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# Angular Momentum

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## Erica Huynh

on 7 June 2013

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#### Transcript of Angular Momentum

Introduction Angular Momentum in Planetary Mechanics Deriving Angular Momentum Sample Problems Magnetic Resonance Imaging (MRI) Thank you for your participation, physics family
ALWAYS REMEMBER:
Physics is Hot and Spicy,
Cool and Icy
:D The World of Angular Momentum What is Angular Momentum? We have all heard of linear momentum, discovered its concepts, and found it to be: The angular analog of linear momentum is angular momentum, and is represented as a vector quantity: For an object to have angular momentum, is rotation necessary?
NO! A particle does not have to be moving in a circular path for angular momentum to be defined. There is angular momentum about any axis displaced from and object’s path. Therefore, an object moving about a straight line has angular momentum. Angular Velocity Recall that angular velocity is the velocity of an object moving in a circle with a radius Angular Acceleration The angular acceleration (or tangential acceleration), is the acceleration of an object rotating about a circle. It can also be defined as the rate of change of angular velocity and the vector is perpendicular to the centripetal acceleration vector. Torque Torque is the tendency of a force to rotate an object about some axis. It is also called the moment of a force. Moment of Inertia Moment of inertia is determined by how Newton’s law of inertia affects the rotation of a body. About an axis of rotation, its resistance is defined to a change in angular velocity. Contents Angular Momentum of a Particle
Angular Momentum of a System of Particles
Angular Momentum of a Rigid Body
Conservation of Angular Momentum
Angular Momentum and Gyroscopes
Angular Momentum in Planetary Mechanics
Quantization of Angular Momentum
Magnetism and Angular Momentum
Medical Resonance Imaging Analysis Angular Momentum of a Particle In any inertial frame of reference, the moment of linear momentum of a particle is known as angular momentum or, angular momentum of a particle is defined as the moment of its linear momentum.
The instantaneous angular momentum L, of a particle relative to the origin O is defined by the cross product of the particle’s instantaneous position vector r and its instantaneous linear momentum, p. The result of the cross product is: Which in component form is: Newton’s second law for translational motion of a particle can be written in terms of linear momentum, so the rotational analog of it is written in terms of angular momentum The SI unit of angular momentum is , which is equivalent to Joules second The magnitude of angular momentum is , where is the angle between r and p. The unit vector for L, is given by:

