Classical Mechanics Review (Louisiana State University Qualifier Exam)

Transcription

1 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel May 8, 7 A particle is dropped into a hole drilled straight through the center of the earth. Neglecting rotational effects, show that the particle s motion is simple harmonic. Compute the period and give an estimate in minutes. Compare your result with the period of a satellite orbiting near the surface of the earth. Figure : A particle dropped from R. Only the mass inside the shell of radius r contributes to the gravitational force. If we assume the Earth have uniform density, then the mass contributing to F g is M 4 3 π r3 M 4 3 π R3 M M R 3 Thus, the force experienced by the particle m from mass M is r 3 F g GmM r Gm M r 3 r R 3 GM m F g r k r R 3

2 Jeff Kissel May 8, 7 Classical Mechanics where the last step absorbs all the constants G, M, R, and m into the constant k, which shows that Eq. is linearly proportional to the radius. Similarly, Hooke s force law for simple harmonic motion states that the force of a simple harmonic oscillator is proportional to the displacement from the equilibrium position. Since the center of the earth would be the particles equilibrium position, r is the displacement, and thus by direct comparison, the particle obeys simple harmonic motion, with spring constant k. To compute the period τ, we use what we know for a spring with resonant frequency ω k/m, τ f π m ω π k R 3 π m GM m τ π R 3 GM τ 4π R 3 GM 3 which is Kepler s 3rd law for planetary motion for a small particle, again indicating that motion is periodic. Finally, one can find an estimate of this period by using Earth s gravitational acceleration of a small particle at the surface, g GM 9.8 m/s π. R τ 4π R 3 GM R 4 π g τ 4R 4 τ m [s].57 7 s τ 569 s 84 min 5 which is approximately equivalent to a satellite in low-earth orbit. At 85 km, i.e. when low-earth orbit becomes stable, the correct period is 88.9 min.

3 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel October, 6 Two identical bodies of mass m are attached by identical springs of spring constant k as shown in the figure. Figure : A system of two masses attached by springs A in equilibrium, and B after some oscillation. a Find the frequencies of free oscillation of this system. The first goal will be to set up the system s characteristic equation, V ω T where V is the potential energy matrix, T is the kinetic energy matrix, and ω are the desired eigenfrequencies frequencies of free oscillation of the system. The potential on the particles is the sum of each individual spring s potentials, V kd + kd + kd 3

4 Jeff Kissel October, 6 Classical Mechanics If we define the displacements between the masses in terms of generalized coordinates η, we can then write the potential as, d x x + η d x x + x x η η d 3 + x x η V kη + kη η + k η kη + kη + η η η η η + kη k η + η η η η η 3 which in the desired matrix form is, V k k V ijη i η j k k 4 The kinetic energy matrix is much easier to find because it is neatly diagonal: T mv + mv T m T ijη i η j m 5 Once we ve got these matrices, we want to find the eigenfrequencies, ω of the charateristic equation V ω T 6 Notice however, that Eq. 4 and Eq. 5 are symmetric, matrices, so we can use some trickery and write them as multiples of the identity matrix or the SU spinor matrices: V k I k ˆσ x T m I Now we can just diagonalize Eq. 6 instead of the usual determinant method, since we know the eigenvalues of I are {,}, and for any ˆσ i are {,-}. I ll also use the fact that a diagonalized ˆσ x is just ˆσ z V ω T k I k ˆσ x mω I D mω ID D k I k ˆσ x D ω D ID k m D ID k m D ˆσ x D

5 Jeff Kissel October, 6 Classical Mechanics ω I k m I k m ˆσ z 7 ω k k m m ω k ± k 8 m So the eigen frequencies for this two mass system are k 3k ω or ω m m We ll need the eigenvectors for part b, so we ll find them now. A big advantage to using this SU group argument is that since Eq. 7 only involes I and ˆσ z, we know immediately that the two normalized eigenvectors for the system are the normalized eigenvectors of those two matrices, for I û for ˆσ z û 9 where û corresponds to when the two masses are sloshing back and forth in phase with each other, and û is when they oscillate out of phase, i.e. if M is moving to the left, M is moving to the right, or vice versa. b M is displaced from its position by a small distance A to the right while M is not moved from its position. If the two masses are released with zero velocity, what is the subsequent motion of M? With the very safe assumption that the system is periodic in time, we know from Fourier s theorem that the positions of each particle as a function of time may be written as a sum of sines and cosines, whose phase is the eigenfrequencies. So, x t Asin ω t + B cosω t + C sin ω t + D cosω t x t E sin ω t + F cosω t + Gsinω t + H cosω t 3 However, we can narrow this down at least a little because of the eigenvectors. k ω m û A E, B F 3k ω m û C G, D H which means Eqs. and 3 and there derivatives simplify a lil bit to x t Asin ω t + B cosω t + C sin ω t + D cosω t 3

6 Jeff Kissel October, 6 Classical Mechanics ẋ t x t ẋ t Aω cosω t Bω sin ω t + Cω cosω t Dω sinω t Asin ω t + B cosω t C sin ω t D cosω t Aω cosω t Bω sin ω t Cω cosω t + Dω sinω t Remember the problem statement said M is diplaced by a small distance A [at t ]? That means the initial conditions where t, so the sine terms are zero, and cosine terms are one are x A B + D x B D B D A ẋ A + C 3 ẋ A C 3 A C So then holy moly with extra cannoli, the positions of M and M at a funcion of time are x t k A cos m t + 3k A cos m t 4 x t k A cos m t k A cos m t 5 4

7 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel May 8, 7 Exam Iteration Question Number Notes/Changes Spring 6 X Fall 6 X Spring 7 #3 Added to bank. A planet is in circular motion about a much more massive star. The star undergoes an explosion where three percent of its mass is ejected far away, equally in all directions. Find the eccentricity of the new orbit for the planet. Initially the planet is in a circular Keplerian orbit, where we know several things. The force and potential are F GMm r For any circular orbit, the radius is given by, V GMm r F GM m r c ma c m v t r c For circular orbits, ṙ v t ṙ + r θ r c θ GM r c θ r c r 3 c GṀ r c θ /3 GṀ θ Also, from Eq 3.57 on p94 of Goldstein, any mass m orbiting a r central

8 Jeff Kissel May 8, 7 Classical Mechanics force will have eccentricity, e + El mk 3 with E as the total energy, l mr θ as the angular momentum, and k is this case as GMm. Since the orbit is circular, the eccentricity is zero, so the initial total energy is + El mk + El mk El mk E mk l 4 And finally, the total energy can also be written as the sum of the kinetic an potential energies, E m ṙ c + r c θ GMm m GṀ θ r c /3 GMm θ GM m θ /3 m m GM θ /3 /3 GṀ θ /3 GM θ E m GM θ /3 5 Phew! Now, just after the explosion, the star is at the same radius, r c. However the star has a new mass, M λm where I ve denoted λ as the percentage of mass loss. The only portion of the energy that is effects is the potential, in which k λk. The angular momentum l mr c θ is independent of M. So, the new energy is E mr c θ GM m m GṀ θ r c /3 G λmm θ /3 GṀ θ GM m θ /3 λ m GM θ /3

9 Jeff Kissel May 8, 7 Classical Mechanics λ m GM θ /3 E + λ m GM θ /3 6 The ratio of initial energy to final energy is then, E E + λ m GM θ /3 m GM θ /3 E E + λ E λ E 7 This in turn affects the eccentricity, which means Eq. 3 becomes, e + E l mk λ El + m λ k λ l + λ mk E Eq. 4: E mk e l λ l λ mk mk l λ λ 8 For λ.3, the new eccentricity is e

10 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel October, 6 A homogeneous cube each edge of which has a length l, is initially in a position of unstable equilibrium with one edge in contact with a horizontal plane. The cube is then given a small displacement and allowed to fall. Find the angular velocity of the cube when one face strikes the plane if: a the edge cannot slip on the plane friction. b sliding can occur no friction. Figure : A cube falling with friction holding one corner stationary, and the same cube sliding on the corner without friction. For both scenarios we can use a conservation of energy arguement to solve for the angular velocity. a The initial kinetic energy, T i is zero, because the the cube is in unstable equilibrium. The potential we can say is simply that of a point mass at the cube s center of mass, a height h above the ground, E i T i + V i mgh

11 Jeff Kissel October, 6 Classical Mechanics V i mgl In this case, the edge that is stationary acts as a pivot around which the cube will rotate. Thus, the kinetic energy of the cube while in motion is only rotational. We can use the parallel axis theorem to find the moment of inertia rotating about an edge of the cube, noting that the moment of inertia for spinning along the center of mass, I CoM 6 ml, and the displacement, d is h l ; I I CoM + md 6 ml + mh l 6 ml + m ml ml The final kinetic energy as the cube hits the plane is then T f I ωf 3 ml ω f 3 ml ω f 3 The final potential is as a point mass at a height, l V f mgl 4 So conservation of energy dictates that V i T f + V f mgl 3 ml ωf + mgl g 3 lω f + g 3 lω f g ω f 3g l 5

12 Jeff Kissel October, 6 Classical Mechanics b In this case, the kinetic energy will be slightly different. Now, because the plane is frictionless, the cube rotates about its center of mass, which is moving only in the direction of gravity. The position and velocity of the center of mass become r CoM y CoM h sin φ + π 4 v CoM ẏ CoM h cos φ + π φ 4 ωl cos φ + π 4 taking φ to be ω. Though the initial kinetic energy is still zero, the final kinetic energy changes to T f m v f + I CoMωf m ωf l cos 4 ml ω f φ f + π 4 + ml ωf + 6 ml ωf 5 4 ml ω f 6 Notice that the initial and final potential energies will remain the same, because the cube starts in the same position as in case a, and ends in the same orientation, just displaced by l. So, using Eq. and 4, the conservation of energy equation is V i T f + V f mgl 5 4 ml ωf + mgl 5 4 lω f g ωf 4g 5l g ω f 5l 7 3

