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 Goals Rotational quantities as vectors Math: Cross Product Angular momentum 1 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 thdihe direct ion o fhf the angu lar quantity. 2 Example: triad of unit vectors showing rotation in x- y plane +z K ω = ωkˆ +y +x ω A disk rotates at 3 rad/s in xy plane as shown above. What is anggyular velocity vector? Torque as a vector ? Torque “vector” is defined from position vector and force vector, using “cross product”. 3 Math: Cross Product 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 product 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 Direction is defined by right hand rule. More About the Cross Product Commutative rule does not apply. GGGGGG G G A×≠×BBA, but AB ×=−× BA, G GGGGG G The distributive rule: A x (BC + ) = ABAC x + x If you are familiar with calculus, the derivative of a cross product obeys the chihain ru le, btbut preserves thdfthtthe orderG of the terms:G ddG GGGA dB AB×= ×+× B A dt() dt dt 4 Cross products using components and unit vectors GG ˆˆ ˆ AB×=()AByz − AB zy i +( AB zx − AB xz) j +( AB xy − AB yx) k _ _ ˆˆˆijk GG AB×=Ax AAyz BBBx yz Or, you can use distrib ut ive rul e and + ˆ ˆ ˆ ˆ ˆ ˆi × ˆi = ˆj × ˆj = kˆ ×kˆ = 0 k = i × j = − j × i i ˆj = kˆ × ˆi = −ˆi ×kˆ ˆi = ˆj ×kˆ = −kˆ × ˆj j k Calculating cross products using unit vectors GG G G Find: AB× Where: A = 23;ˆˆijBij+=−+ ˆˆ 2 5 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 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) ? 6 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 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 G Angular momentum vector: L = Iω 7 Angular momentum of a bowling ball 6.1. A bowling ball is rotating as shown about its mass center axis. Find it’s angular momentum about that axis, in kg.m2/s A) 4 B) ½ C) 7 D) 2 E) ¼ ω = 4d/4 rad/s M = 5 kg r = ½ m I = 2/5 MR2 L = Iω 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) Angular momentum vector for a point particle K G G G L ≡ r × p = m(r × v) 8 Angular momentum of a point particle O G Lrprp= == sinθ rp r r Tp p pT G θ p G G ppp= (, ,0) If rrr= (,xy ,,)0) xyy G L =−(0,0,rpx yyx rp ) Angular momentum for car 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”) ? A) 5.0 × 102 A 6 B) 5.0 × 10 B 4 C) 2.5 × 10 P D) 2.5 × 106 E) 5.0 × 103 5.3. What would the anggpular momentum about point “P” be if the car leaves the track at “A” and ends up at point “B” with the same velocity ? A) Same as above B) Different from above C) Not Enough Information L = r ⊥ p = p ⊥r = pr sin( θ) 9 Net angular momentum GGGG LLLLnet =+++123... 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 10.