where the unit vector is perpendicular to the plane of r and p. Angular Momentum of a System of Particles The total angular momentum of a system of particles about some point can be defined as the vector sum of the angular momenta of the individual particles: Differentiate the above equation (with respect to time): The net external torque acting on a system about some axis passing through an origin in an inertial frame equals the time rate of change of the total angular momentum of the system about the origin.
The resultant torque acting on a system about an axis through the center of mass equals the time rate of change of angular momentum of the system regardless of the motion of the center of mass. Angular Momentum of a Rigid Body Consider a rigid object rotating about a fixed axis containing the z-axis; we will now find the angular momentum Each particle of the object rotates on the xy-plane about the z-axis with an angular speed. Now the magnitude of the angular momentum of a particle with mass about the z-axis is: The vectors for the angular momentum and angular velocity are directed along the z-axis. We assume the angular momentum only contains a z-component. Let's take the total sum of of the angular momentum over all particles Let us see what happens if we take the derivative of both sides of the previous equation (with respect to time); Recall that the moment of inertia is constant. This equation is valid for a rotating rigid body about a moving axis only under these conditions:
(1)The moving axis passes through the centre of mass
(2)The moving axis is a symmetry axis When a symmetrical object rotating about a fixed axis passes through its center of mass, we can write in vector form Conservation of Angular Momentum Recall: This expression works for an isolated system Let us compare this to rotational motion of an isolated system If the isolated system is deformable (in which the mass requires redistribution), the moment of inertia changes. In this case, the angular momentum will change as well This applies to rotation about a fixed axis and rotation about an axis through the centre of mass of a moving system (only if the axis remains fixed in direction); as long as the net torque is zero. Now we have three versions of an isolated system model: Note that just because a system is isolated in one of these quantities, it does not mean that it is isolated in the other(s). Central Force A classic example of the conservation of angular momentum is the motion of a particle in a central force field. When the force and position vectors are parallel, the cross product is zero, and therefore no external torque acts on the particle; this applies to central forces.
A central force field is a field in which the force on a particle is always directed towards or away from a point fixed in an inertial frame; this point is also referred to as the force centre.
A central force acting on a particle at a distance r from the force centre is represented as: Where the unit vector for the radical position can be denoted as ; is a function (only depending on the magnitude of F). The force has no torque around the origin, O. Kepler's Second Law: The Area Law We can make the assumption that angular momentum is conserved around the origin, O. Thus a particle’s angular momentum is a constant vector. Kepler’s second law, or the “area law”, demonstrates the conservation of angular momentum of a planet. Observe an ellipse: We must state the condition that is the perigee, or the smallest value of r, occurring when ;
the apogee is denoted as (largest value of r), and occurs when We use the “cross-product area parallelogram”, where the vector for A (area) is swept out by the position vector r. Take the time derivative: Multiply both sides by m: Planets of the solar system cannot follow Kepler’s second law fully. The reason for this is each planet is not just attracted towards the sun; there is also a gravitational field from each of the other planets, which ultimately disrupts the accuracy of Kepler’s second law.
Simply we will analyze the conservation of angular momentum between perigee and apogee positions: Because the net torque at both perigeee and apogee is zero, so is the rate of change of angular momentum; the angle between the position vectors relative to their corresponding linear momentum vectors is 90°, and therefore: Sample Problem #1 Problem A small mass m block slides down a frictionless surface through the height h and sticks to a uniform rod of length l and mass M. The rod is free to rotate about one end on a frictionless pivot, and its center of mass is raised to a distance h’ above its lowest point after collision as shown in the figure below.

(a)Determine the angle in terms of m, M, h, and l.
(b)If the numerical value of h=20cm, l=40cm, m=50g, and M=100g, find the numerical value of . Solution Conservation of Mechanical Energy: Conservation of Angular Momentum: is found using the Moment of Inertia Table and the parallel axis theorem: Angular velocity for the system immediately after the collision Which means the system has kinetic energy This will turn into gravitational potential energy in the final position, where the block has reached a height h’ (relative to the lowest point) and the center of mass of the stick has increased its height by h’/2. From trigonometric considerations, we note that h’=l (1-cos ). Sample Problem #2 Problem A particle of mass m is shot with an initial velocity that makes an angle of with the horizontal as shown in the diagram.
The particle moves in the gravitational field of the Earth. Find the angular momentum of the particle about the origin when the particle is:
(a)At the origin
(b)At the highest point of its trajectory
(c)Just before it hits the ground
(d)What torque causes its angular momentum to change? Solution Key Ideas: MRI: An Overview Earth travels in a nearly circular orbit around the sun; the earth also rotates about its own axis. Every motion the Earth makes involves both orbital and spin angular momentum, wherefore the total angular momentum is a vector sum of the two.
Magnetic Resonance Imaging constructs images of the interior of the body from signals produced by the precessing protons that constitute the nuclei of the hydrogen atom.
Non-invasive; dependent on powerful magnets and radio waves to construct images of interior of body. The imaging is based on the magnetic properties of atoms.
MRI can distinguish between organs because of the subtle differences in the signals generated from various body tissues; therefore, it can tell between benign and malignant tissue.
Because an MRI relies on magnetic forces, it isn’t hard to understand why certain types of metals around the area of concentration on a body can hinder the effective performance of the machine.
Protons in the hydrogen atoms of different chemical compounds recognize the magnetic field modified slightly by the effect of the surrounding electrons. Each hydrogen-containing compound to exhibit one or more different precession rates. This depends on the number of chemically different sites hydrogen atoms occupy the compound. The Magnetic Dipole Moment This diagram illustrates atomic magnetic moment placed in a magnetic field Magnetic Resonance In Medicine
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