13 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel October, 6 A chain of linear density µ [g/cm] is hanging vertically above a table. Its lowest point is at a height h above the table. The chain is released and allowed to fall. Calculate the force exerted on the table by the chain when a length x of chain has fallen onto the surface. Figure : A chain suspended a height h above the table at t, and then x amount fallen onto the table after time t The force exerted on the table will be equivalent to the sum of the gravitational from whatever part of the chain has already landed on the table and impulse from the part of the chain that has just hit the table forces. The gravitational force from the amount of mass m µx already on the table is simply F g mg µxg

14 Jeff Kissel October, 6 Classical Mechanics During a time interval dt, the mass of the rope equal to µv dt is hitting the table. The change in momentum imparted onto the table is then The impulse force is then dp dm v [µv dt] v µv dt F impulse dp dt µv 3 However we would like to know how velocity v is related to xt. Some one dimensional kinematics should shed light on the matter: v v i a x v gx + h 4 So Eq. 3 becomes F impulse µv µgx + h 5 Finally, the total force of the table from the falling chain is F F g + F impulse µxg + µx + hg 3µxg + µhg µg3x + h 6 which is equivalent to the weight of the length 3x of the rope, plus the correction for the initial height.

15 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel October, 6 A thin circular ring of radius R and mass m is constrained to rotate about a horizontal axis passing through two points on the circumference. The perpendicular distance from the axis to the center of the ring is h. Figure : A thin hoop confined to swing from an off-diameter axis of rotation. a Find a Lagrangian for this object. Looking at the hoop from the side few, it s apparent that this system is akin to a simple rigid pendulum of length h. Since we know the pendulum s bob i.e. the hoop s center of mass is confined the plane of the page or in and out of the page for the front view a convenient generalized coordinate for this system is the angle φ as seen in Figure. The kinetic energy of the hoop is then T I φ mr + mh φ m R + h φ

16 Jeff Kissel October, 6 Classical Mechanics where the moment of inertia I was found using the parallel axis thereom, I I CoM + md noting that d is the distance from the center of mass to the parallel axis, and I CoM mr for a thin hoop. The potential energy will be similar a simple pendulum: V m g r mgh cos φ 3 So using Eq. and 3 the Lagrangian for a thin hoop hung from an off diameter axis of rotation is then L T V m R + h φ + mgh cosφ 4 b Find the period of small oscillations about this axis. Retrieving the equation of motion from the usual Lagrangian formalism will yield the frequency of small oscillations ω, from which we can find the period. d L dt φ L φ d m R + h φ + mgh sinφ dt m R + h φ mgh sinφ φ gh R sin φ 5 + h From here, we make the approximation that sinφ φ so that Eq. 5 becomes a simplified nd order differential equation of the form ẍ φ x, whose general solution is xt x cos φt in which φ is the frequency. Hence, φ gh R + h φ gh φt φ cos R + h t The period of small oscillations is the time it takes Eq. 6 to repeat the initial conditions at t, such that cos φτ π cos. From this we arrive at the familiar expression for the period of a pendulum as expected: 6 τ π φ gh π R + h 7 8

17 Jeff Kissel October, 6 Classical Mechanics Hoop! There it is! c For what value of h is the period minimum? We wish to minimize τ with respect to h, i.e. set the derivative of Eq. 8 equal to zero, τ h π gh R + h gh 3 g R + h + gh R + h h g R + h gh 3 h + gh R + h g R + h R + h + h gh gh 3 R + h gr + h + gh gr gh + gh g h R R h h ±R 9 Only the non-trivial, positive value for h is physical, so the shortest period would result if the axis of rotation which has a distance h R from the center of mass of the hoop. Sanity check: if one wants to minimize the hoop s period, Eq. 7 says that one could maximize the frequency. The axis at which the hoop won t oscillate at all is zero is through center, and an axis with the maximum frequency is that which has the longest pendulum arm, i.e. at the edge of the hoop, or h R. 3

18 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel May 8, 7 An object is dropped from a tower of height h. The tower is located at the equator of the Earth. The rotational speed of the earth is Ω. Figure : Da Erf. Note that +ê x points North, and +ê y points West, so Ω Ω ê x. a If the Earth is treated as perfectly round and uniform and the acceleration due to gravity on a non-rotating earth is g, what is the acceleration of gravity at the tower? With Earth rotating, we must not only consider the acceleration due to gravity, but the centripetal and Coriolis acceleration. At the top of the tower, a grav g ê z, the radial vector is r +h ê z, and the velocity is v v r ê z. The effective acceleration in the rotating frame felt at the tower is then given as a eff a grav + a centrifugal + a Coriolis a grav Ω Ω r Ω v gê z Ωê x Ωê x hê z Ωê x v r ê z g ê z hω ê x ê x ê z Ωv r ê x ê z

19 Jeff Kissel May 8, 7 Classical Mechanics g ê z hω ê x ê y Ωv r ê y g ê z hω ê z + Ωv r ê y a eff Ωv r ê y g hω ê z Yet, at the top of the tower, we assume the object to have no radial velocity, v r. Thus, the effective acceleration is g eff g hω ê z b Even though it is released from rest, this object will not land directly below the point from which it was dropped. Calculate the amount and direction N, E, S, W, or elsewhere of the horizontal deflection of the object. You may assume the deflection is small. The Coriolis effect is the only term that will have an effect in the horizontal ê y, East or West direction, so pulling the first term from Eq. if v r is now g eff t, F Coriolis m a Coriolis mωv r ê y mω g hω t ê y m ÿ mωg hω t ÿ Ωg hω t ẏ Ωg hω t y Ωg hω 6 t3 y Ω 3 g hω t 3 3 This quantity y is the horizontal deflect, get we can be more explicit. The time spent falling t can be found from the free-fall kinematics of a particle released from rest: y v t g eff t h g hω t t h g hω which means the deflection becomes y Ω 3 g hω t / h g hω 4 h g hω 3/ y Ω h3/ 3g hω / 5 The direction of this displacement is in the ê y direction, or to the East.

20 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel October, 6 Consider a pendulum consisting of a uniform disk of radius, R and mass, m suspended from a massless rod that allows it to swing in the plane of the disk. Figure : A circular disk, rotating in its plane on an axis a distance, d away from the center of mass. a Using the parallel axis theorem, or calculating it directly, find the moment of inertia I for the pendulum about an axis a distance d < d < R from the center of the disk is The moment of inertia through the center of a disk of radius R and mass m I CoM mr Using the parallel axis theorem, which states that the moment of inertia around a point parallel to the center of mass a distance d away is I d I CoM + md So, using Eq. the moment of inertia displaced by r d is I d mr + md I d m R + d 3

21 Jeff Kissel October, 6 Classical Mechanics b Find the gravitational torque on the pendulum when displaced by and angle φ. From the definition of rotational torque from a given force: τ r F g r F sinφ d mg sin φ mgd sin φ 4 c Find the equation of motion for small oscillations and give the frequency ω. Further, find the value of d corresponding to the maximum frequency for fixed R and m. If we note that τ Iα, then we know the equation of motion immediately from Eq. 4using the small angle approximation that sin φ φ, I d α α mgd φ mgd φ I d 5 which is a nd order differential equation of the form φ ω φ from which we can pull out ω. Subbing in Eq. 3, ω ω mgd m R + d gd R + d 6 To find the value of d for which ω is maximized, we shall differentiate with respect to d and set to zero, dω d gdr + d dd dd gd g R R + d + d gdr + d 4d R + d g gd R + d 8gd R + d R + d gr + d 8gd gd R + d R + d gr + 4gd 8gd R + d gr d R + d R d

22 Jeff Kissel October, 6 Classical Mechanics R + d R d d ir d ± R 7 But we know the left set is unphysical and R is equivalent to R, so d max R 8 3

23 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel October, 6 A heavy particle of mass m is placed close to the top of a frictionless vertical hoop with radius R and allowed to slide down the hoop. Find the angle at which the particle falls off. We know the point at which the particle falls off the hoop is when the normal force N acting on the particle goes to zero. So, we need an expression for N. Figure : Ball of mass m on a hoop with radius R. Before the particle begins to move, it has only potential energy V i. At the instant it leaves the hoop, it has both potential V f and kinetic T f energies. The particle is allowed to slide, so there assumed to be no friction and therefore no rolling, thus we need no worry about the moment of inertial of the ball. The energy conservation equation is then E i E f V i T f + V f mgr mv + mgr cosθ However, Eq. does not contain the normal force N, which we need. It can be found by assuming the hoop is circular, so that the normal force acting on the

24 Jeff Kissel October, 6 Classical Mechanics particle is just its centripetal acceleration subtracted from its weight, or N mg cosθ ma c mg cosθ m v R Multiplying this by R will yield a more useful equation for the kinetic energy, R N mg cosθ ma c RN mgr cosθ mv mv mgr cosθ RN 3 which we can now plug into the conservation equation Eq. to arrive at explicit equation for N, mgr mgr cosθ RN + mgr cosθ RN 3 mgr cosθ mgr N 3mg cosθ mg 4 Since we ve established that the particle falls off the hoop when N, Eq. 4 becomes, mg3 cosθ cos θ 3 3 cosθ 5 So the angle at which a heavy particle of mass M slides off a circular hoop or radius R is θ 48. 6

