Angular Momentum and Torque: Vector Cross Product and Applications, Slides of Physics

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More Angular Momentum

Torque as a Vector

• Suppose we have a wheel spinning about a fixed

axis: then always points along the axis—so

points along the axis too.

• If we want to write a vector equation

it’s clear that the vector is parallel to the

vector : so points along the axis too!

• BUT this vector , is, remember made of two

other vectors: the force and the place

where it acts!

ω

d ω / dt



τ = I α = Id ω / dt

τ

d ω / dt

τ

τ

F

r

Definition: The Vector Cross Product

• The magnitude C is

AB sin θ , where θ is the

angle between the

vectors.

• The direction of is

perpendicular to both

and , and is your right

thumb direction if your

curling fingers go from

to.

• g

A

B

C

θ

C = A × B

  

A B ,

C

A

B

A

B

The Vector Cross Product in

Components

• Recall we defined the

unit vectors

pointing along the x , y , z

axes respectively, and a

vector can be expressed

as

• Now
• So
  • g

i

j

k

x y z

A = A i + A j + A k

i , j k ,

i × i = 0, i × j = k , i × k = − j ,

x y z x y z

y z z y

A B A i A j A k B i B j B k

i A B A B

× = + + × + +

Angular Momentum and Torque

for a Particle

• Angular momentum about the origin of

particle mass m , momentum at

• Rate of change:
• because

p

r

L = r × p

dL dr dp

p r r F

dt dt dt

= × + × = × = τ

dr

p v mv

dt

× = × =

Torque about the origin

Kepler’s Second Law

As the planet moves, a line from the planet to the center

of the Sun sweeps out equal areas in equal times.

• In unit time, it moves through a

distance.

• The area of the triangle swept out

is ½ rv sin θ (from ½ base x height)

• This is ½ L / m ,.
• Kepler’s Law is telling us the

angular momentum about the

Sun is constant: this is because

the Sun’s pull has zero torque

about the Sun itself.

• A

v

Sun

θ

L = r × p

The base of the thin blue triangle

is a distance v along the tangent.

The height is the perp distance

of this tangent from the Sun.

Guy on Turntable Catches a Ball

• A, of mass m , is standing on the edge

of a frictionless turntable, a disk of

mass 4 m , radius R , at rest.

• B, who’s on the ground, throws a ball

weighing 0.1 m at speed v to A, who

catches it without slipping.

• What is the angular momentum of

turntable + man + ball now?

A. 0.1mvR

B. (0.1/3.1)mvR

C. (0.1/5.1)mvR

• n

A

B

Guy on Turntable Walks In

• A, of mass m , is standing on the

edge of a frictionless turntable,

a disk of mass 4 m , radius R ,

which is rotating at 6 rpm.

• A walks to the exact center of

the turntable.

• How fast (approximately) is the

turntable now rotating?

A. 12 rpm

B. 9 rpm

C. 6 rpm

D. 4 rpm

• n

A

Lots of Particles

• Suppose we have particles acted on by

external forces, and also acting on each other.

• The rate of change of angular momentum of

one of the particles about a fixed origin O is:

• The internal torques come in equal and

opposite pairs, so

int ext

i i i

dL dt = τ +τ

ext

i i

i i

dL dt = dL dt = τ

∑ ∑

Rotational Motion of a Rigid Body

• For a collection of interacting particles, we’ve seen

that

the vector sum of the applied torques, and the

being measured about a fixed origin O.

• A rigid body is equivalent to a set of connected

particles, so the same equation holds.

• It is also true (proof in book) that even if the CM is

accelerating,

i

i

dL dt = τ

∑

L

i

τ

CM

CM

dL / dt = τ

∑

When are Angular Velocity and

Angular Momentum Parallel?

• When the rotating object is

symmetric about the axis of

rotation: if for each mass on one

side of the axis, there’s an equal

mass at the corresponding point on

the other side.

• For this pair of masses,

is along the axis.

• (Check it out!)
  • a

1 1 2 2

r × mv + r × mv