Physics
Classical Angular Momentum
Classical angular momentum refers to the rotational equivalent of linear momentum in classical mechanics. It is a measure of the rotational motion of an object and is defined as the product of the object's moment of inertia and its angular velocity. In classical physics, angular momentum is conserved in the absence of external torques, making it a fundamental quantity in the description of rotational motion.
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11 Key excerpts on "Classical Angular Momentum"
eBook - PDFQuantum Dynamics
Applications in Biological and Materials Systems
- Eric R. Bittner(Author)
- 2009(Publication Date)
- CRC Press(Publisher)
In general, the angular momentum is de-fined as the cross product between a radial vector locating the particle and its linear momentum M = r × p (1.153) Cross products are equivalent to taking the determinant of a matrix M = ˆ i ˆ j ˆ k x y z p x p y p z (1.154) where ˆ i , ˆ j , and ˆ k are the unit vectors along the x , y , z axes. Evaluating the determinant gives M = ˆ i ( yp z − zp y ) − ˆ j ( xp z − zp x ) + ˆ k ( xp y − yp x ) (1.155) = ˆ i M x + ˆ j M y + ˆ kM z (1.156) For motion in the x – y plane, the only term that remains is the M z term, indicating that the angular momentum vector points perpendicular to the plane of rotation, M z = ( xp y − yp x ) = m ( x v y − y v x ) (1.157) Since we have noted that the angular momentum is a constant of the motion, we must have dM z / dt = 0. Let us check: dM z dt = m ( v x v y − v y v x + xa y − ya x ) (1.158) where a x = ˙ v x is the acceleration in x . Thus, dM z dt = ( x F y − yF x ) (1.159) If the force is radial, F x = F r cos( θ ) and F y = F r sin( θ ). Likewise, x = r cos( θ ) and y = r sin( θ ). Putting this into the equations, we have dM z dt = r F r (sin( θ ) cos( θ ) − sin( θ ) cos( θ )) = 0 (1.160) Taking θ = ω t as above where ω is the angular frequency, and using v x = − r ω sin( ω t ) and y y = + r ω cos( ω t ), we can also write M = m ( v x y − v y x ) = m v r ( sin 2 ( ω t ) + cos 2 ( ω t )) = m v r (1.161) 22 Quantum Dynamics: Applications in Biological and Materials Systems 1.4.6 C LASSICAL M OTION OF AN E LECTRON ABOUT A P OSITIVE C HARGE (N UCLEUS ) Now we are ready to describe the motion and energy of a charged particle about a nucleus. This will provide us with a classical description of the hydrogen atom and any hydrogenic (hydrogen-like) species. We shall need these results to begin our discussion of quantum theory.
eBook - PDF- Paul Peter Urone, Roger Hinrichs(Authors)
- 2012(Publication Date)
- Openstax(Publisher)
By now the pattern is clear—every rotational phenomenon has a direct translational analog. It seems quite reasonable, then, to define angular momentum L as (10.90) L = Iω. This equation is an analog to the definition of linear momentum as p = mv . Units for linear momentum are kg ⋅ m/s while units for angular momentum are kg ⋅ m 2 /s . As we would expect, an object that has a large moment of inertia I , such as Earth, has a very large angular momentum. An object that has a large angular velocity ω , such as a centrifuge, also has a rather large angular momentum. Making Connections Angular momentum is completely analogous to linear momentum, first presented in Uniform Circular Motion and Gravitation. It has the same implications in terms of carrying rotation forward, and it is conserved when the net external torque is zero. Angular momentum, like linear momentum, is also a property of the atoms and subatomic particles. Example 10.11 Calculating Angular Momentum of the Earth Strategy No information is given in the statement of the problem; so we must look up pertinent data before we can calculate L = Iω . First, according to Figure 10.12, the formula for the moment of inertia of a sphere is (10.91) I = 2MR 2 5 so that (10.92) L = Iω = 2MR 2 ω 5 . 366 Chapter 10 | Rotational Motion and Angular Momentum This OpenStax book is available for free at http://cnx.org/content/col11406/1.9 Earth’s mass M is 5.979×10 24 kg and its radius R is 6.376×10 6 m . The Earth’s angular velocity ω is, of course, exactly one revolution per day, but we must covert ω to radians per second to do the calculation in SI units. Solution Substituting known information into the expression for L and converting ω to radians per second gives (10.93) L = 0.4 ⎛ ⎝ 5.979×10 24 kg ⎞ ⎠ ⎛ ⎝ 6.376×10 6 m ⎞ ⎠ 2 ⎛ ⎝ 1 rev d ⎞ ⎠ = 9.72×10 37 kg ⋅ m 2 ⋅ rev/d.