25 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel October, 6 A solid sphere of radius R and mass M rolls without slipping down a rough inclined plane of angle θ. Figure : A sphere rolling down a rough ramp. Take the coefficient of static friction to be µ and calculate the linear acceleration of the sphere down the plane, assuming that it rolls without slipping. In this problem, we can use conservation of energy between the top and bottom of the ramp. At the top of the ramp, the ball begins rolling from rest and thus has no initial kinetic energy. This means the initial energy is E i T i + V i m CoM g x sin θ

26 Jeff Kissel October, 6 Classical Mechanics where the origin is set at the top of the ramp. The final energy has only kinetic terms, and is a sum of the motion of the center of mass, and the rotation of the sphere, E f T f + V f m CoM v + I CoM ω Setting Eqs. and equal to each other, plugging in for the moment of inertia of a sphere, I CoM 5 m CoM R and noting that v ẋ, and ω ẋ R, E i E f m CoM g x sin θ 5 m CoM ẋ + ẋ 7 g x sinθ dx dt where I ve defined a constant ξ 5 m CoM R ẋ R ξx 3 7 g sin θ. At this point we integrate both sides to find x as a function of time, which we can then differentiate twice with respect to time to get the acceleration, ẍ a CoM, x dx ξ x ξ dt t x 4 ξ t ẋ ξ t ẍ ξ ẍ 7 g sinθ a CoM 7 5 g sin θ 4 Calculate the maximum value of θ for which the sphere will not slip. The sphere will begin to slip once the x-component of the weight becomes greater than the maximum frictional force providable by µ. The angle at which this occurs, θ c can be found by setting the normal force, F g ˆx equal to the force of friction F f µn, mg sin θ c µn mg sin θ c µ mg cosθ c µ sin θ c cosθ c tanθ c θ c tan µ 5

27 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel October, 6 A rocket is filled with fuel and is initially at rest. It starts moving by burning fuel and expelling gases with the velocity u, constant relative to the rocket. Determine the speed of the rocket at the moment when its kinetic energy is largest. Figure : A free rocket traveling in an inertial frame, without the force of gravity. At some time t the rocket, traveling at a velocity v, has some mass m. An infinitesimal time dt later, the rocket has lost a mass dm, and gained a speed dv. At this time, an infinitessimal amount of fuel, dm is ejected at a velocity u with respect to the rocket. In an inertial frame however, that fuel s velocity is v u. Thus, using conservation of momentum, we can get an expression for the mass of the rocket as a function of time, p rocket p rocket + p fuel

28 Jeff Kissel October, 6 Classical Mechanics m v m dmv + dv + dm v u m v m v + m dv v dm du dv + v dm u dm m dv u dm du dv From here, we note that to first order, the term du dv is extremely small, so it is dropped, leaving u dv m dm Noting that dm dm and integrating both sides give us an express for the mass of the rocket as a function of time: v m u dv m m dm v m u ln m m m e v u 3 Plugging into the expression of the rocket s kinetic energy, T mv m v e v u 4 which we then maximize with respect to v to find the largest speed v max, T v v m v e v u m v e v u m v u e v u m v e v u v u m v e v u v max u 5 Thus, the kinetic energy is maximum when the velocity of the rocket is twice that of the ejected fuel.

29 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel May 9, 7 Exam Iteration Question Number Notes/Changes Spring 6 # Fall 6 # Spring 7 # Added damping. A small lead ball of mass m is attached to one end of a vertical spring with the spring constant k. The other end of the spring oscillates up and down with amplitude A and frequency ω. Determine the motion of the ball after a long period of time. You may assume that the ball is subject to a small amount of damping, with the damping force being given by F γv. Figure : A driven harmonic oscillator sitting in a viscous damping fluid. Hopefully, one can assume that all motion is in the ŷ direction.

30 Jeff Kissel May 9, 7 Classical Mechanics The kinetic energy of the system is T mẏ and the potential, as long as we define the origin to be at the equilibrium point such that the displacement y y y e y, is as follows, This makes the Lagrangian L V ky + mgy T V L mẏ ky mgy 3 We can use the form of the Euler-Lagrange equations which has forces not derivable from a potential, Q, to determine the equation of motion. The forces not derivable from potential are the driving and damping forces, So, the equation of motion is d L L Q i dt ẏ y d mẏ ky mg dt F o A cosωt 4 F d γẏ 5 A cosωt γẏ mÿ + ky + mg A cosωt γẏ mÿ + γẏ + ky A cosωt mg ÿ + γ m ẏ + k m y A m cosωt g Let β γ m, ω k m, and B A m ÿ + β ẏ + ω y B cosωt g Let x g + y y x + g ω ω ẋ ẏ ẍ ÿ ẍ + βẋ + ω x + g ω B cosωt + g ẍ + βẋ + ωx + g B cosωt + g ẍ + βẋ + ω x B cosωt 6 Yes folks, it s the driven harmonic oscillator with damping. Your favorite differential equation to solve.

31 Jeff Kissel May 9, 7 Classical Mechanics We ll start with the homogeneous solution. ẍ H + βẋ H + ω x H The characteristic equation: am ± + bm ± + c m ± + βm ± + ω has roots m p ± iq with p b and q p β and q ω β 4 c b 4 Such that the general solution is e pt C cosq t + D sin q t x H t e βt C cos ω β 4 t + D cos ω β 4 t 7 For the inhomogeneous solution, we can generalize the equation making it complex, x I + β x I + ω x I Be iωt and then take the real part at the end. This way we can assume a complex exponential solution, x I G e iωt, which knocks out the differentiation because x I G e iωt xi iω G e iωt xi ω G e iωt 8 which means all we have to do is find the complex constant G, Be iωt ω G e iωt + iωβ G e iωt + ω G e iωt B ω G + iωβ G + ω G B G ω ω iωβ B G ω ω iωβ B ω ω iωβ G Bω ω + ibωβ ω ω + ω β ω ω + iωβ ω ω + iωβ 9 3

32 Jeff Kissel May 9, 7 Classical Mechanics So the inhomogeneous solution is Bω ω x I t ω ω + ω β e iωt + [ Bω ω x I t Re ω ω + ω β e iωt + [ ] Re e iωt cosωt Re [ ie iωt] sin ωt x I t Bωβ ω ω + ω β ie iωt Bωβ ω ω + ω β ie iωt Bω ω ω ω + ω β cosωt + Bωβ ω ω + ω β sin ωt Yeah. The complete solution for xt is xt e C βt cos ω β 4 t + D cos ω β 4 t + Bω ω ω ω + ω β cosωt + Bωβ ω ω + ω β sin ωt Oh but wait, we have to plug back in for yt xt + g, ω yt yt e βt C cos + ω β 4 t + D cos Bω ω ω ω + ω β cosωt + ω β 4 t Bωβ ω ω + ω β sin ωt g ω And remember: β γ m, ω k m, and B A m γt e k m C cos m γ k 4m t + D cos m γ 4m t + Ak mω k mω + ω γ cosωt + Aωγ mg k mω + ω sinωt γ k Which is the position as a function of time, where C and D are constants determined by initial conditions. OK so what happens after a long time? The exponential damping in the homogeneous term kills off all resonances that would be found from ω, and the spring just oscillates according to the inhomogeneous equation. In other words, the spring just sloshes along with the driving frequency, according to lim yt Ak mω t k mω + ω γ cosωt + Aωγ mg k mω + ω sin ωt γ k 3 ] 4

33 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel May 7, 7 Exam Iteration Question Number Notes/Changes Spring 6 #3 Fall 6 #3 Spring 7 #3 Last sentence changed In movie cameras and projectors film speed is 4 frames/sec. You see on the screen a car that moves without skidding, and know that the real-life diameter of its wheels is m. The wheels on the screen make 3 turns/sec. What is the speed of the car, assuming it is not moving in excess of mi/hr? Figure : A wheel that appears to rotate one eighth of its circumference. In one frame, it appears to the viewer that the wheel has turned and angular velocity ω 3 [turns] [sec] 4 [frames] [sec] 8 turns frame

34 Jeff Kissel May 7, 7 Classical Mechanics However, appearances can be deceiving. To the viewer, measuring the position of the wheel discretely, the wheel has turned only /8 of its circumference. The car could be moving much faster, such that the wheel makes any integer number N of extra turns before reaching the final /8 for example, N in Figure. The angular velocity should then technically be ω N + turns 8 frame Finally, to find the linear speed v, we must convert Eq. into the correct units, ω N + [turns] 8 [f rame] 4 [f rames] π [radians] [sec] [turn] ω 48π N + rads/s 3 8 and then convert angular to linear speed, noting that the wheel has a m diameter, and thus a.5 m radius. v ω r 48π N + rads/s.5 m 8 4π N + m/s 4 8 Thus for N, v 3π m/s 6.7 mi/hr, and N, v 7π m/s 89.7 mi/hr. m/s mi/hr. Though this problem may seem pretty lame, it brings up an important concepts in signal processing: digital sampling of analog signals, which may result in aliasing. Check out the Wikipedia article on it!

