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Angular Momentum

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Angular Momentum Angular Momentum Objective: State why angular momentum is a useful concept; define the magnitude of angular momentum; define the angular momentum vector; correctly calculate a cross product of two vectors and use the right-hand rule. Angular Momentum Suppose that a child running in a direction tangent to a merry-go-round jumps onto the edge. As a result, the merry-go-round will rotate, but how fast? If we apply the momentum principle to the child, we can calculate the force of the merry-go-round on the child, or if we apply the momentum principle to the merry-go-round, we can calculate the force of the ground on the merry-go-round. If we apply the energy principle to the merry-go-round-child system, we can calculate the change in thermal energy due to the collision. However, neither of our fundamental principles so far will help us calculate how fast it rotates. We need a new fundamental principle that will apply to rotating systems–the angular momentum prin- ciple. The quantity, angular momentum, is a quantized quantity for electrons and nuclei and is essential to understanding quantized atomic and nuclear energy levels and allowed atomic energy transitions. The angular momentum of a proton (i.e. Hydrogen nucleus) and its quantized values is used for nuclear magnetic resonance experiments and its associated application in imaging, MRI. To distinguish between momentum and angular momentum, we will call ~p the linear momentum of a particle. Suppose a particle with momentum, ~p moves past a point A, The magnitude of the angular momentum of the particle with respect to point A is |L~ A| = r⊥|~p| (1) Remember that since we get tired of writing magnitudes of vectors with absolute value signs and vector symbols, we will use the more simple notation: LA = r⊥p (2) r⊥ is the shortest distance (or perpendicular distance) between point A and an infinite line passing through the particle and its momentum vector. If we write the position vector from point A to the particle as ~rA, then r⊥ = rAsin(θ). Thus, LA = rA sin(θ)p = rAp sin(θ) (3) where θ is the smallest angle between the position vector and the momentum vector of the particle when drawn tail to tail. But this only gives us the magnitude of the angular momentum vector. To correctly calculate its compo- nents, we must multiply the position vector and the momentum vector of the particle. How can we multiply two vectors? Well, we learned about the scalar product (or dot product), but that gives us a scalar. We need a vector! There’s another method to multiply vectors called the vector product or the cross product. L~ = ~r × ~p (4) Cross Product If A~ × B~ , then |A~ × B~ | = AB sin(θ) (5) A~ × B~ =< (AyBz − AzBy), (AzBx − AxBz), (AxBy − AyBx) > (6) The direction of A~ × B~ is perpendicular to the plane containing A~ and B~ . Its direction relative to that plane is determined by the right-hand-rule. Straighten your fingers point in the direction of A~ and curl them toward B~ (and you can’t curl your hand backwards!). Your thumb points in the direction of A~ × B~ . Multiparticle systems For a multiparticle system, we already know that P~tot = ~p1 + ~p2 + ... = ~pcm (7) and 1 1 1 K = m v2 + m v2 + m v2 + ... = K + K (8) tot 2 1 1 2 2 2 2 3 3 trans rel What about angular momentum? The total angular momentum is the sum of the angular momenta of the individual particles. Thus, ~ ~ ~ LtotA = L1A + L2A + ... (9) However, just as with kinetic energy, there is an easier way to calculate it. The total angular momentum can be equivalently written as the sum of the angular momentum of the center of mass (as if the object is a point particle) plus the angular momentum of the system relative to the center of mass. Therefore, ~ ~ ~ LtotA = LtransA + Lcm + ... (10) Rigid object Often we are analyzing rotating rigid object; that is, an object for which every point has the same angular speed, rotating about the same axis. The points may, however, have different linear speeds depending on their distances from the axis of rotation. The linear speed of a point is related to its angular speed by v = rω (11) In this case, we can define the angular velocity vector ~ω with a direction given by a right-hand rule. Curl your fingers in the direction the object is rotating with your thumb along the axis of rotation. The angular velocity vector is in the direction of your thumb. The angular momentum can then be written as a constant times the angular velocity. That constant I, called the moment of inertia, depends on the total mass of the object and how that mass is distributed about the axis of rotation (this means that where the axis of rotation is located is important!). L~ = I~ω (12) The moment of inertia of a particle is 2 I = mr⊥ (13) Thus, to get the moment of inertia of a rigid body, break it into pieces and add up the moment of inertia of each piece–that’s just an integral! Application 1. A 40 kg child runs at a constant speed of 3.0 m/s in a tangential direction and jumps onto the edge of a merry-go-round of radius 2.0 m. (a) What is the angular momentum of the child just before landing on the merry-go-round? (b) What is her angular momentum 10.0 m before she lands on the merry-go-round? (c) What if she ran in the same direction with the same speed but jumped onto the other side? 2. A 2-D rectangular array of tiny particles, each of mass m and a distance a apart, can rotate clockwise about an axis through the 3 center particles. (a) What is the angular momentum of the system? (b) What if the axis was along 3 of the side particles? 3. What is the moment of inertia of a hollow ring of mass M? 4. Which has a greater moment of inertia, a solid disk of mass M and radius R or a hollow ring of mass M and radius R? 5. What is the angular momentum of the Earth (consider both translational angular momentum and rotational angular momentum)?.
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