an axis. It depends on the object's moment of inertia and angular velocity. The formula for rotational kinetic energy is 1/2 * I * ω^2, where I is the moment of inertia and ω is the angular velocity.
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11 Key excerpts on "Rotational Kinetic Energy"
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2015(Publication Date)
- Wiley(Publisher)
The kinetic energy of the entire rotating body, then, is the sum of the kinetic energies of the particles: Rotational KE 5 S ( 1 2 mr 2 v 2 ) 5 1 2 ( Smr 2 )v 2 u Moment of inertia, I In this result, the angular speed v is the same for all particles in a rigid body and, therefore, has been factored outside the summation. According to Equation 9.6, the term in parenthe- ses is the moment of inertia, I 5 Smr 2 , so the Rotational Kinetic Energy takes the following form: Definition of Rotational Kinetic Energy The Rotational Kinetic Energy KE R of a rigid object rotating with an angular speed v about a fixed axis and having a moment of inertia I is KE R 5 1 2 Iv 2 (9.9) Requirement: v must be expressed in rad/s. SI Unit of Rotational Kinetic Energy: joule (J) Kinetic energy is one part of an object’s total mechanical energy. The total mechan- ical energy is the sum of the kinetic and potential energies and obeys the principle of conservation of mechanical energy (see Section 6.5). Specifically, we need to remember that translational and rotational motion can occur simultaneously. When a bicycle coasts down a hill, for instance, its tires are both translating and rotating. An object such as a rolling bicycle tire has both translational and rotational kinetic energies, so that the total mechanical energy is E 5 1 2 mv 2 1 1 2 Iv 2 1 mgh u Total mechanical energy u Translational kinetic energy u Rotational Kinetic Energy u Gravitational potential energy Figure 9.20 The force F B does work in rotating the wheel through the angle u. F Axis of rotation r Rope θ s B m r 1 r 2 Figure 9.21 The rotating wheel is composed of many particles, two of which are shown. 238 Chapter 9 | Rotational Dynamics Here m is the mass of the object, v is the translational speed of its center of mass, I is its moment of inertia about an axis through the center of mass, v is its angular speed, and h is the height of the object’s center of mass relative to an arbitrary zero level.
eBook - PDF- John Matolyak, Ajawad Haija(Authors)
- 2013(Publication Date)
- CRC Press(Publisher)
161 Rotational Motion © 2010 Taylor & Francis Group, LLC 8.9 Rotational Kinetic Energy The translational kinetic energy of a point-like object rotating in a circular path of radius r with a linear velocity v is K 1 2 mv 2 = . (8.36a) Substituting for v = ω r (see Equation 8.3), the above equation becomes K 1 2 mr 2 2 = ω or K 1 2 I . 2 = ω (8.36b) Equation 8.36b demonstrates an analogy between the Rotational Kinetic Energy of an object and its translational kinetic energy, K, that is equal to (1/2)mv 2 , with the terms I and ω replacing the linear terms m and v, respectively. For a rigid body rotating about an axis through its center of mass, the kinetic energy of each of the particles constituting the rigid body is (K) 1 2 I i i 2 = ω , (8.36c) where I m r i i i 2 = is the moment of inertia of the ith particle about the axis of rotation. The angular velocity ω about the axis of rotation is the same for all constituents of the rigid body, the Rotational Kinetic Energy of the rigid body consisting of discrete particles about its axis of rotation is (K ) 1 2 I , body rot i i 2 = ∑ ω (8.37) while for a continuous distribution of mass, the above relation becomes (K ) 1 2 I . body rot 2 = ω (8.38) 8.10 A RIGID BODY IN TRANSLATIONAL AND ROTATIONAL MOTIONS If a rigid body of mass M executes both translational and rotational motions, it can be shown that its total kinetic energy is always the sum of two contributions: 1. Rotational Kinetic Energy, which results from the rotation of the body about its center of mass. That kinetic energy is given by either Equation 8.37 or Equation 8.38, according to whether the body is a system of discrete particles (Figure 8.3) or a continuous mass distri-bution (Figure 8.4). 2. Translational kinetic energy, which is caused by the motion of the center of mass with a velocity V. This translational term has the velocity of the center of mass V.