35 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel October, 6 A ball of mass m collides with another ball, of mass M am, initially at rest. The collision is elastic and central, i.e. the initial velocity of the ball is along the line connecting the centers of the two objects. Figure : Two balls collide with mass ratio a. a Determine how the energy lost by the moving ball depends on the mass ratio a, and find the value of a for which the energy loss is largest. Describe what happens at that value of the mass ratio. We can solve this problem using conservation of energy and momentum. The problem does not mention any potential, so we can find the velocity of the second ball in terms of the first starting with kinetic energy conservation. E i E f mv i + Mu i mv f + Mu f m vi m vf + a m u f v i v f + au f We can now bring in conservation of momentum to knock out one of the unknowns, p i mv i + Mu i p f mv f + Mu f

36 Jeff Kissel October, 6 Classical Mechanics m v i m v f + a mu f v f v i au f Now, pluggind Eq. into Eq., vi v i au f + au f v i v i + a u f av iu f + av f a a + u f a v i u f a + u f v i u f v i a + Finally, the kinetic energy lost by the first ball is just that that is gained by the second, E Mu am vi a + 3 E amv i a + 4 To find the value for which the energy lost is largest, we can just maximize with respect to a, E mv a a i a a + mvi a + 4mvi a a + 3 mv i a a + a + 3 mvi a + 3 a a + a + a a 5 So, the change in energy is the greatest when the balls have the same exact mass: the first ball s kinetic energy is split evenly between them during the collision such that v f and u f are equal, but in opposite direcetions. b Investigate the limiting cases of heavy and light balls and comment on your result. In the limiting case that a, the first ball will be much more massive than the second. In this case, the first ball will lose virtually none of its energy, but because the second ball is so less massive, it will propel off with u f v i. lim a E amv i a + Mv i

37 Jeff Kissel October, 6 Classical Mechanics lim a v f lim u f a v i au f v i v i a + v i In the case where a, the energy imparted onto the second ball will be neglegible. In addition, since so little velocity is transferred to the second ball, the first will be forced in the opposite direction with a final velocity equal to its initial. lim E amv i a a + lim v f v i av i a a + v i lim u f v i a a + 3

38 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel October, 6 Two particles of mass m and m+ move in circular orbit around each other under the influence of gravity. The period of the motion is T. They are suddenly stopped and then released, after which they fall towards each other. If the objects are treated as mass points, find the time they collide in terms of T. Figure : Left Two masses orbiting about each other suddenly stopped, as a two body problem. Right The same system reduced to a one-body problem using the reduced mass µ. As noted in the picture, the problem can be simplified to a one body problem using the reduced mass, m m µ m + m m + m m + µ m + + m Now a one body problem, we can find the Lagrangian of this more simple system, V k r

39 Jeff Kissel October, 6 Classical Mechanics T µṙ + µ r θ L T V L µ ṙ + r θ + k r where I ve defined the gravitational constant of the system k Gm m. From here we crank the usual Lagrangian formalism, d L L dt ṙ r d dt µṙ µr θ k r µ r µr θ + k r 3 d L dt θ L θ d µr θ dt µr θ θ 4 It appears as though the sooner will be a little more useful than the latter, so let s roll with it. We know for circular motion, not only is Eq. 4 true, but r is also true. Using this fact, we can distill Eq. 3 a bit to get a familiar law from my buddy Kepler: µ r µr θ + k r k r µr θ r 3 r 3 kt 4π µ k µ θ θ ω π T Now that we ve naively re-derived Kepler s 3rd, let s stop the masses. Taking Eq. 5 as our starting radius, r 3, we know stop the angular motion of the masses, i.e. set theta, so the Eq. 3 instead becomes µ r k µr θ + r r k µr r ṙ t r t ṙ r ṙ ṙ r 5

40 Jeff Kissel October, 6 Classical Mechanics ṙ dṙ k µr dr ṙ k µr + C ṙ k µr + C We can use the initial condition that at r r, ṙ, so that C k µr, and finally we can solve for the linear merger time, k ṙ µr k dr dt t k µ k µ µ k µr r r dr r r r dt r r dr 6 which is a horribly disgusting integral, that can be solved analytically using a few u substitutions, let u r, and then u u sec θ but I ll happily just leave it as Eq. 6. Numerically evaluated this turns out to be t T 8 7 3

41 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel October, 6 Three points masses of identical mass are located at a,,,, a, a and, a, a. Find the moment of inertia tensor around the origin, the principal moments of inertia, and a set of principal axes. Figure : A system of three identical masses. The moment of inertia tensor for a discrete mass system is given by I I jk i m i δjk r i x jx k i So for r a ˆx r a r a ŷ + a ẑ r a + 4a 5a r 3 a ŷ + a ẑ r3 4a + a 5a

42 Jeff Kissel October, 6 Classical Mechanics the diagonal elements of I are I m i r i x i x i i I I 33 mr x x + mr x x + mr3 x 3x 3 ma a + m5a + m5a ma m i r i y i y i i mr y y + mr y y + mr3 y 3y 3 ma + m5a a + m5a 4a 6ma 3 m i r i z i z i i mr z z + mr z z + mr3 z 3 z 3 ma + m5a 4a + m5a a 6ma 4 And the off-diagonal elements are I I i m i x i y i m x y + m x y + m x 3 y 3 m + m + m 5 I 3 I 3 m i x i z i i m x z + m x z + m x 3 z 3 m + m + m 6 I 3 I 3 m i y i z i i m y z + m y z + m y 3 z 3 m ma ma 4ma 7 So the moment of inertia tensor for this system is I ma 6ma 4ma 8 4ma 6ma

43 Jeff Kissel October, 6 Classical Mechanics The principle moments of inertia and principle axes are the eigenvalues and eigenvectors respectively of the moment of inertia tensor. So to find eigenvalues, we find the characteristic equation of I [ ] det I λ ma λ 6ma λ 4ma 4ma 6ma λ ma λ 6ma λ6ma λ 4ma + ma λ 36m a 4 λma + λ 6m a 4 ma λ λ λma + m a 4 ma λma λma λ λ ma λ ma λ 3 ma 9 so there is a two-fold degeneracy in the moments of inertia which means the system is akin to a symmetrical top. To find the eigenvectors principle axes, we demand that the principle moments of inertia obey the eigenvalue equation I λ x. So for λ λ ma, 4ma 4ma 4ma 4ma x y z 4ma y 4ma z y z so, picking the simplest eigenvectors, the principle axes for λ & λ are χ and χ And for λ 3 ma, 8ma 4ma 4ma 4ma 4ma x y z 8ma x 4ma y 4ma z 4ma y + 4ma z x, y z and again picking the simplest eigenvector, the principle axis for λ 3 is χ 3 3

44 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel October, 6 Consider a double pendulum consisting of a mass m suspended on a massless rod of length l, to which is attached by a pivot another identical rod with an identical mass m attached at the end, as shown in the Figure. Figure : A double pendulum, with identical masses and string lengths. Using the angles θ and θ as generalized coordinates, a Find a Lagrangian for the system. Using θ and θ as generalized coordinates, the position r, r and velocity v, v vectors become r l sin θ ˆx + l cosθ ŷ v r l θ cosθ ˆx l θ sin θ ŷ r l sin θ + l sin θ ˆx + l cosθ + l cosθ ŷ

45 Jeff Kissel October, 6 Classical Mechanics l sin θ + sin θ ˆx + l cosθ + cosθ ŷ 3 v r l θ cosθ + l θ cosθ ˆx l θ sinθ + l θ sinθ ŷ l θ cosθ + θ cosθ ˆx l θ sinθ + θ sin θ ŷ 4 In the kinetic energy term of the Lagrangian, we ll need the squares of the velocities, which are v v v l θ cosθ + l θ sin θ l θ cos θ + sin θ l θ 5 v v v l θ cosθ + l θ cosθ + l θ sin θ + l θ sinθ l θ cos θ + l θ cos θ + l θ θ cosθ cosθ + l θ sin θ + l θ sin θ + l θ θ sin θ sinθ l θ cos θ + sin θ + l θ cos θ + sin θ + l θ θ cosθ cosθ + sin θ sin θ l θ + l θ + l θ θ cosθ θ 6 Finally, noting that g g ŷ and m m m we can combine the results of Eqs., 3, 5, and 6 to find the kinetic and potential energies: T m v + m v ml θ + l m θ + l θ + l θ θ cosθ θ ml θ + ml θ + ml θ θ cosθ θ 7 V m g r + m g r mgl cosθ mgl cosθ + cosθ mgl cosθ mgl cosθ 8 which we combine to form a Lagrangian for a double pendulum with identical masses and string lengths: L T V ml θ + ml θ + ml θ θ cosθ θ + mgl cosθ + mgl cosθ ml θ + θ + θ θ cosθ θ + mgl cosθ + cosθ 9

46 Jeff Kissel October, 6 Classical Mechanics b Find an approximate Lagrangian that is appropriate for small oscillations and obtain from it the equations of motions when θ & θ. For small oscillations, we can approximate cosx x + Ox 4, and drop those terms which are of higher order than quadratic in θ i and θ i so that Eq. 9 is L ml θ + ml θ + ml θ θ θ θ + mgl θ + mgl θ where what is canceled above will turn out to be quartic in θ i and θ i so it is dropped. Simplified, the approximate Lagrangian is then L ml θ + ml θ + ml θ θ + mgl mglθ + mgl mglθ ml θ + θ + θ θ mgl θ + θ + 3mgl c Assuming that each angle varies as θ, A, e iωt, find the frequencies for small oscillations. Performing the usual Lagrange processes using Eq. to find the equations of motion, we can then use these to solve for the frequencies of small oscillations. In general, Lagrange s equations of motion are defined by d dt L q i L q i where q i are the generalized coordinates of the system, and q i are their respective time derivatives. So the equations of motion for θ and θ are L θ ml θ + ml θ L θ mgl θ ml θ + ml θ + mgl θ l θ + θ + gθ L θ ml θ + ml θ L θ mgl θ ml θ + ml θ + mgl θ l θ + θ + g θ 3 3

