Physics 111 Mechanics Lecture 11 Dale Gary NJIT

Physics 111: Mechanics Lecture 11 Dale Gary NJIT Physics Department

Angular Momentum Vectors q Cross Product q Torque using vectors q Angular Momentum q 05 December 2020

Vector Basics We will be using vectors a lot in this course. q Remember that vectors have both magnitude and direction e. g. q You should know how to find the components of a vector from its magnitude and direction q Ways of writing vector notation y ay q You should know how to find a vector’s magnitude and direction from its components ax x 05 December 2020

Projection of a Vector in Three Dimensions Any vector in three dimensions can be projected onto the x-y plane. q The vector projection then makes an angle f from the x axis. q Now project the vector onto the z axis, along the direction of the earlier projection. q The original vector a makes an angle from the z axis. q z f x 05 December 2020 y

Vector Basics (Spherical Coords) You should know how to generalize the case of a 2 -d vector to three dimensions, e. g. 1 magnitude and 2 directions q Conversion to x, y, z components q z q Conversion from x, y, z components f x q Unit vector notation: 05 December 2020 y

A Note About Right-Hand Coordinate Systems q q A three-dimensional coordinate system MUST obey the righthand rule. Curl the fingers of your RIGHT HAND so they go from x to y. Your thumb will point in the z direction. z y x 05 December 2020

Cross Product The cross product of two vectors says something about how perpendicular they are. q Magnitude: q n n is smaller angle between the vectors Cross product of any parallel vectors = zero Cross product is maximum for perpendicular vectors Cross products of Cartesian unit vectors: y j i x k z i j k 05 December 2020

Cross Product q Direction: C perpendicular to both A and B (right-hand rule) n n n q Place A and B tail to tail Right hand, not left hand Four fingers are pointed along the first vector A “sweep” from first vector A into second vector B through the smaller angle between them Your outstretched thumb points the direction of C First practice 05 December 2020

More about Cross Product The quantity ABsin is the area of the parallelogram formed by A and B q The direction of C is perpendicular to the plane formed by A and B q Cross product is not commutative q q The distributive law The derivative of cross product obeys the chain rule q Calculate cross product q 05 December 2020

Derivation How do we show that q Start with ? q q Then q But q So 05 December 2020

Torque as a Cross Product q The torque is the cross product of a force vector with the position vector to its point of application The torque vector is perpendicular to the plane formed by the position vector and the force vector (e. g. , imagine drawing them tail-to-tail) q Right Hand Rule: curl fingers from r to F, thumb points along torque. q Superposition: Can have multiple forces applied at multiple points. q Direction of tnet is angular acceleration axis q 05 December 2020

Calculating Cross Products Find: Where: Solution: i j Calculate torque given a force and its location Solution: 05 December 2020 k

Net torque example: multiple forces at a single point z 3 forces applied at point r : q y Find the net torque about the origin: x i oblique rotation axis through origin Here all forces were applied at the same point. For forces applied at different points, first calculate the individual torques, then add them as vectors, i. e. , use: j 05 December 2020 k

Angular Momentum q Same basic techniques that were used in linear motion can be applied to rotational motion. n n n F becomes m becomes I a becomes v becomes ω x becomes θ Linear momentum defined as q What if mass of center of object is not moving, but it is rotating? q Angular momentum q 05 December 2020

Angular Momentum I q Angular n n n momentum of a rotating rigid object L has the same direction as * L is positive when object rotates in CCW L is negative when object rotates in CW q Angular momentum SI unit: kg-m 2/s Calculate L of a 10 kg disk when = 320 rad/s, R = 9 cm = 0. 09 m L = I and I = MR 2/2 for disk L = 1/2 MR 2 = ½(10)(0. 09)2(320) = 12. 96 kgm 2/s *When rotation is about a principal axis 05 December 2020

Angular Momentum II q Angular momentum of a particle n n r is the particle’s instantaneous position vector p is its instantaneous linear momentum Only tangential momentum component contribute Mentally place r and p tail to tail form a plane, L is perpendicular to this plane 05 December 2020

Angular Momentum of a Particle in Uniform Circular Motion Example: A particle moves in the xy plane in a circular path of radius r. Find the magnitude and direction of its angular momentum relative to an axis through O when its velocity is v. The angular momentum vector points out of the diagram q The magnitude is L = rp sin = mvr sin(90 o) = mvr q A particle in uniform circular motion has a constant angular momentum about an axis through the center of its path q O 05 December 2020