eBook - PDFWorkshop Physics Activity Guide Module 2
Mechanics II
- Priscilla W. Laws, David P. Jackson, Brett J. Pearson(Authors)
- 2023(Publication Date)
- Wiley(Publisher)
Recall that the kinetic energy of a mass m in translational motion is defined by K tr = 1 2 mv 2 where v is the translational speed (magnitude of the velocity) of the object’s center of mass. In general, kinetic energy is a scalar quantity that represents the energy of motion, and translational kinetic energy is the energy associated with the translational motion of the object as a whole. In our study of rotational motion, we have seen objects that rotate (spin) while having a stationary center of mass (e.g., a spinning record album or a wheel on a fixed axle). Because the center of mass of such an object is not moving, the translational kinetic energy will be zero. However, it should be clear that the object still has motion, and so there will still be kinetic energy. By simply replacing the mass and speed in the definition of translational kinetic energy with their rotational counterparts, we can define an analogous Rotational Kinetic Energy to be K rot = 1 2 I 2 (13.1) where is the rotational speed (magnitude of the rotational velocity) and I is the rotational inertia about the axis of rotation. In general, an object can have 428 WORKSHOP PHYSICS ACTIVITY GUIDE kinetic energy due to its translational motion or its rotational motion, and the total kinetic energy will be the sum of these: K tot = K tr + K rot (13.2) The following activity is designed to give you some practice with the con- cept of Rotational Kinetic Energy. 13.2.2. Activity: An Object Rolling Down a Hill a. Suppose you place a round object of mass m, radius R, and rotational inertia I at the top of a ramp of height h and release it from rest so that it rolls (without slipping) down the ramp. Draw a free-body diagram of the object as it rolls down the hill, placing the tail of each force arrow at the point where the force acts.
eBook - PDF- Michael Tammaro(Author)
- 2019(Publication Date)
- Wiley(Publisher)
The kinetic energy introduced in Chapter 6 was translational kinetic energy—that is, energy associated with motion from place to place. Here in Section 8.6, how- ever, we have another expression for kinetic energy—Rotational Kinetic Energy. Before exploring applications, we discuss the kinetic energy of a rolling object. Rolling Without Slipping and Kinetic Energy As a wheel rolls, the motion of the particles that make up the wheel is complex because the motion of any point on the wheel is due to both translation of the center of mass of the wheel and rotation about an axis through its center. Animated Figure 8.6.1 illustrates this situation. 8.6 ENERGY CONSERVATION FOR ROTATING OBJECTS Learning Objective I N T E R A C T I V E F E A T U R E 228 | Chapter 8 Part (a) shows a wheel that is rotating at a constant angular speed about a fixed axis. A point on the outer edge of the wheel moves in uniform circular motion at a speed given by Equation 8.4.1, v r t ω = . Part (b) shows the same wheel, rotating about the same axis with the same angular speed, but now the wheel is rolling. The motion of a point on the edge of the wheel is complex. If the wheel rolls without slipping, then the point is momentarily at rest when it touches the ground. When the point is on the top of the wheel, it achieves its fastest speed and is moving horizontally. If the wheel does not slip, then the center of the wheel moves with a speed v r ω = over the ground, where r is the radius of the wheel and ω is its angular speed about the (moving) axis through its center. Thus, the speed v is the same as the tangential speed of the point in Animated Figure 8.6.1(a). (Similarly, the linear acceleration a of a rolling object is the same as the tangential acceleration given by Equation 8.4.2, a rα = , where α is the angular acceleration of the wheel about an axis through its center.) Animated Figure 8.6.1 (a) A wheel rotates with constant angular speed about a fixed axis.
