Chapter 11: Angular Momentum

Recall Ch. 7: Scalar Product of Two Vectors If A & B are vectors, their Scalar Product is defined as: A  B ≡ AB cosθ In terms of vector components & unit vectors i,j,k are along the x,y,z axes: A = A x i + A y j + A z k B = B x i + B y j + B z k Using i  i = j  j = k  k = 1, i  j= i  k = j  k = 0 gives A  B = A x B x + A y B y + A z B z Dot Product clearly a SCALAR. Sect. 11.1: Vector Product & Torque

Another kind of product of 2 vectors, useful in physics is called the Vector Product or Cross Product. Look closely at the relationship between a torque τ & the force F which produces it. p Ch. 10: We saw: a force F acting on a body at position r produces a torque with magnitude: Figure: F causes a torque τ that rotates the object about an axis perpendicular to BOTH r AND F. Mathematicians have taught us that the torque τ is a VECTOR in the direction of the axis of rotation & can be written

Vector Product Definition If A & B are vectors, their Vector (Cross) Product is defined as: A third vector C is read as “A cross B” The magnitude of vector C is AB sinθ where θ is the angle between A & B

The magnitude of C, which is AB sinθ is equal to the area of the parallelogram formed by A and B. The direction of C is perpendicular to the plane formed by A and B The best way to determine this direction is to use the right-hand rule Vector Product

The vector product is not commutative! Unlike scalars, the order in which the vectors are multiplied is important By the way it’s defined, If A is parallel to B (θ = 0 o or 180 o ), then If A is perpendicular to B, then The vector product is distributive: Vector Product Properties

The derivative of the cross product with respect to some variable, such as t, obeys the “chain rule” of calculus: Note! It is important to preserve the multiplicative order of A and B Vector Product Derivative Properties

Vector Products of Unit Vectors Contrast with scalar products of unit vectors Signs are interchangeable in cross products

The cross product can be expressed as Expanding the determinants gives Vector Products Using Determinants

Given Find Result is Example 11.1

Given the force F and position r: Find the torque τ produced Result is Example 11.2: Torque Vector

Translation-Rotation Analogues & Connections Translation Rotation Displacementx θ Velocityvω Acceleration aα Force (Torque) Fτ Mass (moment of inertia) mI Newton’s 2 nd Law ∑F = ma ∑τ = Iα Kinetic Energy (KE) (½)mv 2 (½)Iω 2 Work (constant F,τ) Fd τθ Momentummv? CONNECTIONS: v = rω, a t = rα a c = (v 2 /r) = ω 2 r, τ = dF, I = ∑(mr 2 )

Sect. 11.2: Angular Momentum Consider a rigid, vertical pole through a frozen ice pond, as in the figure. As a skater moves in a straight line past it, she grabs & holds the pole & suddenly she is going in a circle around the pole. We can analyze this motion by using the concept of Angular Momentum

Consider a particle of mass m located at the vector position r and moving with linear momentum p. See figure  Find the net torque The instantaneous angular momentum L of a particle relative to the origin O is defined as the cross product of the particle’s instantaneous position vector r and its instantaneous linear momentum p: Angular Momentum

Recall Newton’s 2 nd Law in Momentum Form: ∑F = (dp/dt) The net torque is related to the angular momentum in a manner similar to the way that the net force is related to the linear momentum. That is: The net torque acting on a particle is equal to the time rate of change of the particle’s angular momentum This is the most general form of rotational analog of Newton’s 2 nd Law or Newton’s 2 nd Law for Rotations. ∑τ & L must be measured about the same origin. Can show that this reduces to ∑τ = Iα if I is time independent. Torque and Angular Momentum

SI units of angular momentum are (kg. m 2 )/s Both the magnitude and direction of the angular momentum depend on the choice of origin The magnitude is L = mvr sin   is the angle between p and r The direction of L is perpendicular to the plane formed by r and p. Angular Momentum

Particle of mass m moving in a circular path of radius r. The vector is pointed out of the diagram The magnitude is L = mvr sin 90 o = mvr sin 90 o is used since v is perpendicular to r A particle in uniform circular motion has a constant angular momentum about an axis through the center of its path Ex. 11.3: Angular Momentum of a Particle in Circular Motion

Angular Momentum of a System of Particles = vector sum of the angular momenta of the each particle: Differentiating with respect to time gives Newton’s 2 nd Law for Rotations in a many particle system: System has internal forces & external forces. Can show: All torques coming from internal forces add to zero. So, ∑τ i becomes ∑τ ext where now the sum includes external torques only. Therefore, Newton’s 2 nd Law for Rotations in a many particle system becomes: –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 that origin Angular Momentum of a Particle System

A sphere, mass m 1 & a block, mass m 2 are connected by a light cord passing over a pulley which is a thin ring of radius R & mass M. Block slides on a flat, frictionless surface. Find the acceleration a of the sphere & the block using angular momentum & torque methods. Example 11.4: A System of Objects Angular momentum about pulley rotation axis: Pulley rotates, while other 2 objects translate. At time where m 1 & m 2, moving together, have speed v, angular momentum of m 1 is m 1 vR & that of m 2 is m 2 vR. At that same time, angular momentum of pulley is MvR. Total angular momentum: L tot = m 1 vR + m 2 vR + MvR External torque comes solely from weight of m 1, m 1 g. So, ∑τ ext = m 1 gR. Newton’s 2 nd Law:  m 1 gR = d(m 1 vR + m 2 vR + MvR)/dt = (m 1 + m 2 + M)R(dv/dt) or m 1 g = (m 1 + m 2 + M)a. So, a = (m 1 g)/(m 1 + m 2 + M)