sign in sign up
<<
Home , Angular momentum, Circular motion, Mass, Speed, Velocity

Angular

8.01 W10D1

Today’s Reading Assignment: MIT 8.01 Course Notes: Chapter 19 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: and Angular Momentum

Angular momentum is defined as    LS = rS ×p

Torque about a point S is equal to the of the angular momentum about S .   dL τ = S S dt

When torque about S is zero, angular momentum is constant   L = L S , f S ,i

Angular Momentum of a Point

Point particle of m moving with a v   Momentum p = mv

Fix a point S  Vector r S from the point S to the location of the object

   Angular momentum about the point S LS = rS ×p

2 -1 SI Unit [kg ⋅ m ⋅s ] : Angular Momentum of a    L = r ×p : S S    L = r p sinθ S S a) arm  rS ,⊥ = rS sinθ   LS = rS ,⊥ p b) momentum    L r p pS ,⊥ = p sinθ S = S ⊥ 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 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

Table Problem: Angular Momentum and Cross Product

A particle of mass m moves with a uniform velocity  v = v ˆi + v ˆj x y At time t, the vector of the particle with respect to the point S is  r x ˆi y ˆj S = + Find the direction and the magnitude of the angular momentum about the origin, (the point S) at time t.

Angular Momentum and of a Point Particle:

Fixed axis of : z-axis  ω ≡ ω kˆ z

 Velocity ˆ ˆ v = vθθ = Rω zθ

Angular momentum about the point S

     2 L = r × p = r × mv = Rmv kˆ = mR ω kˆ S S S θ z Table Prob.: Angular momentum Along Axis of Rotation for Circular Motion A particle of mass m moves in a of R at an angular ω about the z axis in a plane parallel to but a h above the x-y plane.

a) Find the magnitude and the  direction of the angular momentum L0 relative to the origin.  b) Find the z component of L 0 .

 Hint: Use r = rrˆ + hkˆ 0 Concept Q.: Direction of Ang. Mom. 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

 1. L is constant. 0

 2.  is constant but direction of is not. L0 L0

  L 3. Direction of L 0 is constant but 0 is not.

 4. L has no z-component. . 0 Worked Example: Angular Momentum of Two Two 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 L 0 relative to the origin.

b) Is this angular momentum relative to the origin constant? If yes, why? If no, why is it not constant? Worked Example: Angular Momentum of Two Particles  r = xˆi + yˆj+ hkˆ = rrˆ + hkˆ 0,1 1 r = (x2 + y2 )1/2  ˆ ˆ v = vθ = rωθ    L = r × mv = (rrˆ + hkˆ ) × mrωθˆ 0,1 0,1 1  2 ˆ L0,1 = mr ωk − hmrωrˆ1 !

   2 L = r × mv = (rrˆ + hkˆ ) × mrωθˆ = mr ωkˆ − hmrωrˆ 0,2 0,2 2 2 rˆ rˆ 1 = − 2   L + L = mr 2ωkˆ − hmrωrˆ + mr 2ωkˆ − hmrωrˆ = 2mr 2ωkˆ 0,1 0,2 1 2 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 L 0 relative to the origin. Divide ring into pairs of small objects with mass   2 L + L = 2Δmr ωkˆ 0,1 0,2  L m r 2 kˆ Mr 2 kˆ 0 = ∑ pair ω = ω pairs Concept Q. Symmetric Body

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

 1. is constant. L0

   2. L is constant but is not. 0 L0 / L0

3.   is constant but  is not. L0 / L0 L0

 4. L 0 has no z-component.

5. Two of the above are true.

Angular Momentum of Cylindrically Symmetric Body

A cylindrically symmetric rigid body rotating about its symmetry axis at a constant angular  velocity ω = ω kˆ with ω > 0 has angular z z momentum about any point on its axis

  L = dL = dm r 2 ω kˆ z ∫ z ∫ dm z body body ⎛ ⎞ = dm r 2 ω kˆ ⎜ ∫ dm ⎟ z ⎝ body ⎠ = I ω kˆ z z Kinetic of Cylindrically Symmetric Body

A cylindrically symmetric rigid body with moment of Iz rotating about its symmetry axis at a constant angular velocity  ω = ω kˆ z ω > 0 z Angular Momentum Lz = Izω z

2 1 2 L K = I ω = z rot 2 z z 2I 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 1 1. MR2ω 2 4. MR2ω 4 4 1 1 2. MR2ω 2 5. MR2ω 2 2 3 3 3. MR2ω 2 6. MR2ω 2 2 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. L 0 is constant.

   2. L 0 is constant but L 0 / L 0 is not.

   3. L 0 / L 0 is constant but L 0 is not.  4. L 0 has no z-component.

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: L = r × p S S  of the angular momentum about S: dLS d   = (rS ×p) dt dt   dLS d   drS   d  Product rule = (rS ×p)= ×p + rS × p dt dt dt dt    dr dr     Key Fact: v = S ⇒ S × mv = v× mv = 0 dt dt Result:  dL  d     S = r × p = r ×F = τ dt S dt S S 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 .

  dLS τ = S dt Concept Q.: Change in Angular Mom.

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

1. rˆ direction

ˆ 2. θ direction

3. kˆ direction Conservation of Angular Momentum  dL Rotational  S τ S = dt  No   dLS 0 = τ S = dt

Change in Angular momentum is zero     ΔL ≡ L − L = 0 S S , f S ,i

Angular Momentum is conserved   L = L S , f S ,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 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 as shown on the sketch. The distance h on the sketch below is called the 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 except the earth. Find the moment arm h for the meteor when it is very far away, (called the impact parameter).