No longer available |Learn more- (Author)
- 2014(Publication Date)
- Learning Press(Publisher)
________________________ WORLD TECHNOLOGIES ________________________ This equation states that the kinetic energy ( E k ) is equal to the integral of the dot product of the velocity ( v ) of a body and the infinitesimal change of the body's momentum ( p ). It is assumed that the body starts with no kinetic energy when it is at rest (motionless). Rotating bodies If a rigid body is rotating about any line through the center of mass then it has Rotational Kinetic Energy ( ) which is simply the sum of the kinetic energies of its moving parts, and is thus given by: where: • ω is the body's angular velocity • r is the distance of any mass dm from that line • is the body's moment of inertia, equal to . (In this equation the moment of inertia must be taken about an axis through the center of mass a nd the rotation measured by ω must be around that axis; more general equations exist for systems where the object is subject to wobble due to its eccentric shape). Kinetic energy of systems A system of bodies may have internal kinetic energy due to the relative motion of the bodies in the system. For example, in the Solar System the planets and planetoids are orbiting the Sun. In a tank of gas, the molecules are moving in all directions. The kinetic energy of the system is the sum of the kinetic energies of the bodies it contains. As already discussed above, this will depend on the reference frame. A macroscopic body that is stationary (i.e. a reference frame has been chosen to correspond to the body's center of momentum) may have various kinds of internal energy at the molecular or atomic level, which may be regarded as kinetic energy, due to molecular translation, rotation, and vibration, electron translation and spin, and nuclear spin. These all contribute to the body's mass, as provided by the special theory of relativity. When discussing movements of a macroscopic body, the kinetic energy referred to is usually that of the macroscopic movement only.
eBook - PDF- Paul Peter Urone, Roger Hinrichs(Authors)
- 2012(Publication Date)
- Openstax(Publisher)
The expression for Rotational Kinetic Energy is exactly analogous to translational kinetic energy, with I being analogous to m and ω to v . Rotational Kinetic Energy has important effects. Flywheels, for example, can be used to store large amounts of Rotational Kinetic Energy in a vehicle, as seen in Figure 10.16. Figure 10.16 Experimental vehicles, such as this bus, have been constructed in which Rotational Kinetic Energy is stored in a large flywheel. When the bus goes down a hill, its transmission converts its gravitational potential energy into KE rot . It can also convert translational kinetic energy, when the bus stops, into KE rot . The flywheel’s energy can then be used to accelerate, to go up another hill, or to keep the bus from going against friction. Example 10.8 Calculating the Work and Energy for Spinning a Grindstone Consider a person who spins a large grindstone by placing her hand on its edge and exerting a force through part of a revolution as shown in Figure 10.17. In this example, we verify that the work done by the torque she exerts equals the change in rotational energy. (a) How much work is done if she exerts a force of 200 N through a rotation of 1.00 rad(57.3º) ? The force is kept perpendicular to the grindstone’s 0.320-m radius at the point of application, and the effects of friction are 360 Chapter 10 | Rotational Motion and Angular Momentum This OpenStax book is available for free at http://cnx.org/content/col11406/1.9 negligible. (b) What is the final angular velocity if the grindstone has a mass of 85.0 kg? (c) What is the final Rotational Kinetic Energy? (It should equal the work.) Strategy To find the work, we can use the equation net W = (net τ)θ . We have enough information to calculate the torque and are given the rotation angle. In the second part, we can find the final angular velocity using one of the kinematic relationships.
No longer available |Learn more- (Author)
- 2014(Publication Date)
- Library Press(Publisher)
________________________ WORLD TECHNOLOGIES ________________________ This equation states that the kinetic energy ( E k ) is equal to the integral of the dot product of the velocity ( v ) of a body and the infinitesimal change of the body's momentum ( p ). It is assumed that the body starts with no kinetic energy when it is at rest (motionless). Rotating bodies If a rigid body is rotating about any line through the center of mass then it has Rotational Kinetic Energy ( ) which is simply the sum of the kinetic energies of its moving parts, and is thus given by: where: • ω is the body's angular velocity • r is the distance of any mass dm from that line • is the body's moment of inertia, equal to . (In this equation the moment of inertia must be taken about an axis through the center of mass and the rotation measured by ω must be around that axis; more general equations exist for systems where the object is subject to wobble due to its eccentric shape). Kinetic energy of systems A system of bodies may have internal kinetic energy due to the relative motion of the bodies in the system. For example, in the Solar System the planets and planetoids are orbiting the Sun. In a tank of gas, the molecules are moving in all directions. The kinetic energy of the system is the sum of the kinetic energies of the bodies it contains. As already discussed above, this will depend on the reference frame. A macroscopic body that is stationary (i.e. a reference frame has been chosen to correspond to the body's center of momentum) may have various kinds of internal energy at the molecular or atomic level, which may be regarded as kinetic energy, due to molecular translation, rotation, and vibration, electron translation and spin, and nuclear spin. These all contribute to the body's mass, as provided by the special theory of relativity. When discussing movements of a macroscopic body, the kinetic energy referred to is usually that of the macroscopic movement only.