eBook - PDF- William Moebs, Samuel J. Ling, Jeff Sanny(Authors)
- 2016(Publication Date)
- Openstax(Publisher)
Angular Momentum of a Rigid Body We have investigated the angular momentum of a single particle, which we generalized to a system of particles. Now we can use the principles discussed in the previous section to develop the concept of the angular momentum of a rigid body. Celestial objects such as planets have angular momentum due to their spin and orbits around stars. In engineering, anything that rotates about an axis carries angular momentum, such as flywheels, propellers, and rotating parts in engines. Knowledge of the angular momenta of these objects is crucial to the design of the system in which they are a part. To develop the angular momentum of a rigid body, we model a rigid body as being made up of small mass segments, Δm i . In Figure 11.12, a rigid body is constrained to rotate about the z-axis with angular velocity ω . All mass segments that make up the rigid body undergo circular motion about the z-axis with the same angular velocity. Part (a) of the figure shows mass segment Δm i with position vector r → i from the origin and radius R i to the z-axis. The magnitude of its tangential velocity is v i = R i ω . Because the vectors v → i and r → i are perpendicular to each other, the magnitude of the angular momentum of this mass segment is l i = r i (Δmv i )sin 90°. Chapter 11 | Angular Momentum 561 Figure 11.12 (a) A rigid body is constrained to rotate around the z-axis. The rigid body is symmetrical about the z-axis. A mass segment Δm i is located at position r → i , which makes angle θ i with respect to the z-axis. The circular motion of an infinitesimal mass segment is shown. (b) l → i is the angular momentum of the mass segment and has a component along the z-axis ( l → i ) z . Using the right-hand rule, the angular momentum vector points in the direction shown in part (b). The sum of the angular momenta of all the mass segments contains components both along and perpendicular to the axis of rotation.
- Joseph C. Amato, Enrique J. Galvez(Authors)
- 2015(Publication Date)
- CRC Press(Publisher)
But spin is only one example of angular momentum. We will now broaden the definition of I L , and show that all bodies, non-rotating ones, even bodies moving along a straight line, may possess angular momentum. Moreover, we will see that the total angular momentum of an isolated system does not change, giving us a new conservation law to add to those of momentum and energy. Conservation of angular momentum is the crux of Kepler’s second law (the equal area law), and we will use it to find the shape of planetary orbits and to predict the trajectories of asteroids, comets, and spacecraft. Like momentum and energy, angu -lar momentum is a powerful tool for studying motion, and it will be employed in Chapters 13 and 14 to study the behavior of rigid bodies in simple lab-scale situations as well as in phenomena that challenge our understanding of the cosmos. 12.2 BROADER DEFINITION OF ANGULAR MOMENTUM Imagine that you are at point A , observing the motion of a small body. Relative to A , the body is located at position I r and moving with momentum I p (see Figure 12.1a ). The angular momentum I L of the body relative to A is formally defined by the cross product I I I L r p ≡ × . (12.1) According to the properties of the cross product, I L is perpendicular to the plane containing I r and I p , and points in the direction determined by the right hand rule. (For a review of cross products, see Section 8.3.) The magnitude of I L is given by L = rp sin θ , where θ is the angle between I r and I p ( θ < 180 ° ). Alternatively, we may express the magnitude as L = rp ⊥ , where p ⊥ = p sin θ is the component of I p perpendicular to I r , or equivalently as L = r ⊥ p (see Figure 12.1b ). EXERCISE 12.1 Show that the angular momentum relative to point A of a body moving with no forces acting on it is a constant, that is, I I I I L r L r ( ( ) ). = ′ A r v r ΄
- David A. B. Miller(Author)
- 2008(Publication Date)
- Cambridge University Press(Publisher)
Chapter 9 Angular momentum Prerequisites: Chapters 2–5. Thus far, we have dealt primarily with energy, position, and linear momentum and have proposed operators for each of these. One other quantity that is important in classical mechanics, angular momentum , is particularly important also in quantum mechanics. Here, we introduce angular momentum, its operators, eigenvalues, and eigenfunctions. If this discussion seems somewhat abstract, the reader can be assured that the concepts of angular momentum will become very concrete in the discussion of the hydrogen atom. One aspect of angular momentum that is different from the quantities and operators discussed previously is that its operators always have discrete eigenvalues. Whereas linear momentum is associated with eigenfunctions that are functions of position along a specific spatial direction, angular momentum is associated with eigenfunctions that are functions of angle or angles about a