Torque and Angular Momentum As Vectors So Far: Simple (Planar

Torque and Angular Momentum As Vectors So Far: Simple (Planar

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 1 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 two vectors Æ a scalar measures the component of one vector along the other b G G G G a Db = abcos(θ) = axbx + ayby + azbz = b D a 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) two vectors Æ a third vector normal to the plane they define measures the component of one vector normal to the other 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 cross ppyproduct of any parallel vectors = zero b cross product is a maximum for perpendicular vectors 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 The direction of C is perpendicular to the plane formed by A and B The right-hand rule shows the direction. G GGGGG G The distributive rule: A x (BC + ) = ABAC x + x 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 The 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 8.

View Full Text

Details

  • File Type
    pdf
  • Upload Time
    -
  • Content Languages
    English
  • Upload User
    Anonymous/Not logged-in
  • File Pages
    8 Page
  • File Size
    -

Download

Channel Download Status
Express Download Enable

Copyright

We respect the copyrights and intellectual property rights of all users. All uploaded documents are either original works of the uploader or authorized works of the rightful owners.

  • Not to be reproduced or distributed without explicit permission.
  • Not used for commercial purposes outside of approved use cases.
  • Not used to infringe on the rights of the original creators.
  • If you believe any content infringes your copyright, please contact us immediately.

Support

For help with questions, suggestions, or problems, please contact us