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# Classical Mechanics PHYS 3210

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This 17 page Class Notes was uploaded by Ubaldo Jacobson on Monday September 21, 2015. The Class Notes belongs to PHYS 3210 at University of Virginia taught by Julian Noble in Fall. Since its upload, it has received 13 views. For similar materials see /class/209752/phys-3210-university-of-virginia in Physics 2 at University of Virginia.

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Date Created: 09/21/15

PHYS 321 Classical Mechanics Lecture 13 27 September 2002 Lagrangian Method Using the Lagrange approach the position ofthe kth particle is described by a set of generalized coordinates that replace the Cartesian coordinates in all of Newton39s equations rk9rkCl1Cl21Cln t The force on the kth particle in a system is the sum of the applied force and the constraint force Fk fkapp fk 141 Because any small distance moved by the kth particle Srk is orthogonal to the constraint force fk Srk O 142 which implies that the constraint force does no work in an idealized situation From Newton39s second law dpkdt Fk fkapp fk 143 9 fk fkapp dpkdt 144 Applying equation 142 to 144 implies fk Srk fkapp dpkdt Srk fk Srk O 145 Summing the forces on all of the particles from k1 to kN where N is the number of particles 3N is the number of Cartesian coordinates required in 3 dimensions and 3N n is equal to the number of constraints on the system gives 2 rkapp dpkdt 6rk o 146 k1 The change in position of the generalized coordinate vector Srk is related to the change in the individual components ofthe vector Sql by the equation 6rk Z M q 5q 147 11 where equation 146 is known as the principle of virtual work Substituting equation 147 into 146 and rearranging the summations yields 3 i fkapp M q i dpkdt M q Sql O 148 N In equation 148 2 fkapp rk q is known as the generalized force k1 labeled QI Because the sum is equal to zero the individual components must all equal zero implying N N QI Z dpkdt rk q Q Z mkdvkdt rk q 0 149 161 k1 The portion inside the sum in equation 149 can be expressed as mkdvkdt rk qddtmkvk M q mkvk ddt rk q 1410 Using the fact that ddtmkvk rk q ddtmkvk vk ql where q39 is the derivative of q equation 149 is rewritten N QI Z ddtmkvk Vk ql mkvk Vk 0 1411 k1 The kinetic energy of the kth particle is mkvk Vk q q12mkvk2 Tk 1412 the sum ofall the individual kinetic energies is T Substituting the kinetic energy relation equation 1412 into equation 1411 produces QI ddt T qi T q O 1413 The generalized force Q is defined as V q where V is the total potential energy ofthe system lfthe applied force is a conservative force implies that ddt V q 0 9 V ql O lfthe Lagrangian is de ned as L T V equation 1413 can be written in the form t 0n When one uses integration by parts this equation becomes 17 b dF 116D Jdt 62 1 110 0 Bq a 3 6q dt Bq But since qa qb are fixed this implies that na nb 0 Thus 6d d M 0 6g dt 6i Which is just the EulerLagrange equation if I is the Lagrangian Hamilton s Principle of Least Action means that the variation of the action A943 equals zero Brachistochrome Problem Envision a bead sliding along a section of downwardcurved wire where a and b are the endpoints and z is the height Then 2a h the height of the wire and 210 Through some simple analysis from last lecture one knows that rvz rfdd di 1quot2 Fltzgt We will let the entire term in the definite integral above be equal to ltIgtz dzdx x From before we know that egg 62 0 62 dX 62 where Z39 dZdX By simply taking the necessary derivatives of the 1 expression inside the definite integral above one can plug the resulting terms into the equation immediately above and get d g 1z2 dX z1Z392 22m Using an integrating factor to simplify z d z z 7 70 21z392dx 21z2 222 The second term is a perfect derivative 1 11 22 dx 2 Meanwhile the first term with the integrating factor is now of the form d d 1 2 XEfXg fXi Thus simplifying the first term in this way while bringing the simplified second term into the derivative gives 1 7 A constant Z 2 1 AZ 1 2 The fraction must be less than one which gives rise to the negative sign on the right side Solving 1 AZ dz 1 AZ 227 7 7 AZ 3dX AZ note that the curve must decrease monotoically with x hence the negative sign Separating variables AZ 1 AZ dX dZ 2 