specific axis. The fact that the eigenvalues are discrete is associated with the fact that for a single-valued spatial function, once we have gone an angle 2 π about a particular axis, we are back to where we started. The wavefunction is presumably continuous and single-valued 1 and, hence, we must therefore have integral numbers of periods of oscillation with angle within this angular range; this requirement of integer numbers of periods leads to the discrete quantization of angular momentum. Another surprising aspect of angular momentum operators is that the operators corresponding to angular momentum about different orthogonal axes (e.g., ˆ x L , ˆ y L , and ˆ z L ) do not commute with one another (in contrast, e.g., to the linear momentum operators for the different orthogonal coordinate directions). We do, however, find that there is another useful angular momentum operator, 2 ˆ L , which does commute with each of ˆ x L , ˆ y L , and ˆ z L separately.
eBook - PDFAngular Momentum
An Illustrated Guide to Rotational Symmetries for Physical Systems
- William J. Thompson(Author)
- 2008(Publication Date)
- Wiley-VCH(Publisher)
A Brief History of Angular Momentum, The historical development of con- cepts in angular momentum and use of symmetry in the physical sciences, emphasiz- ing rotational symmetry, is summarized in Figure 5.14. Scientists and their publi- cations are selected under the criterion that in these works the key ideas are clearly expressed for the first time. The earliest recognition of a conservation condition for rotational motion was probably by Kepler in 1609, with his empirical law of constant areal velocities for the motions of planets about the sun. Newton in Principia Mathematica ( 1 686) showed that this law is a consequence of the central-force nature of gravitational at- traction. Together with the fact that the orbits lie in a plane, we would nowadays call this the constancy of angular momentum for two-body central-force motion. Progress was from an essentially geometrical property (areal velocity) to a dynamical property (central forces). Such a progression is also discussed in Section 4.3.1. Two centuries passed while the followers of Newton piled up layer upon layer of the consequences of his new mechanics, just as hosts of physicists and chemists in the second half of the twentieth century burden the libraries of the world with in- ferences from quantum mechanics. In 1848, Pasteur announced his discovery of a handedness in organic crystals, as discussed in Section 1.2.1. In two critical works in 1883 and 190 1, Emst Mach examined the foundations of mechanics and the rela- tions between space and geometry. This work had significant influence on Ein- stein’s ideas about relativity of motion. About the same time, Pierre Curie presented his principles relating symmetries in causes and effects, as we discuss in Sec- tion 1.1.2. In 1918, Emmy Noether showed the necessary connections between continuous symmetries and conservation laws. Although (as discussed in Sec-
eBook - PDF- David Agmon, Paul Gluck;;;(Authors)
- 2009(Publication Date)
- WSPC(Publisher)
Example 2 Angular momentum for motion in a circle. A particle of moss m moves in a circle of radius r with constant velaeity V (> (a) Calculate its angular momentum L relative to the center O. (h) What is the moment of the force acting on the particle 9 (c) Express L in terms of the angular velocity co. (d ) Express the kinetic energy in terms ofL. Solution la) From its definition, the direction of L is perpendicular to the plane of the circle, out of the page. Since r is perpendicular to V 0 , L = rp sin 90 = mV {) r, (b) Since L is constant in both magnitude and direction, the resultant moment vanishes. We can also see this from the fact that the centripetal force is parallel to is and so r = rxF = 0. (c) L = rn(V)r = m((D- r)r = mrO) = Id) (since / = mr 1 ) I . 1 (m^ J r 1 1 {mV v f 1 Is ! Is id 1 E. = — mV. i = —I m 2mj { r 1 j 2 mr 2 mr 2 I Example 3 Motion under the influence of a central g^ force. Lei a particle move in a central force field of the o form Fir) = fifi r. Prove the following, (a) Its angular momentum with respect to the center offorce is a constant, (h) its motion is confined to a plane, 388 Classical and Relativistic Mechanics / P O / Plane of motion Solution (a) Since the force is parallel to f , its moment about the center of force O vanishes. Therefore the angular momentum relative to O is a constant. (b) By definition, the direction of L is perpendicular to the plane formed by the radius vector and the velocity. But since L is a constant, the velocity and the radius vector must remain in the same plane. Example 4 Kepler's law of areas. Prove thai for a body moving influence of any central force, the radius vector joining it to the centt sweeps out equal areas in equal times. This is a generalization of Kepler's area rule for planets around the Sun, stated in section 10.1. Solution Consider the particle at the time t = 0 and At later. The area M swept out is the shaded area in the diagram.
