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Lecture 5 Torque and Angular Momentum As Vectors SJ 7Thed.: Chap 11.2 to 3

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Lecture 5 Torque and Angular Momentum As Vectors SJ 7Thed.: Chap 11.2 to 3 Physics 106 Lecture 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 “Cross prod uct” represents rotatiilonal quanti iities as vectors: • Cross products point along instantaneous axes of rotation • Directions of rotation axes can be calculated and summed up like other vectors, e.g., G G G G G τi = ri × Fi τi = riFi sin(θi ) τnet = ∑ τi all i Angular momentum - new rotational quantity. • Like linear momentum,G it is conserved for isolated systems G G • Definition: G G G G l i = ri × pi pi = mivi li = ripisin(θi ) Ltot = ∑ l i all i Second Law in terms of conserved quantities: G G G G dp G G • Linear: Fnet = ma ⇒ Fnet = ⇒ p is constant if Fnet = 0 dtG G G G dL G G • Rotational: τ = Iα ⇒ τ = ⇒ L is constant if τ = 0 net net dt net 1 Rotational quantities as vectors: RH Rule Picture applies to: Curl fingers of the right • right-handed coordinates hand in the “sense” of the • displacement θ rotational motion • angular velocity ω Thumb shows direction of a • angular acceleration α rotational vector quantity, • cross product aXb perpendicular to the rotation • torque τ=rxF plane • angular momentum l=rxp Cross product represents • electro-magnetism 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 ˆ ˆ ˆ ˆ ˆ k = i × j = − j × i Both products distinct ˆj = kˆ × ˆi = −ˆi ×kˆ j k from multiplying a ˆi = ˆj ×kˆ = −kˆ × ˆj vector by a scalar 2 More About the Cross Product The quantity AB sin(θ) is equal to the area of the parallelogram formed by A and B The direction of C is perpendicular to the ppylane 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 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 it’s 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 3 Calculating cross products using unit vectors GG G G Find: AB× Where: A = 23;ˆˆijBij+=−+ ˆˆ 2 GG Solution: ABij×=(2ˆˆ + 3 ) ×−+ ( ˆˆij 2 ) =×2()22ˆˆˆˆˆˆˆˆiiijji− +×+× 3()32− +× jj =+04307kkˆˆ + += k ˆ i G Calculate torque Fij=+(2.00ˆˆ 3.00 ) N j k given a force and G rij=+(4.00ˆˆ 5.00 ) m its location: G G G Solution: τ =×=rFij [(4.00ˆˆ + 5.00 )N] × [(2.00 ˆˆij + 3.00 )m] =×+×[(4.00)(2.00)ˆˆii (4.00)(3.00) ˆˆij +×+×(5.00)(2.00)ˆˆji (5.00)(3.00)(5.00)(3.00) ˆˆiˆj × jˆj =⋅2.0kˆ N m More examples: cross product using unit vector expansions Find the magnitude and direction of the torque about the origin for forces applied at point (0, -4.0 m, 3.0 m) due to: a) force F1 with components F1x = 2.0 N and F1y = F1z = 0, and (b) force F2 with components F2x = 0, F2y = 2.0 N, and F2z = 4.0 N? G G G G z τ = r × F r = 0 ˆi − 4 ˆj + 3kˆ G ˆ a) F1 = 2 i G G τ 1 G G ˆ ˆ ˆ ˆ ˆ ˆ ˆ τ 1 = r × F1 = (−4 j + 3k ) × (2 i ) = − (4 x 2) j × i + (3 x 2)k × i ˆ ˆ ˆ ˆ ˆ ˆ kˆ φ1 note: j × i = −k , k × i = + j y G ˆ ˆ ˆ ∴ τ 1 = 6 j + 8k in y-z plane ˆi j G 2 2 | τ 1 |= 6 + 8 = 10 N.m x -1 -1 0 φ 1 ≡ tan (τ 1z /τ 1y ) = tan (8 /6) = 53 K F = 2 ˆj + 4kˆ b) 2 G G G ˆ ˆ ˆ ˆ τ 2 = r × F2 = (−4 j + 3k ) × (2 j + 4k ) = − (4 x 2)ˆj × ˆj − (4 x 4)ˆj × kˆ + (3 x 2)kˆ × ˆj + (3 x 4)kˆ × kˆ note: ˆj × ˆj = kˆ × kˆ = 0, ˆj × kˆ = + ˆi , kˆ × ˆj = − ˆi G ˆ ∴ τ 2 = − 22 i G | τ 2 |= 22 N.m