u and substituting AZ U2 dZ X dll XlJu dull2 A ll u2 Making anotherchange ofvariable U sin 9 dll C039 d9 we find sm39l u 1 7 1 X d931n29731n1U 7U 1 ll2 2 2 So finally we re left with note the two arbitrary constants A and B from the original second order DE X B isin 1 Mm ll AZ Doing this entire process in a much easier way treating 2 as the independent variable yields the same answer 3dZxZ 1dxdzz If one treats the term inside this definite integral as I there is nothing that depends on x so PHYS 321 Lecture 8 09l13l2002 Central Forces Conservation of Angular Momentum 7 F X f7 0 for central forces at 0 for others This is basically Kepler s 2quotd Law equal areas in equal times Newton s 2quotd Law J a 111 F alt2 Change to spherical polar coordinates X rsin 9 cos p y min 9 sin p z r cos 9 Express Newton s 2quotd in spherical coordinates all are time dependent Example X 39rsin 9 costprcos 99 COSp rsin 93m pt39p If we do this for y and 39Z and add all three squared components we get v2 X2y222 i21292rzsin296p2 We can do this another way let us represent the change in the radius vector in spherical coordinates as d dref mBEe rsm Gdtpew This de nes the orthogonal system Er E9 E where Er gtlt E9 E etc We nd that r d A GA e A V E ref 69 rsm pew giving the same thing for v If we try to differentiate v in polar coordinates to get the acceleration we have to do so in two places 1 r 94 for a moving object depend on time 2 The unit vectors 6 E9 E rotate as they move to different points in space so they too depend on time So we skip this hard problem and work instead with the conserved energy For central forces f7 Efr VVr the ability to express f7 as a gradient depends on its form since A A A d V A dV VVr XXy z 7 eff r r r dr dr d dquot We can therefore integrate Newton s 2quotd Law With the integrating factor V y to get E inn2 Vr const Rewriting this 11102 1292 12 sin2 9 2 Vr E Z mrsin9rsin9p const 12 sin2 9 ip const 3 1 Z mif z const Suppose we let p 0 at some initial time then the equation of motion implies ii 01i 0 Therefore p 0 for all time If we choose 3 0 then r2 sinz 9th 0 in the energy equation and we obtain 2 mzr4 62 Substituting back we have 11102 rz z Vr gm z g2 mzf Vr E Which we can solve for 39r rzgzi 2E V 52 I dt m 11122 The equation of an orbit is determined by the function r9 We can eliminate time by diViding 39r by 9 to get dr 39r In2 2E V 2 d9 Z m 11122 That is in general the orbit equation can be found by quadrature PHYS 321 Mechanics by Adam Brown 1 Lecture 4 Sept 4th 2002 Conservation of Linear Momentum In any collision linear momentum must be conserved If an object of mass 11 moving at velocity V collides with and sticks to a stationary object also of mass m the two objects will move together at half the initial velocity V Air track demo That is momentum is conserved p mV hence P 1121an P131221 IIIV10IIIIIIV so in this case V m V 1 V f 2m 1 2 1 39 Application Rockets An example of Newton s 3rd Law Every action has an equal and opposite reaction A rocket of mass m traveling through space in the positive X direction eXpels fuel of mass Am in the negative X direction at a velocity of 11 relative to the rocket The rocket now with mass 111 Am moves with velocity V AV How does AV depend upon Am and 11 Conservation of momentum Initial Momentum Final Momentum HIV m AmVAVAmV 11 HIV m mV VAm mAV VAm uAm This leads to 0 mAV uAm which we can turn into a differential equation letting Am gt dm AV gt dV separating variables and integrating J39m in A V Oh mg m u V0 or PHYS 321 Notes for November 20 2002 Center of Mass Transformation of 2 B0dy Problem 1 a 1 a e a L 1111V12 1112V12 UX1 X2 1111111111 a X 11 22XO P I 11111112 11111112 XX1 XZ P 1111 1112X constant 111 111112 12 m1V11112V2 2P1p2 a 1 m2 1 1 m1 1 X1X XXZX X 11111112 11111112 1 111 a 1 111 L7III1VMZVZEIIIZVM1VZ UX Z Ll111 111 172111111 VOW 111111 170Vl111 5111 312 2 1 Z 1 Z 1 Z 2 1M2 2M quot111112 quot111112 M m1 m2 11111112 1 d 1 d L7M727 72 U 2 dt 2M0 X Heavy Symmetric Top Precession and Nutation 33 0332 nglcose 13 mg p up 510W 3 33 3033 11 c039 u fast 13033 sin 93cos p jsin p g L Igoagsin9 3sinpjc03p d W q M 1375 3033 from Lagrangian 36L 6L d L60 2sinGcosGmgIsin9 Igm3sin9 6L 2 PW f pl1 sm 9 13 c039p pcose const 61 a p 1 sin2 9 13coseoa3 const Ml sinZ 9 Zt39le sine c0399 13033 sine 0 Ion quot sine E 6 P A 116 5mg1 13m3 