ANGULAR MOMENTUM Torque is the rotational counterpart to force. So whereas when a net force is applied, a body accelerates, when a net torque is applied, a body angularly accelerates. Angular momentum is the rotational counterpart to momentum. So whereas a large momentum suggests it will take a relatively large force to stop a body s translational motion in a given amount of time, a large angular momentum suggests it will take a relatively large torque to top a body s angular motion in a given amount of time. And as usual, we can deduce the form of the angular momentum expression know the form of the momentum expression. Specifically, running through a whole series of comparisons: 1.)
Translational motion: momentum p = mv Rotational motion: angular momentum L = Iω or, if we use translational parameters: L =! r x p! For momentum, if all the forces acting on a group of particles are zero or are internal to the system, momentum is said to be conserved. Likewise, if all the torques acting on a group of particles (or just one) are zero or are internal to the system, then angular momentum is said to be conserved. 2.)
Example: A point mass m circles a fixed point at a distance R units out. If its velocity is v, use both angular momentum relationships to determine the body s angular momentum. or L = Iω ( ) = mr 2 = mvr v R R v m L =! r x p! = ( R) ( mv)sin90 o = mvr Shazam! It doesn t matter which approach you use, you get the same value for the body s angular momentum. 3.)
It is true that the rate at which momentum changes is equal to the net force on an object (Newton s Second Law), or F net = Δp Δt from which we get the impulse relationship FΔt = Δp from which came the conservation of momentum relationship p 1,x + F external,x Δt = p 2,x 4.)
It is also true that the rate at which angular momentum changes is equal to the net torque on an object (rotational version of Newton s Second Law), or Γ net = ΔL Δt from which we get the rotational impulse relationship ΓΔt = ΔL from which came the conservation of angular momentum relationship L 1 + Γ ext Δt = L 2 5.)
What is different between the two conservation relationships is that for momentum, it is common to have an external forces acting over a time interval, whereas for angular momentum, there will hardly ever happens that an external torque acts. In other words, with no external torques acting on a system, L 1 = L 2 6.)
Example: An ice skater with arms out have an angular speed of ω 1 and a moment of inertia I 1. She pulls her arms in so her moment of inertia diminishes to I 2. What happens to her angular speed? I 1 I 2 There are no external torques acting on the woman, so her although her angular momentum and speed will change, her angular momentum will not. Mathematically, this comes out to: L 1 + Γ ext Δt = L 2 I 1 ω 1 + 0 = I 2 ω 2 ω 2 = I 1 ω 1 I 2 Note: her angular speed went up with moment of inertia going down so her angular momentum stays the same, BUT because angular speed governs energy ( ), the mechanical energy in the system goes UP. (This is due to chemical energy burned in her body as she pulls her arms inward.) 7.)
Example: Another standard problem is the merrygo-round problem. A merry-go-round, heretofore assumed to be a disk, has mass M and radius R. It also has two kids who start at the outer edge of the m.g.r. and, pushing with their feet, get the m.g.r. up to an angular speed of ω 1,m.g.r. The kids, each of which have a mass of m k, start to walk toward the center of the m.g.r. When they get to within r units of the center, they stop. What is their speed (or angular speed) at that point? kid walk walk kid There is friction acting on the kids, but that does not produce and EXTERNAL torque. In fact, the torque it produces makes the m.g.r. slow down while angularly accelerating the kids to a higher angular speed. With no external torques, the system s angular momentum must be conserved. Mathematically: 8.)
L 1 = L 2 L 1,m.g.r + L 1,kids = L 2,m.g.r + L 2,kids I m.g.r. ω + I 1,kids ω 1 = I m.g.r. ω 2 + I 2,kids ω 2 1 2 MR2 ω 1 ( ) ( ) + 2 m k R 2 ( ) = 1 2 MR2 ω 1 ( ) ω 2 + 2 m k r 2 For the amusement of it, let s assume the final kid position is r =.2R, and the relationship between the kid s mass and the m.g.r. s mass is m =.4M. With that, we get: 1 2 MR2 ω 1 ( ) + 2.4M (( )R 2 ) ω 1 ( ) = 1 2 MR2 1 2 MR2 + 2 (.4M)R 2 ω 1 ω 2 = 1 2 MR2 + 2 (.4M)(.2R) 2 ω 2 =.58.532 ω 1 ω 2 = 1.1ω 1 ω 2 + 2.4M ω 2 (( )(.2R) 2 ) ω 2 9.)
Angular momentum of a body moving in a straight line. θ mv L =! r x p! = r( mv)sinθ Notice that r sinθ is the same as the y coordinate of the motion. This will ALWAYS be true! Apparently, θ r y = r sinθ L = r( mv)sinθ = ( mv)r sinθ = ( mv)y Notice that is the same as the y coordinate of the motion. This will ALWAYS be true! Apparently, 10.)
The math says it s so, but intuitively, how can you justify the idea that something moving in a straight line had angular momentum? θ r θ mv The key is in the fact that you can break any momentum quantity up into a radial component and a tangential (angular) component. This component is shown in the sketch. It is why a mass moving in a straight line can have angular momentum 11.)
So with this in mind, consider a stationary meter stick of mass M sitting on a frictionless table. A puck of mass m p moving with velocity v o hits the stick at its center of mass. What is conserved during the collision? Energy: As is the case with all collisions, energy is not conserved here unless you are told the collision is elastic, which it isn t in this case. m p v o m p v 1 Mv 2 Momentum: There are no external forces, so momentum is conserved. The math: p 1,x + F ext Δt = p 2,x m p v o + 0 = m p v 1 ( ) + Mv 2 Angular momentum: There are no external torques, so angular momentum, which is initially ZERO, is conserved. 12.)
Now, the same collision but with the puck hitting a distance y units from the stick s center of mass. What is conserved now. Energy: Still no! Momentum: Still yes, with: p 1,x + F ext Δt = p 2,x m p v o + 0 = m p v 1 mp v o ( ) + Mv 2 m p v 1 Angular momentum: Still yes, except there is angular momentum about the stick s center of mass due to the puck s straight-line motion. That angular momentum equation becomes: L 1 + Γ ext Δt = L 2 y m p v o y + 0 = m p v 1 y + I stick,cm ω ω Mv 2 13.)
What s unfortunate, at least from the perspective of problem solving, is that the velocity of the center of mass of the stick is NOT related to the angular velocity about the stick s center of mass by v 2 = rω = yω That s why you usually see this problem in the following alternate form: The stick is pinned at its center of mass, and when the puck hits, it stops dead. In that case: y mp v o pin ω v 1 = 0 v 2 = 0 Momentum is no longer conserved (an external force acts at the pin) but angular momentum is conserved with the final angular momentum being that of the stick only (the puck is not moving). With that, the stick s final angular velocity can be calculated using the conservation of angular momentum equation, or: m p v o y = I stick,cm ω 14.)