Goals Rotational quantities as vectors. Math: Cross Product. Angular momentum

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Physics 106 Week 5

Torque and Angular Momentum as Vectors

SJ 7thEd.: Chap 11.2 to 3

• Rotational quantities as vectors • Cross product

• Cross product

• Torque expressed as a vector • Angular momentum defined • Angular momentum as a vector • Newton’s second law in vector form

1

Goals

Rotational quantities as vectors

Angular momentum

q

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2 So far: simple (planar) geometries

Rotational quantities Δθ, ω, α, τ,etc… represented by positive or negative numbers

Rotation axis was specified simply as CCW or CW

Problems were 2 dimensional with a perpendicular rotation axis

Now: 3D geometries

rotation represented in full vector form

Angular displacement, angular velocity, angular

acceleration as vectors, having direction

The angular displacement, speed, and acceleration

( θ, ω, α )

are vectors with direction. The directions are given by the right-hand rule:

Fingers of right hand curl along the angular direction (See Fig.) Then, the direction of thumb is

h di i f h l

the direction of the angular quantity.

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Example: triad of unit vectors showing rotation in

x-y plane

_+z

kˆ

ω

=

ω

K

+x +y

ω

A disk rotates at 3 rad/s in xy plane as shown above.

What is angular velocity vector?

g

y

Torque “vector” is defined from

position vector and force vector,

Torque as a vector ?

position vector and force vector,

using “cross product”.

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4

Cross Product (Vector Product)

two vectors Æa third vectornormal to the plane they define measures the component of one vector normal tothe other θ= smaller angle between the vectors

)

sin(

ab

b

a

c

G

≡

G

×

G

=

θ

a

G

×

b

G

=

−

b

G

×

a

G

c

G

=

a

G

×

b

G

Math: Cross Product

)

sin(

ab

b

a

c

≡

×

θ

b

G

a

G

b

a

c

×

a

b

a

×

=

−

×

cross product of any parallel vectors = zero

cross product is a maximum for perpendicular vectors cross products of Cartesian unit vectors:

0 kˆ kˆ jˆ jˆ iˆ iˆ× = × = × = jˆ kˆ kˆ jˆ iˆ kˆ iˆ iˆ kˆ jˆ iˆ jˆ jˆ iˆ kˆ × × × − = × == × =− ×

i

k

j

j k k j i= × =− ×

Direction is defined by right hand rule.

More About the Cross Product

Commutative rule does not apply.

G

x ( + ) = x + x

A

G

B

G

C

G

A

G

B

G

A

G

C

G

The distributive rule:

If you are familiar with calculus, the derivative of a cross product obeys th h i l b t th d f th t

, but

,

A B

× ≠ ×

B A

A B

× = − ×

B A

(

)

d

dt

×

=

dt

× + ×

dt

A

B

A B

B

A

G

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ˆ

i

j

k

(

)

(

)

(

)

× =

−

+

−

+

−

G

ˆ

y z z y z x x z x y y x

A B

i

j

k

Cross products using components and unit vectors

_

× =

G

x

y

z

x

y

z

A

B

i

j

k

A B

+

di

ib i

l

d

+

0 kˆ kˆ jˆ jˆ iˆ iˆ× = × = × = jˆ kˆ kˆ jˆ iˆ kˆ iˆ iˆ kˆ jˆ iˆ jˆ jˆ iˆ kˆ × − = × = × =− × == × =− ×

i

k

j

Or, you can use distributive rule and

Calculating cross products using unit vectors

ˆ

2 3 ;

2 =

+

= − +

A

G

i

j B

G

i

j

×

A B

G

Find:

_Where:

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6 Torque as a Cross Product

_τ

G

₌

_r

G

_×

_F

G

• The torque is the cross product of a force vector with the position vector to its point of application.

⊥ ⊥

=

θ

=

τ

r

F

sin(

)

r

F

r

F

• 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) • Right Hand Rule: curl fingers from r to

F, thumb points along torque.

Superposition

:

sum)

(vector

F

r

_i i all i i all i net

G

×

=

τ

=

τ

∑

p p

• Can have multiple forces applied at multiple points.

• Direction of τnet is angular acceleration axis

5.1. A particle located at the position vector (in meters) has a force N acting on it. The torque in N.m about the origin is?

Finding a cross product

) jˆ iˆ ( rG= + ) jˆ iˆ ( Fˆ= 2 +3 kˆ A) 1 kˆ 1 C) kˆ B) 5 kˆ D) 5 jˆ 3 iˆ E) 2 +

What if ?

Fˆ=(3iˆ+3jˆ)

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i

Net torque example: multiple forces at multiple points

1 1 2 2

ˆ

2 N i applied at R = -2m j

ˆ

4 N k applied at R

3m i

F

=

G

k

j

Find the net torque about the origin:

Angular momentum – concepts & definition

-

Linear momentum:

p = mv

- Angular (Rotational) momentum:

L = moment of inertia x angular velocity = I

ω

inertia speed linear rotational m v I ω p=mv L=Iω linear momentum rigid body angular momentum

Angular momentum vector:

_L

=

_I

ω

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8 6.1. A bowling ball is rotating as shown about its mass center

axis. Find it

’

s angular momentum about that axis, in kg.m

2

_/s

Angular momentum of a bowling ball

A) 4 B) ½ C) 7 D) 2 E) ¼

4 d/

I L= ω

ω

= 4 rad/s

M = 5 kg

r = ½ m

I = 2/5 MR

2

Angular momentum of a point particle

2

sin( )

L

=

I

ω

=

mr

ω

=

mv r

_⊥

=

mvr

ϕ

=

mvr

_⊥

= ×

r

G G

p

P

r

G

r

v

_⊥

=

ω

r

G

r

⊥

G

: moment arm

⊥

v

G

r

φ

v

G

⊥

r

Note: L = 0 if v is parallel to r (radially in or out)

Angular momentum vector for a point particle

)

v

r

(

m

p

r

L

K

≡

×

G

=

G

×

G

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Angular momentum of a point particle

O

r

G

r

L

=

rp

T

=

sin

rp

θ

=

r p

p

If

p

=

(

p p

x

,

y

, 0)

G

( , , 0)

_x _y

r

G

=

r r

T

p

r

θ

p

G

p

r

y

( , , )

_x _y

(0, 0,

_x

_y

_x

)

L

G

=

r p

−

r p

5.2. A car of mass 1000 kg moves with a speed of 50 m/s on a circular track of radius 100 m. What is the magnitude of its angular momentum (in kg •m2_/s)

relative to the center of the race track (point “P”) ?

Angular momentum for car

A) 5.0 ×102 A

B) 5.0 ×106 C) 2.5 ×104 D) 2.5 ×106 E) 5.0 ×103

5.3. What would the angular momentum about point “P”be if the car leaves the P

B

g p

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10 Net angular momentum

1

2

3 ...

net

L

G

=

L

G

+

L

G

+

L

G

+

Example: calculating angular momentum for particles

**PP10602-23*:**

Two objects are moving as shown in the figure . What is their total angular momentum about point O?

m

₂