sin e 713033 6 E 11 7 13033 3033 2 sin 957 1 3111571 13 cos sin 9 d2 WWW39miq wimp P DP 0313 30COS OJLI 2 57m DZLX 0 PHYS 321 Lecture 22 23 October 2002 The relativistic rocket We can treat the rocket in free space relativistically in two ways 1 Imagine we are in the in stantaneous rest frame of of the rocket and that a D P small chunk of mass Am dm gt 0 is ejected Am rearward at velocity 11 Then we nd mdv uAm udm and apply the velocity addition formula m VdV1 VZCZ V dV 1 VdVCZ to nd that 2 dm 2 dV dv 1 VZC u71 V2c m This equation can be integrated to obtain u m C X tanh iln C Mf 2 The second method is to work in the lab system always In that frame we have from conservation of momentum mV 2 V u mm Am 1 VdV Vl VZCZ l UVCZZ V UZC2A V1 ZCZJ1 VdVZCZ m Am dV mV V u V m A 1V2cz 1V2CZ1uzcz m 41 11 1 v2c2 HZc2 We recognize that in the lab frame Am quot1 112 C2 so conservation of momentum simplifies to dm dV w zm dm PHYS 321 Classical Mechanics Lecture 28 8 November 2002 Inertia Tensor In a noninertial reference frame the kinetic energy of any body is represented by the equation N d a M Z TE1 mk dfQgtltrkj 11 For a rigid body in a rotating frame the magnitude of 7 remains constant implying that the time derivative of F1 is zero The resulting energy equation for a rigid body in a rotating frame is TEm 2xg 2xa 12 This expression is simplified through the use of several properties ofthe cross product 9 X r39k 9X r39k r39kx Q r39kx Q r39k 9X r39kx 9 13 Using the vector triple product 9 X r39kx Q r39kQ Q Q r39k Q 9X r39k 9X r39k r39k 9 9 r39k 9 r39k 9 14 Substituting equation 14 into 12 yields 71 x71 TE2mk397 2 2 15 The radial vector r39k is represented by the sum ofall of the components ofthe vector in all possible directions r39k r39k u 16 141 where all represents the unit vector in the udirection The variable u is limited to the values 1 2 3 due to the fact that there are 3 perpendicular directions in Cartesian coordinates eg X y z Using relation 16 3 3 1 E2 21122 21m 17 v1 1 Substituting these expressions into equation 15 and rearranging the summations which is possible due to the fact that all of the indices of summation are finite note that even the summation going from 1 to N is finite as N represents the number ofall particles that compose a rigid body and is on the order of magnitude of Avogadro39s Number 1023 produces T zzs2zmkiltrrgt6wwwij v 18gt u v 111 where 8W is the Kronecker delta 1 when u v and 0 when Hi v An alternative representation ofthe energy ofthe rigid body in a rotating frame is given by T 112 21 I by 19 where I v is termed the inertia tensor Comparing equations 18 and 19 the inertia tensor is N 1 Z mkr39k25 r39k r39kV 110 161 Equation 110 implies for a continuous mass distribution 1 dmc39k25 r39k r39kV 111 There are two important observations to note about the inertia tensor 1 the inertia tensor is real 2 the inertia tensor is symmetric 3 X 3 matrix which indicates that a rotation matrix 3 x 3 orthogonal matrix can diagonalize the inertia tensor the inertia tensor can be transformed by rotations Example Compute the inertia tensor of a rod extended along the yaxis from L2 to L2 Solution All 9 ofthe elements ofthe 3 x 3 matrix are computed using equation 111 with r39k2 x2 y2 22 I 1139 1 dx f dy dz 6x6zxy0 PHYS 321 Lecture 30 13 November 2002 Center of Percussion Example Bat hitting a ball Let mass ofbat m distance from hands to the centerofmass COM a distance from COM to where ball hits s 7tas force of ball s impact F force hands must apply to bat if Then dV F f 7 mdt F fmad dt Torque hFId dt do AF or 77 dt I Combining these f F K F 1 quotE I I Notice that for the right 7 the terms in the brackets goes to zero happens at x i ma For a thin rod like a meter stick I 1 mL2 3 and L a m i Z we hold it at the end so the sweet spot is located a distance W gL 6 Sweet spot m 3 from the end Free Rotation of a Rigid Body Rigid Body is merely a convenient approximationithere is no such thing in nature d Conservatlon of angular momentum 3 N Transforming to body coordinates we have 9 9 x7N dt space j11 2123112 22 2 13933 d7 Look at say th1rd component of y 1323 9192 12 Il N3 and cyclic permutations These are the Euler Equations General form of Euler Equations 1ka 814th NR l 1171 123 where 81 0 k10r m l 1111 odd perm of 123

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