Engineering Physics : Lecture 12 Angular Momentum: Definitions & Derivations What does it mean? Rotation about a fixed axis L = I Example: Two disks Student on rotating stool Angular momentum of a freely moving particle Bullet hitting stick Student throwing ball

Example Rotations A girl is riding on the outside edge of a merry-go-round turning with constant w. She holds a ball at rest in her hand and releases it. Viewed from above, which of the paths shown below will the ball follow after she lets it go? (a) (b) (c) (d) w

Solution Just before release, the velocity of the ball is tangent to the circle it is moving in. w

Solution After release it keeps going in the same direction since there are no forces acting on it to change this direction. w

Angular Momentum: Definitions & Derivations We have shown that for a system of particles Momentum is conserved if What is the rotational version of this?? p = mv The rotational analogue of force F is torque  Define the rotational analogue of momentum p to be angular momentum

Definitions & Derivations... First consider the rate of change of L: So (so what...?)

Definitions & Derivations... Recall that  EXT Which finally gives us: Analogue of !!

What does it mean? where and In the absence of external torques Total angular momentum is conserved

Angular momentum of a rigid body about a fixed axis: Rolling chain Consider a rigid distribution of point particles rotating in the x-y plane around the z axis, as shown below. The total angular momentum around the origin is the sum of the angular momenta of each particle: (since ri and vi are perpendicular) v1 We see that L is in the z direction. m2 j Using vi =  ri , we get r2 r1 m1 i v2  r3 m3 v3 I  Analogue of p = mv!!

Angular momentum of a rigid body about a fixed axis: In general, for an object rotating about a fixed (z) axis we can write LZ = I  The direction of LZ is given by the right hand rule (same as ). We will omit the Z subscript for simplicity, and write L = I  z 

Example: Two Disks A disk of mass M and radius R rotates around the z axis with angular velocity i. A second identical disk, initially not rotating, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity f. z f z i

Example: Two Disks First realize that there are no external torques acting on the two-disk system. Angular momentum will be conserved! Initially, the total angular momentum is due only to the disk on the bottom: z 2 1 0

Example: Two Disks First realize that there are no external torques acting on the two-disk system. Angular momentum will be conserved! Finally, the total angular momentum is due to both disks spinning: z 2 1 f

Example: Two Disks Since Li = Lf An inelastic collision, since E is not conserved (friction)! z f z Li Lf f

Example: Rotating Table A student sits on a rotating stool with his arms extended and a weight in each hand. The total moment of inertia is Ii, and he is rotating with angular speed i. He then pulls his hands in toward his body so that the moment of inertia reduces to If. What is his final angular speed f? i f Ii If

Example: Rotating Table... Again, there are no external torques acting on the student-stool system, so angular momentum will be conserved. Initially: Li = Iii Finally: Lf = If f i f Ii If Li Lf

Example: Angular Momentum A student sits on a freely turning stool and rotates with constant angular velocity w1. She pulls her arms in, and due to angular momentum conservation her angular velocity increases to w2. In doing this her kinetic energy: (a) increases (b) decreases (c) stays the same w1 w2 I1 I2 L L

Solution (using L = I) L is conserved: I2 < I1 K2 > K1 K increases! w1 w2 I2 I1 L

Solution Since the student has to force her arms to move toward her body, she must be doing positive work! The work/kinetic energy theorem states that this will increase the kinetic energy of the system! w1 w2 I1 I2 L L

Angular Momentum of a Freely Moving Particle We have defined the angular momentum of a particle about the origin as This does not demand that the particle is moving in a circle! We will show that this particle has a constant angular momentum! y x v

Angular Momentum of a Freely Moving Particle... Consider a particle of mass m moving with speed v along the line y = -d. What is its angular momentum as measured from the origin (0,0)? y x d m v

Angular Momentum of a Freely Moving Particle... We need to figure out The magnitude of the angular momentum is: Since r and p are both in the x-y plane, L will be in the z direction (right hand rule): y x r d p=mv  

Angular Momentum of a Freely Moving Particle... So we see that the direction of L is along the z axis, and its magnitude is given by LZ = pd = mvd. L is clearly conserved since d is constant (the distance of closest approach of the particle to the origin) and p is constant (momentum conservation). y x d p

Sample problem: Torque, Penguin Fall Calculations: The magnitude of l can be found by using The perpendicular distance between O and an extension of vector p is the given distance D. The speed of an object that has fallen from rest for a time t is v =gt. Therefore, To find the direction of we use the right-hand rule for the vector product, and find that the direction is into the plane of the figure. The vector changes with time in magnitude only; its direction remains unchanged. (b) About the origin O, what is the torque on the penguin due to the gravitational force ? Calculations: Using the right-hand rule for the vector product we find that the direction of t is the negative direction of the z axis, the same as l. Sample problem: Torque, Penguin Fall 24

Example: Bullet hitting stick A uniform stick of mass M and length D is pivoted at the center. A bullet of mass m is shot through the stick at a point halfway between the pivot and the end. The initial speed of the bullet is v1, and the final speed is v2. What is the angular speed F of the stick after the collision? (Ignore gravity) M F D m D/4 v1 v2 initial final

Example: Bullet hitting stick... Conserve angular momentum around the pivot (z) axis! The total angular momentum before the collision is due only to the bullet (since the stick is not rotating yet). M D m D/4 v1 initial

Example: Bullet hitting stick... Conserve angular momentum around the pivot (z) axis! The total angular momentum after the collision has contributions from both the bullet and the stick. where I is the moment of inertia of the stick about the pivot. F D/4 v2 final

Example: Bullet hitting stick... Set Li = Lf using M F D m D/4 v1 v2 initial final

