Lagrangian Dynamics: Virtual Work and Generalized Forces

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1 Admssble Varatons/Vrtual Dsplacements J/1.053J Dynamcs and Control I, Sprng 2007 Paula Echeverr, Professor Thomas Peacock 4/4/2007 Lecture 14 Lagrangan Dynamcs: Vrtual Work and Generalzed Forces Readng: Wllams, Chapter 5 All q are scalars. q : Generalzed Coordnates L: Lagrangan Q : Generalzed Forces L = T V d L L = Q dt q q Admssble Varatons/Vrtual Dsplacements Vrtual Dsplacement: Admssble varatons: hypothetcal (not real) small change from one geometr cally admssble state to a nearby geometrcally admssble state. Bead on Wre Fgure 1: Bead on a wre. Fgure by MIT OCW. Cte as: Thomas Peacock and Ncolas Hadjconstantnou, course materals for 2.003J/1.053J Dynamcs and Control I, Sprng MIT OpenCourseWare ( Massachusetts Insttute of Technology.

2 Admssble Varatons/Vrtual Dsplacements 2 Both δ x and δ y are admssble varatons. Hypothetcal geometrc confguraton dsplacement. dx mples t nvolved. δ = d δx = dx Generalzed Coordnates y = f(x) df dy = dx dx df(x) δy = δx dx Mnmal, complete, and ndependent set of coordnates s s referred to as complete: capable of descrbng all geometrc confguratons at all tmes. s s referred to as ndependent: If all but one coordnate s fxed, there s a contnuous range of values that the free one can take. That corresponds to the admssble system confguratons. Example: 2-Dmensonal Rod Fgure 2: 2D rod wth fxed translaton n x and y but free to rotate about θ. Fgure by MIT OCW. Cte as: Thomas Peacock and Ncolas Hadjconstantnou, course materals for 2.003J/1.053J Dynamcs and Control I, Sprng MIT OpenCourseWare ( Massachusetts Insttute of Technology.

confguratons at all tmes. s s referred to as ndependent: If all but one coordnate s fxed, there s a contnuous range of values that the free one can take.

3 Vrtual Work 3 If we fx x and y, we can stll rotate n a range wth θ. # degrees of freedom = # of generalzed coordnates: True for 2.003J. True for Holonomc Systems. Lagrange s equatons work for Holonomc systems. Vrtual Work = forces act at that locaton W = f dr Actual Work δw = f δr Vrtual Work f = f appled + f constraned Constraned: Frcton n roll. Constrant to move on surface. Normal forces. Tenson, rgd body constrants. δw = f app δr = 0 at equlbrum No work done because no moton n drecton of force. No vrtual work. f = 0 Cte as: Thomas Peacock and Ncolas Hadjconstantnou, course materals for 2.003J/1.053J Dynamcs and Control I, Sprng MIT OpenCourseWare ( Massachusetts Insttute of Technology.

Vrtual Work = forces act at that locaton W = f dr Actual Work δw = f δr Vrtual Work f = f appled + f constraned Constraned: Frcton n roll. Constrant to move on surface.

4 Vrtual Work 4 Example: Hangng Rgd Bar Fgure 3: Hangng rgd bar. The bar s fxed translatonally but s subject to a force, F. It therefore can dsplace tself rotatonally about ts pvot pont. Fgure by MIT OCW. Dsplacement: Forces: Two forces appled: = 2 δy = aδθĵ A δy = lδθĵ B F = Fĵ R = Rĵ δw = Flδθ Raδθ = 0 Fl R = at equlbrum a Could also have taken moments about O. Cte as: Thomas Peacock and Ncolas Hadjconstantnou, course materals for 2.003J/1.053J Dynamcs and Control I, Sprng MIT OpenCourseWare ( Massachusetts Insttute of Technology.

Dsplacement: Forces: Two forces appled: = 2 δy = aδθĵ A δy = lδθĵ B F = Fĵ R = Rĵ δw = Flδθ Raδθ = 0 Fl R = at equlbrum a Could also have

5 Vrtual Work 5 Example: Tethered Cart Fgure 4: Tethered cart. The cart s attached to a tether that s attached to the wall. Fgure by MIT OCW. df (x) Usng δy = dx δx c δw = Fδy B Rδx c = 0 y B = l sn θ δy B = l cos θδθ δx c = 2l sn θδθ ( Fl cos θ + 2R sn θ)δθ = 0 F Fl cos θ + 2R sn θ = 0 R = 2 tan θ at equlbrum Fgure 5: Applcaton of Newton s method to solve problem. The ndcated extra forces are needed to solve usng Newton. Fgure by MIT OCW. Cte as: Thomas Peacock and Ncolas Hadjconstantnou, course materals for 2.003J/1.053J Dynamcs and Control I, Sprng MIT OpenCourseWare ( Massachusetts Insttute of Technology.

at equlbrum Fgure 5: Applcaton of Newton s method to solve problem. The ndcated extra forces are needed to solve usng Newton. Fgure by MIT OCW.