47 Jeff Kissel October, 6 Classical Mechanics If we assume θ j A j e iωt, then θ j ω θ j, and the equations of motion yield four oscillation frequencies: l ω θ ω θ + g θ ω l A e iωt + A e iωt ω l A + A ga + ga e iωt l ω θ ω θ ω l A e iω t + A e iω t ω ga la + A ga ω ± ± la + A + g θ + ga e iω t ω l A + A ga ω ga la + A ω ± ga ± la + A 4 5 So der you go. 4

48 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel October, 6 A particle of mass m. at rest initially, slides without friction on a wedge of angle θ and and mass M that can move without friction on a smooth horizontal surface. Figure : A mass m slides in the +ˆx direction down a frictionless wedge M, which can slide across a frictionless plane in the ˆx direction. Note, this picture does not represent the actual motion; realistically Ẋ is in the ˆx direction as will be shown later. a What is the Hamiltonian? First we must find the Lagrangian L T V, to determine whether it is explicitly independent of time. If this is so, then the Hamiltonian is simply the total energy H T + V. We can define a coordinate system where the origin is the top of the wedge at some initial time as in Figure, such that the position vector for the top corner of the wedge at time, t will be and particle s position vector will be R X ˆx + ŷ r R + x ˆx + y ŷ r X + l cosθ ˆx + l sinθ ŷ

49 Jeff Kissel October, 6 Classical Mechanics where l is the distance the mass has traveled down the ramp. While the particle is on the ramp, it is constrained to move along l, or tanθ y x sin θ y Y cosθ x X x Xsinθ y cosθ f : x Xsinθ y cosθ 3 which is an equation of constraint related to the normal force. The kinetic energy of the system is that of two free particles, since there is no friction: T mẋ + ẏ + MẊ 4 The only potential in the system is gravitational, with g g ŷ. We can ignore the potential on the wedge because it is constant, so the system s potential is just the gravitational potential of the particle, V g m g r V mgy 5 The Lagrangian is then potential subtracted from the kinetic energy, L T V mẋ + ẏ + MẊ + mgy L mẋ + mẏ + MẊ + mgy 6 which does not explicitly depend on time, so the Hamiltonian is the sum of the kinetic and potential energies, H mẋ + mẏ + MẊ mgy 7 b Derive the equation of motion from the Lagrangian. With Lagrange multipliers, The Euler-Lagrange equation in general is d L L λ f + µ f 8 dt q i q i q i q i yet the second term on the right-hand side is zero because Eq. 3 is not a function of any q i. From Eq. 8, the equations of motion for each coordinate are, d mẋ λ sin θ dt

50 Jeff Kissel October, 6 Classical Mechanics mẍ λ sin θ 9 d mẏ mg dt λ cosθ mÿ mg λcosθ d mẋ dt mẍ λ sin θ λsin θ Immediately, if we add Eq. to Eq. 9, we see why Figure has the direction of motion for the wedge incorrect, mẍ + MẌ ẍ M m Ẍ Also, this tells us that the ˆx position of the system s center of mass stays stationary because the only external force on the system is gravity, which is in the ŷ direction. We ve assumed that the particle s initial position is at the origin, but if we also assume that the initial velocity of the center of mass is also zero then from Eq., x M X t t m t t x t t M X t m t x x M m X X x M m X or X m M x 3 giving us an equation relating X in terms of x. We can use the equation of constraint Eq. 3 to get y in terms of x, x Xsinθ y cosθ x + m xsin θ y cosθ M from which only y and x are a function of time, so y x + m tanθ 4 M ẏ ẋ + m M tanθ ÿ ẍ + m tanθ 5 M 3

51 Jeff Kissel October, 6 Classical Mechanics OK, OK, let s finally get the equations of motion. Starting with Eq. 9, mẍ λsin θ mÿ g Eq. : mÿ mg λcosθ λ cosθ mẍ mg ÿtanθ 6 Eq. 5: ÿ ẍ + m M tanθ mẍ + mẍtan θ + mẍ m M tan θ mg tan θ m + tan θ + m M tan θẍ mg tanθ cos θ cos θ + sin θ cos θ + m M mẍ mg tanθ mẍ + m M tan θ sin θ sin θ cos ẍ g θ cosθ cos θ + sin θ + m/msin θ ẍ g sinθ cosθ ẍ g sin θ cosθ + m/msin θ ẍ M sinθ cosθ g M + m sin θ 7 Since the acceleration in the ˆx direction is a constant in time, a x Eq. 7 has the usual -D kinematic solution, xt x + v,x t + a xt set x, and v,x xt g M sin θ cosθ M + m sin θ t 8 For the ŷ equation of motion, we plug Eq. 7 back into Eq. 5, ÿ ẍ + m M tanθ M sin θ cosθ g M + m sin θ ÿ g M + msin θ M + m sin θ M M + m sinθ cosθ 9 which is also constant, so with the same initial conditions y, y,y we get a similar free-fall equation, y M + msin g θ M + m sin t θ 4

52 Jeff Kissel October, 6 Classical Mechanics Finally we get the solution for X plugging Eq. 7 into Eq., Ẍ m M ẍ m M g M sinθ cosθ M + m sin θ m sin θ cosθ Ẍ g M + m sin θ X, and v,x X m sin θ cosθ g M + m sin t θ One could also solve for the normal force λ via Eq. 9, but the problem does not ask for it, so it is left as an exercise to the reader. Ew. I feel so dirty saying that. c What are the constants of motion? To find the constants of motion, i.e. conserved quantities, we need to express the Lagrangian Eq. 6, in terms of independent coordinates. We can use the constraint equation Eq. 3 to solve for X, x Xsinθ y cosθ x X y tan θ X x y tanθ Ẋ ẋ ẏ tanθ L mẋ + mẏ + MẊ + mgy L mẋ + mẏ + M ẋ ẏ + mgy tan θ Because this Lagrangian is explicitly independent of x, then the L x term in the Euler-Lagrange equation will be zero. Thus, the canonical momentum in the ˆx direction is a conserved quantity. p x L ẋ p x mẋ + M ẋ ẏ tanθ m + Mẋ Mẏ tanθ 3 General Rule: If the Lagrangian is independent of any generalized coordinate, that coordinate is cyclic, and therefore it s respective canonical momentum p i L/ q i is conserved. 5

53 Jeff Kissel October, 6 Classical Mechanics Also, the Lagrangian does not explicitly depend on time, so as previously stated, the total energy is conserved which is why we could write the Hamiltonian as the sum of kinetic and potential energies. d Describe the motion of the system. Figure : An alternate view of the particle-wedge system at times t,, 3 and 4. Since friction is not involved in the system, the ˆx coordinate of the center of mass between the two objects will remain stationary as has been shown in part b. The ŷ coordinate will accelerate as though in slow free-fall a la Eq., until the particle reaches the surface on which the wedge slides. From then on, it will simply travel according to Eq. 8. Thus, as in Figure from a reference point following the center of mass, the wedge and particle will fly apart from each other. If M m, it will not move much Ẍ, and the system approximates that of a mass sliding down a fixed incline. 6

54 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel October 6, 6 A uniform rod slides with its ends inside a smooth frictionless vertical circle of radius a. The rod of uniform density and mass m subtends an angle of at the center of the circle. Figure : A rod, with length L which is the length of a chord subtending is restricted to slide inside a frictionless circular pipe. a Compute the center of mass moment of inertia of the rod I CoM in terms of m and a not the length of the rod. In most general terms, the moment of inertia tensor is calculated as I CoM ρrδ ij r x i x j d 3 r However, we assume the rod is only has one dimension so we need only integrate along dl instead of d 3 r; there are no off-diagonal terms i.e. x i x j ; and the rod has constant density, so ρr m/l, which means Eq. becomes considerable less intimidating, I CoM m L +L/ L/ l dl m L L/ l dl

55 Jeff Kissel October 6, 6 Classical Mechanics m L l 3 3 L/ m L 3 3 L 4 m L3 L I CoM ml Finally, because we know angle which the rod subtends is, it forms an isosceles triangle whose sides are a : a : L, and interior angles are : 3 : 3, which means L/ a cos3, or L 3 a. Thus, in terms of m and a, the center of mass moment of inertia is I CoM 4 ma 3 b Obtain the potential energy, the kinetic energy, and the Lagrangian Lθt, θt; m, a, g for this system, where θt the dynamical variable, is the instantaneous angular position relative to its equilibrium position and m, a, and g are constant parameters of the system. As the question implies later, once we have the center of mass moment of inertia, we can treat the rod like a pendulum, as long as we displace I CoM to the center of the circle, a distance d a sin3. So, following the parallel axis theorem, I d I CoM + md 4 ma + ma sin 3 4 ma + m 4 ma + 4 ma a I d ma 4 Now the kinetic energy T is only rotational, the potential V is only gravitational since there is no friction, and the Lagrangian is L T V, T I d θ ma θ T 4 ma θ 5 V mga cos θ 6 L 4 ma θ mga cosθ 7