Example: Throwing ball from stool A student sits on a stool which is free to rotate. The moment of inertia of the student plus the stool is I. She throws a heavy ball of mass M with speed v such that its velocity vector passes a distance d from the axis of rotation. What is the angular speed F of the student-stool system after she throws the ball? M v F d I I top view: initial final

Example: Throwing ball from stool... Conserve angular momentum (since there are no external torques acting on the student-stool system): Li = 0 Lf = 0 = IF - Mvd M v F d I I top view: initial final

Angular Momentum A student is riding on the outside edge of a merry-go-round rotating about a frictionless pivot. She holds a heavy ball at rest in her hand. If she releases the ball, the angular velocity of the merry-go-round will: (a) increase (b) decrease (c) stay the same w2 w1

Solution The angular momentum is due to the girl, the merry-go-round and the ball. LNET = LMGR + LGIRL + LBALL Initial: w v R m same Final: w v m

Solution Since LBALL is the same before & after, w must stay the same to keep “the rest” of LNET unchanged. w w

Conceptual answer Since dropping the ball does not cause any forces to act on the merry-go-round, there is no way that this can change the angular velocity. Just like dropping a weight from a level coasting car does not affect the speed of the car. w2 w

Where applicable, provide a numerical answer with appropriate units. Example Door Below is a closed swinging door leading to a restaurant kitchen, seen from above. A meatball has been thrown at the door. The door has a mass of M = 4.0 kg, a length of L = 1.0 m, and a rotational inertia of I = ⅓ML2 about the hinge. The m = 0.2 kg meatball is about to strike the door at a point ¾ down the length, also shown below. The meatball is moving horizontally toward the door and has a speed of 15.0 m/s on impact. (Neglect its vertical motion due to gravity.) The meatball strikes the door and sticks to it. Both the meatball and the door fly open. Door jamb Where applicable, provide a numerical answer with appropriate units.

p = mv = (0.2 kg)(15 m/s) = 3 kg∙m/s Example Door Door: M = 4.0 kg, L = 1.0 m, and a rotational inertia of I = ⅓ML2 Meatball: m = 0.2 kg strikes the door at a point ¾ down the length, with speed of 15.0 m/s . Door jamb a. What are the initial linear momentum, angular momentum about the hinge, and kinetic energy of the meatball? (Before it strikes the door.) p = mv = (0.2 kg)(15 m/s) = 3 kg∙m/s L = r×p = rmv = (3L/4)(m)(v) = (0.75 m)(0.2 kg)(15 m/s) = 2.25 kg∙m2/s K = ½mv2 = (0.5)(0.2 kg)(15 m/s)2 = 22.5 J

b. Which of these (p, L, K) are conserved in the collision? Explain. Example Door Door: M = 4.0 kg, L = 1.0 m, and a rotational inertia of I = ⅓ML2 Meatball: m = 0.2 kg strikes the door at a point ¾ down the length, with speed of 15.0 m/s . Door jamb b. Which of these (p, L, K) are conserved in the collision? Explain. Only angular momentum is conserved. There is no net torque here. There is a net force applied by the hinge, and there is energy transferred out of the system to heat, sound, deformation, etc.

So ωf = Li / If = (2.25 kg∙m2/s) / (1.45 kg∙m2) = 1.6 rad/s. Example Door Door: M = 4.0 kg, L = 1.0 m, and a rotational inertia of I = ⅓ML2 Meatball: m = 0.2 kg strikes the door at a point ¾ down the length, with speed of 15.0 m/s . Door jamb c. What is the angular velocity of the door/meatball combination after the collision? Li = Lf = Ifωf = (Imeatball + Idoor)ωf =[(0.2 kg)(0.75 m)2 + 1/3(4 kg)(1 m)2)]ωf = (1.45 kg∙m2)ωf So ωf = Li / If = (2.25 kg∙m2/s) / (1.45 kg∙m2) = 1.6 rad/s. d. How long after the collision is it until the door hits the door jamb (that is, until θ = 90°)? θ = ωt (no angular acceleration) so t = θ / ω = (π/2) / (1.6 rad/s) = 1 second.

Gyroscopic Motion: Suppose you have a spinning gyroscope in the configuration shown below: If the left support is removed, what will happen?? pivot support  g

Gyroscopic Motion... Suppose you have a spinning gyroscope in the configuration shown below: If the left support is removed, what will happen? The gyroscope does not fall down! pivot  g

Gyroscopic Motion... ... instead it precesses around its pivot axis ! Bicycle wheel ... instead it precesses around its pivot axis ! This rather odd phenomenon can be easily understood using the simple relation between torque and angular momentum we derived in Lecture 22. pivot 

Gyroscopic Motion... d L pivot  mg The magnitude of the torque about the pivot is  = mgd. The direction of this torque at the instant shown is out of the page (using the right hand rule). The change in angular momentum at the instant shown must also be out of the page! d L pivot  mg

Gyroscopic Motion... L(t) dL d pivot L(t+dt) top view Consider a view looking down on the gyroscope. The magnitude of the change in angular momentum in a time dt is dL = Ld. So where is the “precession frequency” L(t) dL d pivot L(t+dt) top view

Gyroscopic Motion... Toy Gyroscope So In this example,  = mgd and L = I: The direction of precession is given by applying the right hand rule to find the direction of  and hence of dL/dt. d  L pivot  mg

Recap of today’s lecture Angular Momentum: Definitions & Derivations What does it mean? Rotation about a fixed axis L = I Example: Two disks Student on rotating stool Angular momentum of a freely moving particle Bullet hitting stick Student throwing ball Skip: Ch. 12 Equilibrium (topics embedded in previous chapters), Ch. 13 Gravitation, Ch. 14 Fluids