eBook - PDF- Yoni Kahn, Adam Anderson(Authors)
- 2018(Publication Date)
- Cambridge University Press(Publisher)
The analogues of the equations p = mv and F = dp/dt are only used on the GRE in their scalar forms: L = I ω, (1.19) τ = dL dt . (1.20) Problems involving angular momentum can also be con- ceptual, asking for the configuration of momentum, velocity, and acceleration vectors for a system involving rotational motion. The key point to remember is that the angular 14 Classical Mechanics momentum vector L is generally parallel 3 to the angular velocity vector ω, which points along the axis of rotation, just like an object’s linear momentum is parallel to its velocity. The direction of L is determined by the right-hand rule: curl the fingers of your right hand in the direction of rotation, and your thumb gives the direction of L. More advanced classical mechanics texts will discuss rotat- ing reference frames, which are mostly beyond the scope of the GRE. All you need to know is that a reference frame rotating at constant angular velocity is not inertial, but, nonetheless, one can write a formula resembling Newton’s second law F = ma at the price of introducing “fictitious” forces, which only appear because of the noninertial choice of coordinates: F centrifugal = −m 2 r, (1.21) F Coriolis = −2m × v. (1.22) The centrifugal force (which we emphasize once again is not a real force!) is the apparent force on an object in a uni- formly rotating frame that pushes it away from the axis of rotation. The Coriolis force vanishes if the object is station- ary in the rotating frame, but often appears in the context of the Earth’s rotation, which defines a rotating frame. Unless the motion of the object is defined with respect to a rotat- ing frame, in which case you typically need to use the Coriolis force, we recommend sticking with inertial frames to avoid confusion. 1.4.3 Moment of Inertia As we’ve seen, an object’s moment of inertia is analogous to its mass in the context of rotation, but, unlike mass, it depends on the distance from the center of rotation.
- Raymond Serway, John Jewett(Authors)
- 2018(Publication Date)
- Cengage Learning EMEA(Publisher)
Consider a rigid object rotating about a fixed axis that coincides with the z axis of a coordinate system as shown in Figure 11.7. Let’s determine the angular momentum of this object. Each particle of the object rotates in the xy plane about the z axis with an angular speed v. The magnitude of the angular momentum of a particle of mass m i about the z axis is m i v i r i . Because v i 5 r i v (Eq. 10.10), we can express the magni- tude of the angular momentum of this particle as L i 5 m i v i r i 5 m i ( r i v ) r i 5 m i r i 2 v The vector L S i for this particle is directed along the z axis, as is the vector v S . We can now find the angular momentum (which in this situation has only a z component) of the whole object by taking the sum of L i over all particles: L z 5 o i L i 5 o i m i r i 2 v 5 S o i m i r i 2 D v L z 5 Iv (11.16) where we have recognized o i m i r i 2 as the moment of inertia I of the object about the z axis (Eq. 10.19). Notice that Equation 11.16 is mathematically similar in form to Equation 9.2 for linear momentum: p S 5 m v S . Now let’s differentiate Equation 11.16 with respect to time, noting that I is con- stant for a rigid object: dL z dt 5 I dv dt 5 Ia (11.17) where a is the angular acceleration relative to the axis of rotation. Because dL z / dt is equal to the net external torque (see Eq. 11.15), we can express Equation 11.17 as o t ext 5 Ia (11.18) That is, the net external torque acting on a rigid object rotating about a fixed axis equals the moment of inertia about the rotation axis multiplied by the object’s angular acceleration relative to that axis. This result is the same as Equation 10.18, which was derived using a force approach, but we derived Equation 11.18 using the concept of angular momentum. As we saw in Section 10.5, Equation 11.18 is the mathematical representation of the rigid object under a net torque analysis model.
