Angular Momentum

8.01

W10D1

Today’s Reading Assignment: MIT 8.01 Course Notes:

Chapter 19 Angular Momentum Sections 19.1-19.6

Announcements

Problem Set 7 due Week 10 Tuesday at 9 pm in box outside 26-152 Math Review Week 10 Tuesday at 9 pm in 26-152 Exam 3 Tuesday Nov 26 7:30-9:30 pm Conflict Exam 3 Wednesday Nov 27 8-10 am, 10-12 noon Nov 27 Drop Date

Overview: Torque and Angular Momentum

Angular momentum is defined as

Torque about a point S is equal to the time derivative of the angular momentum about S .

When torque about S is zero, angular momentum is constant

SSddt

= L

τ

LS , f =

LS ,i

S S= ×L r p

Angular Momentum of a Point Particle Point particle of mass m moving with a velocity Momentum Fix a point S Vector from the point S to the location of the object Angular momentum about the point S SI Unit

v

m=p v

Sr

S S= ×L r p

[kg ⋅m2 ⋅ s-1]

Cross Product: Angular Momentum of a Point Particle

Magnitude: a)  moment arm

b)  perpendicular momentum

S S= ×L r p

sinS S θ=L r p

, sinS Sr θ⊥ = r

,S Sr ⊥=L p

, sinSp θ⊥ = p S S p⊥=L r

Angular Momentum of a Point Particle: Direction

Direction: Right Hand Rule

Concept Q. Mag. of Angular Momentum In the above situation where a particle is moving in the x-y plane with a constant velocity, the magnitude of the angular momentum about the point S (the origin) 1)  decreases then increases

2)  increases then decreases

3)  is constant

4)  is zero because this is not circular motion

Table Problem: Angular Momentum and Cross Product

A particle of mass m moves with a uniform velocity At time t, the position vector of the particle with respect to the point S is Find the direction and the magnitude of the angular momentum about the origin, (the point S) at time t.

v = vx i + vy j

rS = x i + y j

Angular Momentum and Circular Motion of a Point Particle:

Fixed axis of rotation: z-axis Angular velocity Velocity Angular momentum about the point S

v = vθθ = Rω zθ

ω ≡ ω z k

LS =

rS ×p = rS × mv = Rmvθ k = mR2ω z k

Table Prob.: Angular momentum Along Axis of Rotation for Circular Motion

A particle of mass m moves in a circle of radius R at an angular speed ω about the z axis in a plane parallel to but a distance h above the x-y plane. a) Find the magnitude and the direction of the angular momentum relative to the origin.

b) Find the z component of .

Hint: Use

0L

0L

r0 = rr + hk

Concept Q.: Direction of Ang. Mom.

1. is constant.

2. is constant but direction of is not. 3. Direction of is constant but is not. 4. has no z-component. .

0L

0L

L0

L0

0L

0L

A particle of mass m moves in a circle of radius R at an angular speed ω about the z axis in a plane parallel to but above the x-y plane. Relative to the origin

Worked Example: Angular Momentum of Two Particles

Two identical particles of mass m move in a circle of radius r, 180º out of phase, at an angular speed ω about the z axis in a plane parallel to but a distance h above the x-y plane. a) Find the magnitude and the direction of the angular momentum relative to the origin.

b) Is this angular momentum relative to the

origin constant? If yes, why? If no, why is it not constant?

0L

Worked Example: Angular Momentum of Two Particles

r0,1 = xi + yj+ hk = rr1 + hk

v = vθ = rωθ r = (x2 + y2 )1/2

L0,1 =

r0,1 × mv = (rr1 + hk)× mrωθ

L0,1 = mr 2ωk − hmrω r1

!

L0,2 =

r0,2 × mv = (rr2 + hk)× mrωθ = mr 2ωk − hmrω r2

L0,1 +

L0,2 = mr 2ωk − hmrω r1 + mr 2ωk − hmrω r2 = 2mr 2ωk

r1 = −r2

Angular Momentum of a Ring

A circular ring of radius R and mass M rotates at an angular speed ω about the z-axis in a plane parallel to but a distance h above the x-y plane. Find the magnitude and the direction of the angular momentum relative to the origin. Divide ring into pairs of small objects with mass

0L

L0,1 +

L0,2 = 2Δmr 2ωk

L0 = mpairr

2ωkpairs∑ = Mr 2ωk

Concept Q. Symmetric Body

A rigid body with rotational symmetry body rotates at a constant angular speed ω about it symmetry (z) axis. In this case

1. is constant.