6 Generalzed forces for Holonomc Systems 6 Generalzed forces for Holonomc Systems In an holonomc system, the number of degrees of freedom equals the number of generalzed coordnates. = number of appled forces: 1 to n j = number of generalzed coordnates δw = f δr = Q δq j r = r (q 1, q 2,... q j ) r : Poston of pont where force s appled Substtute: δr = m r j δq j q j n m m n r r f δq j = f q j q j q j j j n r Q j = f Generalzed Forces q j f = f NC + f CONS. f CONS. : Gravty, Sprng, and Buoyancy are examples; Potental Functon Exsts. Example: V g = mgz, r = zĵ f = mg z ĵ = mgĵ g dz f CONS. V = r r V r V f cons. = = q r q j q j The conservatve forces are already accounted for by the potental energy term n the Lagrangan for Lagrange s Equaton Cte as: Thomas Peacock and Ncolas Hadjconstantnou, course materals for 2.003J/1.053J Dynamcs and Control I, Sprng MIT OpenCourseWare ( Massachusetts Insttute of Technology.

.. q j ) r : Poston of pont where force s appled Substtute: δr = m r j δq j q j n m m n r r f δq j = f q j q j q j j j n r Q j = f Generalzed Forces q j f = f NC + f CONS.

7 Example: Cart wth Pendulum, Sprngs, and Dashpots 7 Lagrange s Equaton n Q NC NC r j = f qj d L L = Q NC j dt q j q j Q NC j = nonconservatve generalzed forces L q j V q j contans. Example: Cart wth Pendulum, Sprngs, and Dashpots Fgure 6: The system contans a cart that has a sprng (k) and a dashpot (c) attached to t. On the cart s a pendulum that has a torsonal sprng (k t ) and a torsonal dashpot (c t ). There s a force appled to m that s a functon of tme F = F (t) We wll model the system as 2 partcles n 2 dmensons. Fgure by MIT OCW. Cte as: Thomas Peacock and Ncolas Hadjconstantnou, course materals for 2.003J/1.053J Dynamcs and Control I, Sprng MIT OpenCourseWare ( Massachusetts Insttute of Technology.

On the cart s a pendulum that has a torsonal sprng (k t ) and a torsonal dashpot (c t ).

8 Example: Cart wth Pendulum, Sprngs, and Dashpots 8 4 degrees of freedom: 2 constrants. Cart moves n only 1 drecton. Rod fxes dstance of the 2 partcles. Thus, there are a net 2 degrees of freedom. For 2.003J, all systems are holonomc (the number of degrees of freedom equals the number of generalzed coordnates). q 1 = x q 2 = θ Fgure 7: Forces felt by cart system. Fgure by MIT OCW. F 1 : Damper and Sprng n x drecton F 2 : Two torques: F 3 : (kx + cẋ)î τ = (k t θ + c t θ kˆ F 3 = F 0 sn ωtî r A = xî = q 1 î r 1 r 2 r B = r A + r B/A = (x + l sn θ)î l cos θĵ r 3 = θkˆ (Torque creates angular dsplacement) = q 2 kˆ Cte as: Thomas Peacock and Ncolas Hadjconstantnou, course materals for 2.003J/1.053J Dynamcs and Control I, Sprng MIT OpenCourseWare ( Massachusetts Insttute of Technology.

F 1 : Damper and Sprng n x drecton F 2 : Two torques: F 3 : (kx + cẋ)î τ = (k t θ + c t θ kˆ F 3 = F 0 sn ωtî r A = xî = q 1 î r 1 r 2 r B = r A + r B/A = (x + l sn θ)î l cos θĵ r 3 = θkˆ (Torque

