ROTATIONAL DYNAMICS Newton's 3rd Law – A Closer Look ● Consider 2 particles with action/reaction pair of forces: – rd = − Newton's 3 Law → F 1 on 2 F 2 on 1 2 – Puts no restrictions on the direction of F 1 on 2 r 12 ● Symmetry Considerations 1 – Attempt to write a formula for the force F 1 on 2 – No “universal” xyz directions → What can F 1 on 2 depend on? – Relative position vector r 12 and relative velocity vector v 12 only ● If F 1 on 2 depends only on r 12 mathematically: – Direction of r 12 → only “defined” direction in space – F must point in direction of r (or opposite direction) 1 on 2 12 – Forces of this type are called “central” forces Central Forces and Torque 2 ● × = Mathematical definition of central force: r 12 F 1 on 2 0 r 12 – True in every reference frame! y 1 ● Calculating in a particular reference frame S: x × r × F = 0 The quantity r i F i is called the “torque” 12 1 on 2 th ( i) on the i particle − × = r 2 r 1 F 1 on 2 0 Internal central forces produce zero net torque × − × = r 2 F 1 on 2 r 1 F 1 on 2 0 on a system × × = r 2 F 1 on 2 r 1 F 2 on 1 0 External forces and non-central internal forces can exert non-zero net torque on a system ● Examples: – Central forces: gravity, electric – Non-central force: magnetic (depends on position and velocity) Static Equilibrium ● “Equilibrium” → Fnet = 0 → aCM = 0 ● “Static Equilibrium” → ai = 0 for every particle – Examples: buildings, bridges → (not perfectly static!) – = = ∑ × = Requires: F net , external 0 and net , external r i F i 0 i – Useful for calculating structural loads and stresses ● Examples: d – 3d/4 Shelf → calculate tension in cable m 2m – Calculate force of wall on plank d – Door of width w and height h → draw direction of each Fhinge Angular Momentum ● Consider the net torque on a system of particles d p d = ∑ × = ∑ × i = ∑ r × p net r i F i net r i net i i i i dt dt i – × th r i p i is referred to as “angular momentum”( L i ) of i particle – Internal, central forces exert zero net torque – Net torque must be provided by external forces: d L = total (Similar to Newton's 2nd Law ) net , external dt – If zero net external torque → angular momentum is conserved ● Both and L depend on choice of origin – Unlike force and momentum (only depend on xyz directions) – However, the equation above is true in all reference frames Examples ● Pendulum at some instant (angle θ, speed v) θ d – Using top of string as origin: – Calculate torque and angular momentum m – Plug in to Tnet = dL/dt Repeat, using mass's lowest point as origin d/4 ● Wooden board falls off table d – Mass m, starting from rest – Using edge of table as origin: – Calculate Tnet and aright edge of board at t=0 – (Assume board stays rigid → v proportional to r) ● Why does a helicopter need a tail rotor? CM Frame of a Physical System ● Empty space → homogeneous and isotropic – All points are physically the same → no point is “special” – All directions physically the same → no direction is “special” y y' y' x' y' vCM x' x' x ● System of particles → breaks this symmetry – A “special” reference frame can be defined for the system: – Origin = CM of system (called the “CM frame”) – x'y'z' axes → defined by particle positions (“principal axes”) – Angular velocity vector CM → defined by particle velocities Angular Momentum in CM Frame N = = 1 2 ∑ 1 2 psystem M system v CM KE system M system vCM mi v ' i 2 i = 1 2 ● For a system of particles measured in the CM frame: – Total momentum must be zero, but total KE can be non-zero! ● Angular Momentum (calculated in an inertial frame S): = ∑ [ × ] = ∑ [ × ] Lsystem r i pi rCM r i ' mi vCM vi ' i i = ∑ [ × × × × ] Lsystem mi r CM v CM r i ' v CM r CM v i ' r i ' vi ' i ∑ = ∑ = mi ri ' 0 mi vi ' 0 i i = ∑ × Lsystem LCM mi r i ' v i ' “spin” angular momentum i “orbital” angular momentum (zero in CM frame) Rotating CM Frame ● “Special” reference frame for a system of particles: – Must allow for rotating axes to account for angular momentum ● Let S be an inertial (non-rotating) CM frame – Let S' be a rotating CM frame with matrix R(t) [ ] – Recall: v ' = R v − × R r = ∑ × = Lsystem ' mi r i ' vi ' 0 For some “special” i = ∑ [ × − × ] = Ω Lsystem ' mi R r i R vi R r i 0 i = ∑ [ × ] − ∑ [ × × ] = Lsystem ' mi R r i R v i mi R r i R r i 0 i i = [ − ∑ [ × × ]] = Lsystem ' R Lspin , inertial mi r i r i 0 i = ∑ [ × × ] This can be used to calculate the “special” Lspin , inertial mi r i r i i for a given system (which makes L' = 0) Inertia Tensor = ∑ [ × × ] Lspin , inertial mi r i r i i = − ∑ [ × × ] Lspin , inertial mi r i r i i − − 0 z y 0 z y x i i i i = − ∑ − − L m [ z 0 x z 0 x ] spin , inertial i i i i i y i − − yi xi 0 yi xi 0 z 2 2 − − y z x y z x x i i i i i i = ∑ − 2 