56 Jeff Kissel October 6, 6 Classical Mechanics c Compare the dynamics of this system with that of a simple pendulum with mass M and length L exactly, without any approximations such as the small oscillation approximation. Compare the parameters L and M to a and m. With out making any approximations I promise! we can find the equation of motion, and compare those to that of a simple pendulum. Let s get Lagrangimiphysical! d dt d L dt θ ma θ L θ + mga sinθ ma θ mg asin θ θ g sin θ a 8 which is indeed comparable to a simple pendulum s equation of motion which is different only by the factor of if L a and M m. For a pendulum of length L, and mass M, T ML θ V MgL cosθ L ML θ MgL cosθ d L dt θ L θ d ML θ + MgL sinθ dt ML θ Mg Lsin θ θ g sin θ L 9 c Find the frequency of small oscillations of the system. Oh, NOW it s OK to make approximations, huh? Fine. Gosh. In using the small angle approximation, sinθ θ, which means Eq. 8 becomes a second-order differential equation of the form ẍ ω x which has the general solution xt x cosωt, where ω is the frequency small of oscillations. θ g a θ g ω a 3

57 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel May 8, 7 Exam Iteration Question Number Notes/Changes Spring 6 X Fall 6 X Spring 7 # Added to bank. A mass m is attached to a spring of spring constant k that can slide vertically on a pole without friction, and moves along a frictionless inclined plane as shown in Fig. After an initial displacement along the plane, the mass is released. Find an expression for the x and y position of the mass as a function of time. The initial displacement of the mass is x. You may assume that the object never slides down the ramp so far that it strikes the floor. Figure : A box of undisclosed materials oscillating on a ramp. If one sets up the coordinates as in Fig, where the equilibrium is at the origin, then two important simplifications occur: y ax + b y ax x x x e x x With these simplifications, the kinetic energy is T m ẋ + ẏ 3

58 Jeff Kissel May 8, 7 Classical Mechanics and the potential energy is the sum of that from the spring and that from gravity, V k x + mgy Because of Eq., x x V kx + mgy 4 Making the Lagrangian for the system in terms of x and y, L T V m ẋ + ẏ kx + mgy L mẋ + mẏ kx mgy 5 However, there is a fixed relationship between x and y, namely the Eq., so which makes Eq. 5 a function only of x and ẋ, y ax ẏ aẋ 6 L mẋ + maẋ kx mgax mẋ + ma ẋ kx mgax L m + a ẋ kx mgax 7 Using the Euler-Lagrange equation, we can then find an expression for xt, d L L dt ẋ x d m + a ẋ kx mga dt m + a ẍ + kx + mga ẍ k m + a x ga m + a Let ω k m + a, and R ga m + a ẍ ω + R Let x ω x + R ẋ ω ẋ ẍ ω ẍ x R ẍ ω ω ω + R

59 Jeff Kissel May 8, 7 Classical Mechanics ẍ ω x ẍ ω x x t Acos ωt 8 Where the last step we know comes from the typical harmonic oscillator solution, with A as some constant. Plugging back in for x, and using the initial conditions that x x, ω xt + R Acosωt xt A ω cosωt R ω x x A ω R ω x + R ω A ω A ω x R ω x R ω cosωt R ω x + Rω cosωt R ω R ω ga m + a m + a k k ω m + a xt x ga k cos k m + a t + ga k ga k 9 and using Eq., we can get yt, yt axt yt a x ga k cos k m + a t + ga k 3

60 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel May 7, 7 Exam Iteration Question Number Notes/Changes Spring 6 X Fall 6 X Spring 7 # An astronaut is on the surface of a spherical asteroid of radius r and mean density similar to that of the earth. On the earth this astronaut can jump about a height h of.5 m. If he jumps on this asteroid, he can permanently leave the surface. a Taking the radius of the earth as m, find the largest radius the asteroid can have. With the initial velocity on earth v, the astronaut jumps a height h. If he can leave the asteroid with this same velocity, we can treat v as the escape veloctiy for the asteroid v esc. We can determine his velocity on Earth from -D kinematics using the usual trick that his final velocity on the way down is equal to his initial velocity on the the way up, v f v i a x v g h v g h The escape velocity on the asteroid occurs when the astronaut s total kinetic energy equals the potential felt from the asteroid, i.e. his total energy is zero, E m av GM Am a r A m av esc GM Am a m avesc GM A m a r A r A

61 Jeff Kissel May 7, 7 Classical Mechanics v esc GM A v esc r A GMA r A G r A r A ρ A MA V A M A ρ A V A M A 4π 3 r3 A ρ A 4π 3 r3 A ρ A G 4πρ A 3 ρa ρ M 4π 3 R3 ρ A r A 4π G 3 GM v esc r A R 3 g g v r A R r g A g h M 4π 3 R3 GM R, v esc v R r A R h m.5 m r A m b How fast could the asteroid rotate and not have the astronaut be flung away from the surface? Here, the maximum rotation speed would be when the astronaut s weight on the asteroid exactly equals the asteroid s centripetal acceleration, F a m ag A g A m a a c vt,a m a r A v t ωr ω max r A

62 Jeff Kissel May 7, 7 Classical Mechanics ω max ga r A 3 Yet, we don t know what gravity is like on this asteroid. We can find out, because its density is equal to Earth s, g A GM A r A G 4π ra 3 r3 Aρ A Gr A 4π M 3 4π 3 r A GM R R R 3 g A r A R g 4 So the maximum rotation the asteroid can have with out flinging the astronaut into space is ω max r A g R r A g ω max 5 R 9.8 m/s m ω max.3 3 rads/s 6 3

63 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel May 8, 7 Exam Iteration Question Number Notes/Changes Spring 6 #X Fall 6 #4 Added to question bank. Spring 7 #4 In 956, G. Kuzmin introduced the simple axisymmetric potential GM Φ K R, z R + a + z as an approximate description of the potential of a disk-like galaxy. a Show that this potential can be though of as the result of two point masses placed at R, z ±a, generating the potential in the half-space z > and z < respectively. In cylindrical polar coordinates, a point mass placed at the origin generates the following potential, Φ PM R, z GM R + z where R x + y. If the point mass is displaced from the origin a positive distance z, without moving radially, then Φ PM R, z becomes Φ PM R, z + z GM R + z z GM 3 R + z z where I m able to flip the quantity in parentheses because it is squared. In the region of z <, The Kuzmin potential takes the form GM Φ K R, z 4 R + a z

64 Jeff Kissel May 8, 7 Classical Mechanics and hence by direct comparison with Eq. 3, for z <, Eq. 4 is the potential of a point mass centered at z a, and R. Similarly, the potential of a point mass shifted from the origin a negative z again keeping R R is Φ PM R, z z GM R + z z GM 5 R + z + z In the region of z >, the Kuzmin potential is GM Φ K R, z 6 R + a + z so, again by direct comparison with 5, Eq. 6 is the potential of a point mass centered at z a, and R. b Show that all the mass must be located on the z plane and distributed according to the surface density Σ k R am πr + a 3/ 7 If ΦR, z below the z plane acts as a point mass at z a, then above the z plane, Poisson s Equation Φ 4πGρ contains no mass density ρ. Poisson s Equation then becomes Laplace s Equation: Φ. Similarly, if ΦR, z above the z plane acts as a point mass at z a, then the potential also satisfies Laplace s Equation in the region above the z plane. If ΦR, z > ΦR, z <, assuming the mass must be somewhere, then all the mass must lie in an infinitely thin disk at z, of mass density Σ. Taking a Gaussian pill-box surrounding the sheet of mass Σ, we know the potential at z must satisfy Gauss Law of flux, Φ K R, d a 4πGΣ K A 8 pill-box where the A is the area of the disk. We can simplify the left-hand side a bit, Φ K R, d a ΦK da ΦK pill-box z z da A A z Φ K 9 z A However, the partial derivative is discontinuous across z, so we can take the difference between the partial derivative coming from the top and from the bottom, 4πGΣ K A A z Φ KR, +z z z Φ KR, z z GM z R + a + z z

65 Jeff Kissel May 8, 7 Classical Mechanics GM z R + a z z GM R + a + z / z z + GM R + a z / z [ z GM ] R + a + z 3/ a + z z [ + GM ] R + a z 3/ a z z a + GM + R + a + + a 3/ R + a 3/ z a 4π GΣ K GM R + a 3/ Ma Σ K π R + a 3/ c What is the total mass generating Φ K? We can integrate the surface density Σ over the disk to find the total mass: S S Σ da π π Ma R dr dφ R + a 3/ π Ma R dr π R + a 3/ du R dr let u R + a du R dr a, Ma [ Ma u 3/ du Ma a + ] a [ u /] Σ da M Thus, the total mass gererating Φ K is M. a 3

66 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel May 9, 7 Exam Iteration Question Number Notes/Changes Spring 6 X Fall 6 X Spring 7 #5 Added to bank. A particle of mass m moves in a force field given by a potential energy V Kr s where both K and s may be positive or negative. a For what values of K and s do stable, circular orbits exist? These results are a product of Goldstein Section 3.6 pp89-9. Circular orbits occur when the effective potential, has an extremum, i.e. when V eff r V rr r V eff V + l mr + l rr r mr rr Fr V r V r FR + l mr 3 FR l mr 3 l mr ω FR mω R 3 where I ve denoted R as the circular radius, of which Ṙ. However, these extrema can be either a maxima or minima. Circular orbits are only stable when the the extrema of the effective potential is a minima, i.e.