9 Example: Cart wth Pendulum, Sprngs, and Dashpots 9 Q 1 : r 1 q 1 = 1î, r 2 q 1 = 0, r 3 q 1 = 1î r 1 q 2 = 0, r 2 q 2 = 1ˆk, r 3 q 2 Q 1 = cq 1 + F 0 sn ωt = l cos q 2 î + l sn q 2 ĵ Q 2 = c t q 2 + F 0 sn ωt l cos q 2 Wth the generalzed forces, we can wrte the equatons of moton. Knematcs M: r M ṙ M r M = xî = ẋî = ẍî m: r m = (x + l sn θ)î l cos θĵ ṙ m = ( ẋ + l cos θ θ )î + l sn θθ ĵ r m = (ẍ + l(cos θ) θ l(sn θ)θ 2 )î + (l(sn θ) θ + l(cos θ)θ 2 )ĵ Generalzed Coordnates: q 1 = x and q 2 = θ. Lagrangan L = T V T = T M + T m 1 T M = 2 M( ṙ M ṙ M ) = 1 2 M ẋ 2 T m = = 1 2 m( ṙ m ṙ m ) 1 2 m( ẋ 2 + 2l ẋ cos θθ + l 2 θ 2 ) (1) (2) Cte as: Thomas Peacock and Ncolas Hadjconstantnou, course materals for 2.003J/1.053J Dynamcs and Control I, Sprng MIT OpenCourseWare ( Massachusetts Insttute of Technology.

Knematcs M: r M ṙ M r M = xî = ẋî = ẍî m: r m = (x + l sn θ)î l cos θĵ ṙ m = ( ẋ + l cos θ θ )î + l sn θθ ĵ r m = (ẍ + l(cos θ) θ l(sn θ)θ 2 )î + (l(sn θ) θ + l(cos θ)θ 2 )ĵ Generalzed Coordnates: q

10 Example: Cart wth Pendulum, Sprngs, and Dashpots 10 T = 1 2 M ẋ m( ẋ 2 + 2l ẋ cos θθ + l 2 θ 2 ) V = V M,g + M M,k + V m,g + V m,kt = Mg(0) k( ṙ M ṙ 1 M ) + mg( l cos θ) + 2 k tθ 2 (3) (4) Symbol V M,g V M,k V m,g V m,kt Potental Energy Gravty on M Sprng on M Gravty on m Torsonal Sprng on m Substtute n L = T V V = 1 2 kx2 + ( mgl cos θ) k tθ 2 L = 1 2 M ẋ m( ẋ 2 + 2l ẋθ cos θ + l 2 θ 2 ) 1 2 kx2 + mgl cos θ 1 2 k tθ 2 Equatons of Moton Use d L L dt q q = Ξ to derve the equatons of moton. Ξ = Q. From before, Ξ x = F 0 sn ω 0 t c ẋ and Ξ θ = F 0 (sn ωt)l cos θ c t θ. For Generalzed Coordnate x δx = 0 and δθ = 0. Unts of Force. L ẋ L x = kx = (M + m) ẋ + ml(cos θ)θ L = (M + m)ẍ + ml θ ẋ cos θ + ml( sn θ)θ 2 d dt d L L = (M + m)ẍ + ml θ (cos θ) + ml( sn θ)θ 2 + kx = F 0 sn ωt cẋ dt ẋ x Cte as: Thomas Peacock and Ncolas Hadjconstantnou, course materals for 2.003J/1.053J Dynamcs and Control I, Sprng MIT OpenCourseWare ( Massachusetts Insttute of Technology.

ẋθ cos θ + l 2 θ 2 ) 1 2 kx2 + mgl cos θ 1 2 k tθ 2 Equatons of Moton Use d L L dt q q = Ξ to derve the equatons of moton. Ξ = Q.

11 Example: Cart wth Pendulum, Sprngs, and Dashpots 11 For Generalze Coordnate θ δx = 0 and δθ = 0. Unts of Torque. L = ml ẋθ ( sn θ) mgl sn θ kt θ θ L = ml ẋcos θ + ml 2 θ θ d L = ml ẋ( sn θ)θ + mlẍcos θ + ml 2 θ dt θ d L L = ml ẋθ ( sn θ)+mlẍcos θ+ml 2 θ ml ẋθ ( sn θ)+mgl sn θ+k t θ dt θ θ d L L = mlẍcos θ + ml 2 θ + mgl sn θ + kt θ = F 0 (sn ωt)l cos θ c t θ dt θ θ Cte as: Thomas Peacock and Ncolas Hadjconstantnou, course materals for 2.003J/1.053J Dynamcs and Control I, Sprng MIT OpenCourseWare ( Massachusetts Insttute of Technology.

ml ẋθ ( sn θ)+mgl sn θ+k t θ dt θ θ d L L = mlẍcos θ + ml 2 θ + mgl sn θ + kt θ = F 0 (sn ωt)l cos θ c t θ dt θ θ Cte as: Thomas Peacock and

Rotation Kinematics, Moment of Inertia, and Torque

Rotation Kinematics, Moment of Inertia, and Torque Rotaton Knematcs, Moment of Inerta, and Torque Mathematcally, rotaton of a rgd body about a fxed axs s analogous to a lnear moton n one dmenson. Although the physcal quanttes nvolved n rotaton are qute