eBook - PDF- James D Louck(Author)
- 2008(Publication Date)
- World Scientific(Publisher)
There are, fortunately, general invariance properties of the Schr¨ odinger equation and its solution state functions for a complex composite phys-ical systems that can be used to classify the quantum states of physical systems into substates available to the system. Our focus here is on the properties of the total angular momentum of a physical system, which is a quantity L that has a vector expression L = L 1 e 1 + L 2 e 2 + L 3 e 3 in the right-handed frame ( e 1 , e 2 , e 3 ) and the expression L = L 1 e 1 + L 2 e 2 + L 3 e 3 in a second rotated right-handed frame ( e 1 , e 2 , e 3 ) . At a given instant of time, necessarily L = L , since these quantities are just redescriptions of the total angular momentum of the system at a given time. The total angular momentum is a conserved quantity; that is, d L /dt = 0 , for all time t, and this property makes the total angular momentum an important quantity for the study of the behavior of complex physical systems. For a system of n point particles, the total angular momentum relative to the origin of the reference frame ( e 1 , e 2 , e 3 ) is obtained by vector addition of that of the individual parti-cles by L = ∑ n i =1 L i , where L i is expressed by the vector cross product L i = x i × p i in terms of the vector position x i = x 1 i e 1 + x 2 i e 2 + x 3 i e 3 and the vector linear momentum p i = p 1 i e 1 + p 2 i e 2 + p 3 i e 3 of the particle labeled i. While angular momentum can be exchanged between interact-ing particles, the total angular momentum remains constant in time for an isolated physical system of n particles. The quantum-mechanical op-erator interpretation of such classical physical quantities is obtained by Schr¨ odinger’s rule p i → − i ∇ i , = h/ 2 π, where h is Planck’s constant. The reference frame vectors ( e 1 , e 2 , e 3 ) remain intact. 1.1. BACKGROUND AND VIEWPOINT 13 The viewpoints of Newtonian physics and quantum physics may be contrasted in many ways.
No longer available |Learn morePhysics for Scientists and Engineers
Foundations and Connections, Extended Version with Modern Physics
- Debora Katz(Author)
- 2016(Publication Date)
- Cengage Learning EMEA(Publisher)
All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-300 13-7 Conservation of Angular Momentum 379 All content on this page is © Cengage Learning. CHECK and THINK Because the CD slows down and because its angular velocity is in the negative z direction, the change in its angular momentum is in the positive z direction. In Problem 12.82, you should find that the angular acceleration is also in the positive z direction. Again, it is not a coincidence. 13-7 Conservation of Angular Momentum We found in Chapter 10 that when no net force acts on a system, the system’s trans- lational momentum is conserved, and this principle gave us a way to analyze com- plicated situations such as the decay of a nucleus. In this section, we will learn when and how to apply the principle of conservation of angular momentum. How- ever, to better understand conservation of angular momentum, we consider a system whose angular momentum is changing and derive another expression for Newton’s second law. DERIVATION Another Expression of Newton’s Second Law Another way to express Newton’s second law for a system is dL u tot dt 5 a t u ext (13.27) where t u ext is the total external torque exerted and L u tot is the system’s total angular momentum. Equation 13.27 holds for a single particle, a collection of particles, or an object. Our derivation is made simpler if we consider a single particle moving along a straight path (Fig. 13.28). A net force F u tot acts on the particle so that it speeds up along that path. We will show that for this particle, dL u dt 5 t u tot (13.28) Because the particle’s speed increases, its translational momentum also increases. The particle’s path is along the straight line shown, so r ' does not change, but because the translational momentum p increases, the particle’s angular momentum L around the z axis increases also.
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