- Raymond Serway, John Jewett(Authors)
- 2018(Publication Date)
- Cengage Learning EMEA(Publisher)
Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 274 Chapter 10 Rotation of a Rigid Object About a Fixed Axis Using v CM 5 Rv, this equation can be expressed as K 5 1 2 I CM v 2 1 1 2 Mv CM 2 (10.31) The term 1 2 I CM v 2 represents the Rotational Kinetic Energy of the object about its center of mass, and the term 1 2 Mv CM 2 represents the kinetic energy the object would have if it were just translating through space without rotating. Therefore, the total kinetic energy of a rolling object is the sum of the Rotational Kinetic Energy about the center of mass and the translational kinetic energy of the center of mass. This statement is consistent with the situation illustrated in Figure 10.25, which shows that the velocity of a point on the object is the sum of the velocity of the center of mass and the tangential velocity around the center of mass. Energy methods can be used to treat a class of problems concerning the roll- ing motion of an object on a rough incline. For example, consider Figure 10.26, which shows a sphere rolling without slipping after being released from rest at the top of the incline. Accelerated rolling motion is possible only if a friction force is present between the sphere and the incline to produce a net torque about the center of mass. Despite the presence of friction, no loss of mechanical energy occurs because the contact point is at rest relative to the surface at any instant. (On the other hand, if the sphere were to slip, mechanical energy of the sphere–incline–Earth system would decrease due to the nonconservative force of kinetic friction.) In reality, rolling friction causes mechanical energy to transform to internal energy.
eBook - PDF- Raymond Serway, Chris Vuille(Authors)
- 2017(Publication Date)
- Cengage Learning EMEA(Publisher)
The x -, y -, and z- of an object’s center of gravity are given by Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-202 254 TOPIC 8 | Rotational Equilibrium and Dynamics Unless otherwise noted, all content on this page is © Cengage Learning. where i and i and i f refer to initial and final values, respectively. f refer to initial and final values, respectively. f When nonconservative forces are present, it’s necessary to use a generalization of the work–energy theorem: W nc nc 5 DKE t 1 DKE r 1 DPE [8.17] 8.6 Angular Momentum The angular momentum of a rotating object is given by L ; Iv v [8.18] Angular momentum is related to torque in the following equation: a t 5 change in angular momentum time interv rv r al 5 DL Dt [8.19] If the net external torque acting on a system is zero, the total angular momentum of the system is constant, L i 5 L f L f L [8.20] and is said to be conserved. Solving problems usually involves substituting into the expression I i i v i 5 I f I f I v f v f v [8.21] and solving for the unknown. 8.5 Rotational Kinetic Energy If a rigid object rotates about a fixed axis with angular speed v, its Rotational Kinetic Energy is is KE r r 5 1 2 Iv v 2 [8.15] where I is the moment of inertia of the object around the I is the moment of inertia of the object around the I axis of rotation. A system involving rotation is described by three types of energy: potential energy PE, translational kinetic energy , translational kinetic energy KE t t , , and Rotational Kinetic Energy KE r (Fig. 8.40). All these (Fig. 8.40). All these forms of energy must be included in the equation for con- servation of mechanical energy for an isolated system: (KE t t 1 KE r r 1 PE ) i 5 (KE t t 1 KE r r 1 PE ) f [8.16] M u h v S R v Figure 8.40 A ball rolling down an incline converts potential energy to translational and Rotational Kinetic Energy.