67 Jeff Kissel May 9, 7 Classical Mechanics when the second derivative of the effective potential is positive. This restricts both K and s, V eff rr r Veff r r < r F eff rr < F + l r mr 3 rr F < 3l r rr mr rr F r rr F V skr s r F ss Kr s r ss KR s ss KR s ss KR s < F r s < +3 s l l mr 4 < 3 R mr 3 Eq. : FR l mr 3 < 3 FR R F V skr s r skr s < 3 R < +3 s > 3 where canceling out K demands that it s positive so KR s s > 4 K < 5 b What is the relation between the period P of the orbit and the radius R of the orbit? We ve already shown that F V r skr s 6 We can set this equal to Eq. 3, using r R and ω π/p because the orbit is circular, skr s mω R

68 Jeff Kissel May 9, 7 Classical Mechanics ω P π mr skr s m skr s P 4π m sk R s P 4π m sk R s/ 7 For a quick sanity check, let s make sure we get Kepler s 3rd Law when K GMm and s, L chayim! P Kepler 4π m GMm R / 4π m GM m R+/ P Kepler 4π GM R3/ 8 3

69 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel November, 6 All parts of this question call for quick and brief answers. a A bead of mass m slides along a wire bent into a parabolic shape. The wire is pivoted at the origin and is spinning around the vertical. Identify suitable generalized coordinates and constraints to describe the bead s motion. Figure : A bead confined to a parabolic wire that is spinning about the ẑ axis. A suitable set of generalized coordinates would be cylindrical coordinates, r, φ, and z, as shown in Figure where the equation of constraint would be that z r. b A system with generalized coordinates q, q and q 3 is described by the Lagrangian L q, q, q 3, q 3 ; that is, the Lagrangian is independent of q, q and time explicitly. What are the conserved quantities of this motion? Short answer: The conserved quantities are energy, p, and p. Thorough answer: The canonical momentums are defined as p i L q i

70 Jeff Kissel November, 6 Classical Mechanics Which means if the Lagrangian is independent of any generalized coordinate q i, then its Euler-Lagrange equation will yield a constant of motion, d dt L q i q i d dt p i p i constant L Thus, since the Lagrangian is independent of q and q, p and p are constants of motion. Also, if the Lagrangian is independent of time then the energy function h is constant of motion, i.e. energy is conserved, and the Hamiltonian is H T +V. The proof is as follows, L dq i + L L d q i + q i i q i i t d L dq i + L d q i dt q i i q i i d L q i d L dt q i dt q i d L q i dt q i d L q i dl dt q i i dt d L q i L dt q i dl dt dl dt Constant h i i dq i + L d q i q i dt q i L q i L 3 c If the kinetic energy is a quadratic function of generalized velocities, k,l a kl q k q l, express k q k T/ q k in terms of T. Euler s theorem states if a function f is a an homogeneous function of degree n in variables x i, then f x i nf 4 x i Since T is quadratic in terms of q, T has degree, so k i q k T q k T 5

71 Jeff Kissel November, 6 Classical Mechanics d A hollow and a solid sphere, made of different materials so as to have the same masses M and radii R, roll down an inclined plane, starting from rest at the top. Which one will reach the bottom first? Short Answer: The solid sphere will reach the bottom faster because it has a smaller moment of rotational inertia, i.e. less energy is put into rotating the solid sphere so more energy can by used in linear velocity of the center of mass. Thorough answer: The moments of inertia for a solid and hollow sphere abbreviated as SS and HS, respectively are I SS 5 MR I HS 3 MR 6 Remember v ωr, so each sphere s kinetic energy is T SS Mv CoM,SS + I SSω Mv CoM,SS + 5 M R vcom,ss R Mv CoM,SS + 5 Mv CoM,SS T SS 7 Mv CoM,SS 7 T HS Mv CoM,HS + I HSω Mv CoM,HS + 3 M R vcom,hs R Mv CoM,HS + 3 Mv CoM,HS T HS 5 6 Mv CoM,HS 8 Energy is conserved in this system, so the gravitational potential at the top of the ramp will equal the kinetic energy at the bottom, so V i T f Mgh 7 Mv CoM,SS Mgh 5 6 Mv CoM,HS vcom,ss 7 gh v CoM,HS 6 gh 9 5 which means the solid sphere will arrive first, since it is traveling at a faster linear velocity /7 > 6/5 down the ramp. 3

72 Jeff Kissel November, 6 Classical Mechanics e Write down the Hamiltonian for a free particle in three-dimensional spherical-polar coordinates. Short Answer: In physics spherical-polar coordinates where θ is the angle from the ẑ axis, and φ is the angle from the ˆx axis in the xy plane, H p r m + p θ mr + p φ mr sin θ is Thorough Answer: The Lagrangian in free space in spherical coordinates L mṙ + r θ + r sin θ φ and Hamiltonian can be found from and is in fact defined as a Legendre transformation between q i and p i, H i q i p i L where p i L q i So, turning the crank on Eq., and plugging back into Eq., p r L ṙ mṙ ṙ p r m p θ L θ p θ mr θ θ mr p φ L φ mr sin θ φ φ the Lagrangian as a function of q i and p i is then L m pr m + r p θ mr + r sin θ p φ mr sin θ p φ mr sin θ L p r m + and the Hamiltonian is pr H m p θ mr + p φ mr sin θ pθ p r + mr p φ p θ + mr sin θ p φ 3 p r m p r m + p θ mr + p θ mr p φ mr sin θ p φ mr sin θ p r m H p r m + p φ mr + p φ mr sin θ p θ mr p φ mr sin θ 4 4

73 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel November, 6 All parts of this question call for quick and brief answers. a In a bound Keplerian orbit, what is the ratio of the average kinetic energy to the average potential energy? Short Answer: The only force experienced by a particle in a bound Keplerian orbit is the central, gravitational inverse square force a power law of order n. This gravitational force is derivable from a scalar is derivable from a scalar potential V. Any system satisfying these conditions follows the Virial Theorem: the average kinetic energy is twice that of the average potential, or T V Thorough Answer: An useful scalar quantity known as the Virial, or the Virial of Clausius, is defined as Gt i p i r i where the summation is over all particles in a given system. Newton s equations for each particle in the system are F i d p i dt which gives us the inspiration to take the full time derivative of the Virial, dgt dt i i d p i dt r i + i F i r i + i p i d r i dt m i ri r i 3 i F i r i + i mṙ i dgt dt i F i r i + T 4

74 Jeff Kissel November, 6 Classical Mechanics Since we re interested in the time averaged kinetic energy, T, we ll integrate Eq. 4 over time, from to some time τ and normalize by the same time τ, τ τ dgt dt dt Gτ G τ i i F i r i + T F i r i + T 5 If we chose τ to be the period and motion is cyclic as in a Keplerian orbit, the left-hand side vanishes Gτ G leaving T i F i r i T i F i r i 6 which is the general form of what is known as the Virial Theorem. If the forces are derivable from a potential, as in a Keplerian orbit i.e. F V then the theorem becomes T i V r i 7 and for a single particle moving under a central force as in a particle in a Keplerian orbit, this reduces to T V r r 8 If V is a power law as in a Keplerian orbit, where the exponent is chosen so that the force law goes as r n, then V r r n + V V ar n+ T n + V 9 which means for a Keplerian orbit whose force follows and inverse square law, n, the average kinetic energy and average potential have the following constant of proportionality T V

75 Jeff Kissel November, 6 Classical Mechanics b Write down the Lagrangian and the Hamiltonian of a Free Particle of mass m in three dimensional polar coordinates. Short Answer: The Lagrangian and Hamiltonian in polar spherical coordinates are L mṙ + r θ + r sin θ φ H p r m + p θ mr + p φ mr sin θ Thorough Answer: A free particle is that which is traveling at some velocity v r, without the the influence of any potentials. Thus, the Lagrangian L T V will only contain the kinetic term T. The Cartesian components of the particle s position in spherical-polar coordinates where we use the physics definitions of θ and φ: θ is the angle from the ẑ axis, and φ is the angle from the ˆx axis in the xy plane are, x y z r sin θ cosφ r sin θ sin φ r cosθ so the components of velocities are, ẋ ṙ sin θ cosφ + r cosθ cosφ θ r sinθ sin φ φ ẏ ṙ sin θ sin φ + r cosθ sinφ θ + r sin θ cosφ φ ż ṙ cosθ r sinθ θ The kinetic energy T of any given particle can always be written as T m ẋ + ẏ + ż 3 So, we can plug in the spherical-polar versions of ẋ, ẏ, and ż from above, T [ m ṙ sin θ cos φ + r cos θ cos φ θ + r sin θ sin φ φ +rṙ sin θ cosθ cos φ θ {}} { rṙ sin θ sin φcosφ φ r sin θ cosθ sin φcosφ θ φ + ṙ sin θ sin φ + r cos θ sin φ θ + r sin θ cos φ φ +rṙ sin θ cosθ sin φ θ {}} + { rṙ sin θ sinφcos φ φ + r sin θ cosθ sin φcosφ θ φ + ṙ cos θ + r sin θ θ ] rṙ sin θ cosθ θ 3

76 Jeff Kissel November, 6 Classical Mechanics ṙ sin θsin φ + cos φ + r cos θsin φ + cos φ θ + r sin θsin φ + cos φ φ rṙ sinθ cosθsin φ + cos φ θ + ṙ cos θ + r sin θ θ rṙ sin θ cosθ θ sin α + cos α ṙ sin θ + r cos θ θ }{{} +r sin θ φ + rṙ sin θ cosθ θ + ṙ cos θ + r } sin {{ θ θ } rṙ sinθ cosθ θ ṙ sin θ + cos θ + r sin θ + cos θ θ }{{} +r sin θ φ T m ṙ + r θ + r sin θ φ T m ṙ + r θ + r sin θ φ 4 Now as previously mentioned, the potential V for a free particle is zero, so the Lagrangian L T V as a function of q i and q i is L m ṙ + r θ + r sin θ φ 5 The Hamiltonian can be found from and is in fact defined as a Legendre transformation between q i and p i, H i q i p i L where p i L q i 6 So, turning the crank on Eq. 5, and plugging back into Eq. 6, p r L ṙ mṙ ṙ p r m p θ L θ p θ mr θ θ mr p φ L φ mr sin θ φ φ the Lagrangian as a function of q i and p i is then L m pr m + r p θ mr + r sin θ p φ mr sin θ p φ mr sin θ L p r m + p θ mr + p φ mr sin θ 7 4

