Physics 106 Week 5 Torque and Angular Momentum as Vectors SJ 7thEd.: Chap 11.2 to 3

• Rotational quantities as vectors • Cross product • Torque expressed as a vector • Angular momentum defined • Angular momentum as a vector • Newton’s second law in vector form

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So far: simple (planar) geometries Rotational quantities Δθ, ω, α, τ, etc… represented by scalars 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

1 Rotational quantities as vectors: RH Rule

Curl fingers of the right hand in the “sense” of the rotational motion Thumb shows direction of a rotational vector quantity, perpendicular to the rotation plane Cross product represents this computationally

Example: triad of unit vectors showing rotation in x-y plane

z K ω = ωkˆ

y

x ω Right Hand Rule applied to cross product

Multiplying Vectors (Review) Dot Product (Scalar Product) G ƒ two vectors Æ a scalar ƒ measures the component of one vector along the other b G G G G a Db = abcos(θ) = axbx + ayby + azbz = b D a ƒ dot products of Cartesian unit vectors: θ G ˆi D ˆj = ˆi Dkˆ = ˆj Dkˆ = 0 ˆi D ˆi = ˆj D ˆj = kˆ Dkˆ = 1 a

Cross Product (Vector Product) ƒ two vectors Æ a third vector normal to the plane they define ƒ measures the component of one vector normal to the other ƒ θ = smaller angle between the vectors G G G G G G G G G G c ≡ a ×b = ab sin(θ) a ×b = −b × a c = a ×b G ƒ cross ppyproduct of any parallel vectors = zero b ƒ cross product is a maximum for perpendicular vectors ƒ cross products of Cartesian unit vectors: θ ˆ ˆ ˆ ˆ ˆ ˆ G i × i = j × j = k ×k = 0 i a kˆ = ˆi × ˆj = −ˆj × ˆi ˆj = kˆ × ˆi = −ˆi ×kˆ j k ˆi = ˆj ×kˆ = −kˆ × ˆj

2 More About the Cross Product

ƒThe direction of C is perpendicular to the plane formed by A and B ƒThe right-hand rule shows the direction.

G GGGGG G ƒThe distributive rule: A x (BC + ) = ABAC x + x

ƒ Calculate cross products using A & B written in terms of the unit vectors (just multiply the terms out, or use determinants). K G ˆ ˆ ˆ A × B = (AyBz − AzBy ) i + (AzBx − AxBz ) j + (AxBy − AyBx ) k

ƒThe derivative of a cross product obeys the chain rule, but preserves the order of the terms: G G ddG GGGA dB AB×= ×+× B A dt() dt dt

Calculate cross products using components and unit vectors GG ˆˆ ˆ AB×=()AByz − AB zy i −( AB xz − AB zx) j +( AB xy − AB yx) k _ _ ˆˆˆijk GG AB×=Ax AAyz

BBBx yz

+

3 Calculating cross products using unit vectors GG G G Find: AB× Where: A = 23;ˆˆijBij+=−+ ˆˆ 2

G G G Torque as a Cross Product τ = r × F

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

τ = r F sin( θ) = 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. Suppperposition: G G G G τnet = ∑τi = ∑ri ×Fi (vector sum) all i all i

• Can have multiple forces applied at multiple points.

• Direction of τnet is angular acceleration axis

4 Finding a cross product G ‰ 5.1. A particle located at the position vector r = ( ˆi + ˆj ) (in meters) has a force Fˆ = (2ˆi + 3ˆj) N acting on it. The torque in N.m about the origin is?

A) 1 kˆ

B) 5 kˆ

C) - 1 kˆ

D) - 5 kˆ

E) 2ˆi + 3ˆj

What if Fˆ = (3 ˆi + 3 ˆ j) ?

Net torque example: multiple forces at multiple points GG ˆˆ F11 = 2 N i applied at R = -2m j GG F == 4 N kˆˆ applied at R 3m i 22i

Find the net torque about the origin: j k

5 Angular momentum – concepts & definition

- Linear momentum: p = mv

- Angular (Rotational) momentum: L = moment of inertia x angular velocity = Iω

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

G ω G L = the angular momentum of a rigid L body relative to a selected axis about which I and ω are measured: G G • units: [kg.m2 /s] L ≡ Iω

Example: Calculating Angular Momentum for a Rigid Body

Calculate the angular momentum of a 10 kg disc when: ω = 320 rad / s, r = 9 cm = 0.09 m, m = 10 kg

6 Angular momentum of a point particle 2 G G L== Iωω mr = mv⊥⊥ r = mvrsin( ϕ ) = mvr =× r p

v⊥ = ω r P G G : moment arm G r G r⊥ r⊥ v G ⊥ v φ

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

K G G G L ≡ r × p = m(r × v)

Net angular momentum of particles G G K G G G Lnet =L 1 +L 2 +...+L n = ∑ L i = ∑ r i ×p i all i all i

7 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?

m2

m1

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