eBook - PDFEngineering Mechanics
Problems and Solutions
- Arshad Noor Siddiquee, Zahid A. Khan, Pankul Goel(Authors)
- 2018(Publication Date)
- Cambridge University Press(Publisher)
However, moment of momentum is also known as angular momentum for a rotating body and given by I ω. 16.2.2 Torque and angular momentum Newton’s second law for rotary motion states that the rate of change of angular momentum is directly proportional to the external torque ( T ) applied on the body and lies in the direction of the torque. i.e., Torque ∝ rate of change of angular momentum T d dt I T I d dt T I ∝ ∝ ∝ w w α ( ) ( ) T = C . I α where C is a constant value of proportionality and its value remain unity. Finally, the relation between torque and angular acceleration is given by T = I α 16.3 Kinetic Energy of a Body in Translatory and Rotary Motion Consider a rigid body of mass m moving in general plane motion with velocity v . If the mass of each particle is dm then kinetic energy ( K . E .) of particle during translatory motion will be 710 Engineering Mechanics K E of particle dm v Translatory . . ( ) = 1 2 2 The kinetic energy ( K . E .) of rigid body during translatory motion will be K E of body dm v K E of body m Translatory Translatory . . . . ( ) = ( ) = ∫ 1 2 1 2 2 v 2 When a body rotates about its axis with angular velocity ω then the kinetic energy ( K . E .) in rotary motion for a particle will be K E of particle dm v v r K E of body Rotary Rotary . . . . ( ) = = × ( ) 1 2 2 Since w = = × ( ) = ( ) = ∫ ∫ 1 2 1 2 1 2 2 2 2 2 dm r dmr K E of body I Rotary w w w . . Thus K.E . of rigid body under general plane motion = ( ) + ( ) = + = K E of body K E of body mv I Translator y R otary . . . . 1 2 1 2 1 2 2 2 w mv mk 2 2 2 1 2 + ( ) w where k is the radius of gyration of body. 16.4 Principle of Conservation of Energy According to this principle, when a rigid body has general plane motion under the influence of conservative forces then sum of potential energy and kinetic energy of a rigid body remains constant. ( P . E . + K . E .) = constant Thus, ( P . E . + K . E .) 1 = ( P . E . + K . E .) 2
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
Here, we are told that her angular velocity doubles. Thus, ω f = 2ω 0 , and KE R f _____ KE R 0 = 2ω 0 ___ ω 0 = 2 . While her angular momentum during this maneuver is conserved, her Rotational Kinetic Energy is not. Where does the extra Rotational Kinetic Energy come from? It comes from the work that she does internally to move her arms closer to the axis of rotation. (a) (b) FIGURE 9.24 (REPEATED) (a) A skater spins slowly on one skate, with both arms and one leg outstretched. (b) As she pulls her arms and leg in toward the rotational axis, her angular velocity ω increases. 268 CHAPTER 9 Rotational Dynamics Concept Summary 9.1 The Action of Forces and Torques on Rigid Objects The line of action of a force is an extended line that is drawn colinear with the force. The lever arm ℓ is the distance between the line of action and the axis of rotation, measured on a line that is perpen- dicular to both. The torque of a force has a magnitude that is given by the magni- tude F of the force times the lever arm ℓ. The magnitude of the torque τ is given by Equation 9.1, and τ is positive when the force tends to produce a counterclockwise rotation about the axis, and negative when the force tends to produce a clockwise rotation. Magnitude of torque = Fℓ (9.1) 9.2 Rigid Objects in Equilibrium A rigid body is in equilibrium if it has zero translational acceleration and zero angular accelera- tion. In equilibrium, the net external force and the net external torque acting on the body are zero, according to Equations 4.9a, 4.9b, and 9.2. Σ F x = 0 and ΣF y = 0 (4.9a and 4.9b) Στ = 0 (9.2) 9.3 Center of Gravity The center of gravity of a rigid object is the point where its entire weight can be considered to act when calculating the torque due to the weight. For a symmetrical body with uniformly distributed weight, the center of gravity is at the geometrical center of the body. When a number of objects whose weights are W 1 , W 2 , .
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