2. is constant but is not.

3. is constant but is not.

4. has no z-component.

5. Two of the above are true.

0L

0L

0 0/L L

0 0/L L

0L

0L

Angular Momentum of Cylindrically Symmetric Body

A cylindrically symmetric rigid body rotating about its symmetry axis at a constant angular velocity with has angular momentum about any point on its axis

Lz = d

Lz

body∫ = dm rdm

2

body∫ ω z k

= dm rdm2

body∫

⎛

⎝⎜

⎞

⎠⎟ω z k

= Izω z k

ω =ω z k ω z > 0

Kinetic Energy of Cylindrically Symmetric Body

A cylindrically symmetric rigid body with moment of inertia Iz rotating about its symmetry axis at a constant angular velocity

Angular Momentum

Kinetic energy

Krot =

12

Izω z2 =

Lz2

2Iz

ω =ω z k

ω z > 0

Lz = Izω z

Concept Q.: Angular Momentum of Disk

A disk with mass M and radius R is spinning with angular speed ω about an axis that passes through the rim of the disk perpendicular to its plane. The magnitude of its angular momentum is:

1.

14

M R2ω 2

2.

12

M R2ω 2

3.

32

M R2ω 2

4.

14

M R2ω

5.

12

M R2ω

6.

32

M R2ω

Concept Q.: Non-Symmetric Body

A body rotates with constant angular speed ω about the z axis which is not a symmetry axis of the body. Relative to the origin

1. is constant.

2. is constant but is not.

3. is constant but is not.

4. has no z-component.

0L

0L

0 0/L L

0 0/L L

0L

0L

Table Problem: Angular Momentum of a Two Particles

About Different Points Two point like particles 1 and 2, each of mass m, are rotating at a constant angular speed about point A. How does the angular momentum about the point B compare to the angular momentum about point A? What about at a later time when the particles have rotated by 90 degrees?

Conservation of Angular

Momentum

Time Derivative of Angular Momentum for a Point Particle

Angular momentum of a point particle about S: Time derivative of the angular momentum about S: Product rule Key Fact: Result:

( )SS

d ddt dt

= ×L r p

( )S SS S

d dd ddt dt dt dt

= × = × + ×L rr p p r p

SS S S

d ddt dt

= × = × =L r p r F

τ

S Sd d m mdt dt

= ⇒ × = × =r rv v v v 0

LS = rS ×

p

Torque and the Time Derivative of Angular Momentum: Point Particle

Torque about a point S is equal to the time derivative of the angular momentum about S .

S

Sddt

= L

τ

Concept Q.: Change in Angular Mom.

A person spins a tennis ball on a string in a horizontal circle with velocity (so that the axis of rotation is vertical). At the point indicated below, the ball is given a sharp blow (force ) in the forward direction. This causes a change in angular momentum in the

1.  direction

2.  direction

3.  direction

r

θ

k

F

ΔL

v

Conservation of Angular Momentum Rotational dynamics No torques Change in Angular momentum is zero Angular Momentum is conserved

τS =

dLS

dt

ΔLS ≡

LS , f −

LS ,i =

0

0 =τS =

dLS

dt

LS , f =

LS ,i

Demo: Rotating on a Chair

A person holding dumbbells in his/her arms spins in a rotating stool. When he/she pulls the dumbbells inward, the moment of inertia changes and he/she spins faster.

Concept Question: Figure Skater

A figure skater stands on one spot on the ice (assumed frictionless) and spins around with her arms extended. When she pulls in her arms, she reduces her rotational moment of inertia and her angular speed increases. Assume that her angular momentum is constant. Compared to her initial rotational kinetic energy, her rotational kinetic energy after she has pulled in her arms must be

1.  the same. 2.  larger. 3.  smaller. 4.  not enough information is given to decide.

Table Problem: Meteor Encounter

A meteor of mass m is approaching earth as shown on the sketch. The distance h on the sketch below is called the impact parameter. The radius of the earth is Re. The mass of the earth is me. Suppose the meteor has an initial speed of v0. Assume that the meteor started very far away from the earth. Suppose the meteor just grazes the earth. You may ignore all other gravitational forces except the earth. Find the moment arm h for the meteor when it is very far away, (called the impact parameter).