Angular momentum III q Angular n n momentum of a system of particles angular momenta add as vectors be careful of sign of each angular momentum for this case: 05 December 2020

Example: calculating angular momentum for particles Two objects are moving as shown in the figure. What is their total angular momentum about point O? m 2 m 1 Direction of L is out of screen. 05 December 2020

Angular Momentum for a Car q What would the angular momentum about point “P” be if the car leaves the track at “A” and ends up at point “B” with the same velocity ? A) 5. 0 102 B) 5. 0 C) 2. 5 104 D) 2. 5 A 106 B P E) 5. 0 103 05 December 2020

Recall: Linear Momentum and Force Linear motion: apply force to a mass q The force causes the linear momentum to change q The net force acting on a body is the time rate of change of its linear momentum q 05 December 2020

Angular Momentum and Torque Rotational motion: apply torque to a rigid body q The torque causes the angular momentum to change q The net torque acting on a body is the time rate of change of its angular momentum q and are to be measured about the same origin q The origin must not be accelerating (must be an inertial frame) q 05 December 2020

Demonstration q Start from q Expand using derivative chain rule 05 December 2020

What about SYSTEMS of Rigid Bodies? Total angular momentum of a system of bodies: • • individual angular momenta Li all about same origin ti = net torque on particle “i” internal torque pairs are included in sum BUT… internal torques in the sum cancel in Newton 3 rd law pairs. Only External Torques contribute to Lsys net external torque on the system Nonisolated System: If a system interacts with its environment in the sense that there is an external torque on the system, the net external torque acting on a system is equal to the time rate of change of its angular momentum. 05 December 2020

Example: A Non-isolated System A sphere mass m 1 and a block of mass m 2 are connected by a light cord that passes over a pulley. The radius of the pulley is R, and the mass of the thin rim is M. The spokes of the pulley have negligible mass. The block slides on a frictionless, horizontal surface. Find an expression for the linear acceleration of the two objects. 05 December 2020

Masses are connected by a light cord. Find the linear acceleration a. • Use angular momentum approach • No friction between m 2 and table • Treat block, pulley and sphere as a nonisolated system rotating about pulley axis. As sphere falls, pulley rotates, block slides • Constraints: • • • I Ignore internal forces, consider external forces only Net external torque on system: Angular momentum of system: (not constant) same result followed from earlier method using 3 FBD’s & 2 nd law 05 December 2020

Isolated System q Isolated system: net external torque acting on a system is ZERO n n no external forces net external force acting on a system is ZERO 05 December 2020

Angular Momentum Conservation Here i denotes initial state, f is the final state q L is conserved separately for x, y, z direction q For an isolated system consisting of particles, q q For an isolated system that is deformable 05 December 2020

First Example A puck of mass m = 0. 5 kg is attached to a taut cord passing through a small hole in a frictionless, horizontal surface. The puck is initially orbiting with speed vi = 2 m/s in a circle of radius ri = 0. 2 m. The cord is then slowly pulled from below, decreasing the radius of the circle to r = 0. 1 m. q What is the puck’s speed at the smaller radius? q Find the tension in the cord at the smaller radius. q 05 December 2020

Angular Momentum Conservation m = 0. 5 kg, vi = 2 m/s, ri = 0. 2 m, rf = 0. 1 m, vf = ? q Isolated system? q Tension force on m exert zero torque about hole, why? q 05 December 2020

Isolated System Moment of inertia changes 05 December 2020

Controlling spin (w) by changing I (moment of inertia) In the air, tnet = 0 L is constant Change I by curling up or stretching out - spin rate w must adjust Moment of inertia changes 05 December 2020

Example: A merry-go-round problem A 40 -kg child running at 4. 0 m/s jumps tangentially onto a stationary circular merry-goround platform whose radius is 2. 0 m and whose moment of inertia is 20 kg-m 2. There is no friction. Find the angular velocity of the platform after the child has jumped on. 05 December 2020

The Merry-Go-Round The moment of inertia of the system = the moment of inertia of the platform plus the moment of inertia of the person. n Assume the person can be treated as a particle q As the person moves toward the center of the rotating platform the moment of inertia decreases. q The angular speed must increase since the angular momentum is constant. q 05 December 2020

Solution: A merry-go-round problem I = 20 kg. m 2 VT = 4. 0 m/s mc = 40 kg r = 2. 0 m 0 = 0 05 December 2020