along negative x - axis 4 Net torque example: multiple forces at a single point G z F 3 forces applied at point r : 2 G G G F r ≡ r cos(θ) ˆi + 0 ˆj + r sin(θ) kˆ 1 F3 G G G G ˆ ˆ ˆ o r F1 = 2 i F2 = 2 k F3 = 2 j r = 3 θ = 30 θ y Find the net torque about the origin: G G G G x o τnet = r × Fnet = r × ( F1 + F2 + F3 ) set rx ≡ rsin(θ) = 3sin(30 ) = 1.5 ˆ ˆ ˆ ˆ ˆ o = (rx i + rzk) × (2i + 2j + 2k) rz ≡ rcos(θ) = 3cos(30 ) = 2.6 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ = 2rx i × i + 2rx i × j + 2rx i ×k + 2rzk × i + 2rzk × j + 2rzk ×k ˆ ˆ ˆ ˆ τnet = 0 + 2rxk + 2rx (−)j + 2rz j + 2rz (−)i + 0 i ˆ ˆ oblique rotation axis ∴ τnet = − 3i − 222.2 j + 525.2k through origin j k Here all forces were applied at the same point. For forces applied at different points, first calculate the individual torques, then add them as vectors, G i.e., use: G G G τnet = ∑τi = ∑ri ×Fi (vector sum) all i all i 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) ? 5 Angular momentum – concepts & definition How much linear or rotational “staying power” does a moving object have? - Form the product of an inertia measure and a speed measure. - Linear momentum: p = mv (linear). What if mass center of object is not moving, but it’s rotating? - Rotational (angular momentum): - moment of inertia x angular velocity = Iω = L - linear momentum x moment arm abbtout some axis for siilmple cases 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 Angular momentum – concepts & definition Example: angular momentum of a point mass moving in a straight line choose point P as a rotation axis 2 P G L = Iω = mr ω = mvr = pr v = ωr r ⊥ ⊥ ⊥ ⊥ ⊥ G L = linear momentum X moment arm v Note: L = 0 if moment arm =0 L is the same for all point along line of v Example: same as above, but with velocity not perpendicular to r; v does not affect L rad G G P G L = Iω = mr2ω = mv r = mvr sin(φ) = r ×p ω = v /r r G ⊥ ⊥ v G ⊥ v L = linear momentum X moment arm φ Note: L = 0 if v is parallel to r (radially in or out) Example: angular momentum of a rotating hoop about symmetry axis through “P” L = Iω If it’s a hoop: I = mr2 v = ωr L = mr2ω = mvr = pr G P r same as a point particle, for hoop G v L = linear momentum X moment arm 1 2 1 2 If it’s a disc: I = mr L = mr ω 2 2 6 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 Solution: 1 2 L = Iω I = mr for a dis k 2 1 2 1 2 −4 +2 L = m r ω = × 10 × 9 ×10 × 3.2 ×10 2 2 2 L = 12.96 ≈ 13 Kg m / s What angular speed would a 10 kg SOLID SPHERE have if it’s angular momentum is the same as above? 2 L = 12.96 Kg m /s , r = 9 cm = 0.09 m, m = 10 kg Solution: 2 2 ω = L / I I = m r for a sphere 5 ω = 12.96 × 5 × 10 × 92 ×10−4 2 ω = 400 rad / s Angular momentum defined using cross product • The instantaneousG angular momentum L of a particle relative to the origin O is defined as the cross product of the G G particle’ s instantaneous position L ≡ r × p vector r and its instantaneousG linear momentum p K G G G L ≡ r × p = m(r × v) 7 Angular momentum of a single particle - definition H θ • extension of linear momentum p = mv 90o p • depends on chosen rotation axis (here along z) p • “moment of momentum” ⊥ θ L = r p = p r = pr sin( θ) y ⊥ ⊥ prad same picture as for torques θ line of action use moment arm r ⊥ = r sin(θ)….or..
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