2 − ≡ L [ m ] I spin , inertial i xi yi xi zi yi zi y i −z x − y z x2 y 2 i i i i i i z “Inertia Tensor” – fully describes the distribution of Reference frame for a system of mass in a system particles is almost complete: Diagonal elements are called “moments of inertia” 1) origin → CM 2) angular velocity → using L and I Off-diagonal elements are called “products of inertia” 3) need to find “principal axes” Inertia Tensor for Mass Distributions y 2 z2 −x y −z x i i i i i i ● = ∑ − 2 2 − For a system of particles: I mi xi yi xi zi yi zi i − − 2 2 zi xi yi zi xi yi ● Inertia tensor extends naturally to mass distributions: 2 2 − − y z x y z x = ∫ − 2 2 − I x , y , z dxdydz x y x z y z −z x − y z x 2 y 2 ● Example: Calculate I for a flat disk in the xy-plane M 2 2 x y R 1 2 2 M R 0 0 x , y , z = z { R 4 = 1 2 z 0 elsewhere I 0 M R 0 4 1 2 0 0 M R 2 x y Principal Axes y 2 z2 −x y −z x I I I i i i i i i xx xy xz = ∑ − 2 2 − = I mi xi yi xi zi yi zi I yx I yy I yz i − − 2 2 I I I zi xi yi zi xi yi zx zy zz ● Elements of inertia tensor depend on choice of xyz axes – Is there a “special” set of xyz axes for a given system? – Yes! Possible (but difficult) to find “principal axes” such that L products of inertia are zero → in this reference frame: 2 2 = y z 0 0 L x I xx x i i = ∑ 2 2 = I mi 0 xi zi 0 L y I yy y i 0 0 x 2 y2 = i i Lz I zz z Note: L and can point in different ● For a system which is symmetric about an axis: directions! – The symmetry axis is one of the principal axes of the system Rigid Body Rotation ● Physics definition of “rigid body” – System of particles which maintains its shape (no deformation) – i.e. velocity of particles in CM frame comes from rotation only – Notation: ω = angular velocity of rigid body (inertial CM frame) ω Fcent ● View from rotating CM frame Tension – Every particle stands still in equilibrium (I is constant) – Centrifugal forces balanced by internal stresses ● Rigid body model for solid objects → reasonably good – In reality, solids can 1) deform and 2) dissipate energy as heat – Rigid body model → not applicable to fluids, orbits, stars, etc. Example: Rotating Skew Rod ω y ● Rigid massless rod (length 2d): – Rotates as shown with mass m at each end φ x – At the instant shown: – Calculate Lsystem, using 2 different methods: – × 1) r i p i , and 2) calculate the inertia tensor – Notice L and ω don't point same direction! ● Is an external torque needed to sustain this motion? – If so, calculate it ● Use symmetry to guess the principal axes – Verify guess by calculating inertia tensor in principal frame Fixed-Axis Rotation d L ● ≠ 0 Common practice → hold ω constant (not L) → dt – If L and ω not parallel → “axle” must exert an external torque! ● Example: Rotating ceiling fan – If mass is symmetrically distributed: – L and ω parallel → fan turns smoothly – If mass distribution has asymmetry (poor alignment, etc.): – L and ω not parallel → fan wobbles ● High rotation speeds → wheels, lathes, etc. – Mounting must be able to exert external torque – Enough to handle a “tolerable” amount of asymmetry Parallel-Axis Theorem ● Often: fixed axis does not pass through CM of system – ω Example: door hinge – how to calculate Ldoor? – = × Recall Lsystem M total vCM rCM I principal – = × For rigid body rotation: v CM r CM d = × × Plug in: Lsystem M total r CM r CM I principal = × × Lsystem M total r CM r CM I principal = Lsystem I orbital I principal ● Parallel-Axis Theorem: – Let d be the perpendicular distance from CM to fixed axis = 2 = 2 I orbital , axis M d I axis I principal M d = 1 2 2 = 1 2 I M 2 d M d M 2 d door about hinges 12 door door 3 door Rotational KE ● For a rotating rigid body (in inertial frame) : = ∑ 1 2 = ∑ 1 × ⋅ × KE system mi vi mi r i r i i 2 i 2 = 1 ⋅ ∑ × × KE system [ mi r i r i ] (Using vector identity) 2 i = 1 ⋅ = 1 ⋅ KE system I L (maximum angle between and L is 90º) 2 2 (For non-rigid bodies, must also include motion of particles toward/away from CM) = 1 2 1 2 1 2 ● KE I I I In principal axis frame: principal 2 xx x 2 yy y 2 zz z ● Example: “average” frisbee toss – Estimate KEtranslation and KErotation in Joules Rolling Without Slipping ● System's v CM and : independent of each other – No relation between translational / rotational motion in general – However, by using a force → the 2 can be coupled ω v ● Example: friction contact F – Drop a ball spinning at angular velocity ω on the floor – Relative velocity of ball's surface / floor causes kinetic friction – This force has 2 effects: – 1) pushes the ball to the right (affecting v CM ) – 2) exerts a CCW torque on the ball (affecting ) ● = Rolling without slipping → condition where vCM R – vcontact=0 → fixed-axis rotation about contact pt.