77 Jeff Kissel November, 6 Classical Mechanics and the Hamiltonian is pr H m pθ p r + mr p φ p θ + mr sin θ p φ p r m p r m + p θ mr + p θ mr p φ mr sin θ p φ mr sin θ p r m H p r m + p φ mr + p φ mr sin θ p θ mr p φ mr sin θ 8 c A system with generalized coordinates q, q and q 3 is described by the Lagrangian L q, q, q 3, q 3 ; that is, the Lagrangian is independent of q, q and time explicitly. What are the conserved quantities of this motion? Short answer: The conserved quantities are energy, p, and p. Thorough answer: The canonical momentums are defined as p i L q i 9 Which means if the Lagrangian is independent of any generalized coordinate q i, then its Euler-Lagrange equation will yield a constant of motion, d dt L q i q i d dt p i p i constant L Thus, since the Lagrangian is independent of q and q, p and p are constants of motion. Also, if the Lagrangian is independent of time then the energy function h is constant of motion, i.e. energy is conserved, and the Hamiltonian is H T +V. The proof is as follows, dl dt i i d dt L dq i + L L d q i + q i q i i t d L dq i + L d q i dt q i q i i L q i d L q i dt q i dq i + L d q i q i dt 5

78 Jeff Kissel November, 6 Classical Mechanics dl dt L q i q i d dt d dt i d dt Constant h i i q i L q i dl dt q i L q i L q i L q i L To recap, we ve decided that h energy, p, and p are the conserved quantities of this system. d A hollow and a solid sphere, made of different materials so as to have the same masses M and radii R, roll down an inclined plane, starting from rest at the top. Which one will reach the bottom first? Short Answer: The solid sphere will reach the bottom faster because it has a smaller moment of rotational inertia, i.e. less energy is put into rotating the solid sphere so more energy can by used in linear velocity of the center of mass. Thorough answer: The moments of inertia for a solid and hollow sphere abbreviated as SS and HS, respectively are I SS 5 MR I HS 3 MR Remember v ωr, so each sphere s kinetic energy is T SS Mv CoM,SS + I SSω Mv CoM,SS + 5 M R vcom,ss R Mv CoM,SS + 5 Mv CoM,SS T SS 7 Mv CoM,SS 3 T HS Mv CoM,HS + I HSω Mv CoM,HS + 3 M R vcom,hs R Mv CoM,HS + 3 Mv CoM,HS T HS 5 6 Mv CoM,HS 4 6

79 Jeff Kissel November, 6 Classical Mechanics Energy is conserved in this system, so the gravitational potential at the top of the ramp will equal the kinetic energy at the bottom, so V i T f Mgh 7 Mv CoM,SS Mgh 5 6 Mv CoM,HS vcom,ss 7 gh v CoM,HS 6 gh 5 5 which means the solid sphere will arrive first, since it is traveling at a faster linear velocity /7 > 6/5 down the ramp. e A satellite initially at a distance R E radius of the Earth above the Earth s surface is moved out to a distance R E. How does its orbital time period change? A satellite orbiting around the Earth can be approximated quite accurately by a Keplerian orbit. Kepler s 3rd law says that such an orbit s period P is proportional to the radius of orbit a taken to the 3/ s power, so P a 3 P a 3 P 8a 3 P 8P 6 The satellite s period increases by a factor of 8, if the radius of orbit is doubled. 7

80 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel May 8, 7 A particle of mass m moves under the influence of an attractive force F k r, where r is the distance from the force center. What are the conserved quantities of this three dimensional motion? If the Lagrangian of the system is independent of any generalized coordinate q i i.e. cyclic in q i, the its Euler-Lagrange equation demands that the respective canonical momentum p i is a constant of motion, d L L dt q i q i p i L q i d dt p i p i constant In order to determine if this system has any of its canonical momentums conserved, we must find the Lagrangian, L T V. The force of this system F k r is derivable from a potential V in which we need, F V V r k r dr V kr The kinetic energy T is that of any moving particle, which spherical polar coordinates where θ is the angle from the ẑ axis and φ is the angle from the ˆx to the project of the vector onto the xy plane is T m ṙ + r θ + r sin θ φ 3 Combining Eqs. and 3 yields the Lagrangian, L m ṙ + r θ + r sin θ φ kr 4

81 Jeff Kissel May 8, 7 Classical Mechanics Since Eq. 4 is independent of φ, p φ is a constant of motion. Also, if the Lagrangian is independent of time then the energy function h is a constant of motion. Because of this energy is conserved, and the total energy E of the system is the Hamiltonian E H T + V. The proof is as follows, dl dt dl dt constant i i i d dt L dq i q i dt + i d L dt q i d L dt L q i q i d dt dqi L d q i q i dt + L t L d q i q i dt dt + i dqi dt + L d q i q i dt L q i d dt L q i q i i d L q i dl dt q i i dt d L q i L dt q i i p i L q i d q i p i L dt i i q i dqi dt + L d q i q i dt q i p i L h 5 Write down an expression for the total energy. As shown in part a, the total energy E is conserved, thus it is the Hamiltonian E H T + V. From Eqs. and 3, the total energy as a function of q i and q i is then E H m ṙ + r θ + r sin θ φ + kr 6 Either from b or otherwise, derive an orbit equation relating r to θ. As stated previously, the Lagrangian for the system cyclic of φ, so the system is spherically symmetric. The total angular momentum vector j r p 7

82 Jeff Kissel May 8, 7 Classical Mechanics is conserved. The definition of the cross product requires then that j is perpendicular to r and p. This can only be true if r lies in a plane whose normal is parallel to j. Thus, motion is restricted to a plane, and φ must not only be constant but zero. If φ, then we can reduce the Lagrangian Eq. 4 to just a function of r, ṙ, and θ, E m ṙ + r θ + kr 8 and p θ is now a constant of motion because Eq. 8 is cyclic in θ. We can find this constant denoted as p θ l, and therefore θ, p θ L θ mr θ l θ l mr 9 This means the total energy after killing the φ term becomes E ṙ m + r θ + V l m ṙ + r mr E mṙ + l mr + V From here, we can solve Eq. for ṙ, and then rθ, mṙ E V l mr mr l ṙ E m V m l m r E ṙ m V / m l m r ṙ dr dθ dr θ dθ dt dθ dθ dr dθ / E m V m l m r l E mr m V m m E r l m kr m l m r l dr mr dθ / dr / l m r dr me l mkr l / r r dr { du r let u r dr r u 3

83 Jeff Kissel May 8, 7 Classical Mechanics me l mk / u l u du dv u dv let v u u v / me dθ l mve l du mk / vl v mk l v v / dv u / dv / dv let η me l and ǫ mk dθ ηv ǫ l v / dv ηv ǫ l v ηv ǫ η 4 ǫ l l v + η 4 η 4 v ηv + η η 4 ǫ l v η η 4 ǫ l v η / dv let y v η dy dv let a η 4 ǫ l dθ a y / dy dθ a y dy θ sin y + γ a θ γ sin v η sin γ θ η 4 ǫ l u me l v η η 4 ǫ l m E l 4 mk / l sin γ θ sin A B sinacosb r me l + l m E mkl sin γ θ cosγ θ where γ is an arbitrary integration/phase constant. *Phew*. That s my final answer. The orbit equation for a simple harmonic oscillator potential. 4 4

84 Classical Mechanics Review Louisiana State University Qualifier Exam Jeff Kissel November 5, 6 All parts of this question call for quick and brief answers. a Positronium is a bound state of an electron and a positron. What is the effective mass of positronium? Short Answer: For classical mechanics purposes, the effective mass of a system of particles is its reduced mass. An electron and a positron differ in charge but have the same mass, so positronium s effective mass is m Ps µ Ps m e m e + m e + m e + m e m e m Ps m e Thorough Answer: Effective mass is defined by analogy with Newton s second law F m a. Using quantum mechanics it can be shown that for an electron in an external electric field E, a h d ǫk dk q E where a is acceleration, h is reduced Planck s constant, h h/π, k is the wave number often loosely called momentum since k p/ h, ǫk is the energy as a function of k, or the dispersion relation as it is often called. From the external electric field alone, the electron would experience a force of F qe, where q is the charge. Hence under the model that only the external electric field acts, effective mass m becomes: d m h ǫk dk 3 For a free particle, the dispersion relation is a quadratic, and so the effective mass would be constant and equal to the real mass. In a crystal, the situation is

85 Jeff Kissel November 5, 6 Classical Mechanics far more complex. The dispersion relation is not even approximately quadratic, in the large scale. However, wherever a minimum occurs in the dispersion relation, the minimum can be approximated by a quadratic curve in the small region around that minimum. Hence, for electrons which have energy close to a minimum, effective mass is a useful concept. In energy regions far away from a minimum, effective mass can be negative or even approach infinity. Effective mass, being generally dependent on direction with respect to the crystal axes, is a tensor. However, for most calculations the various directions can be averaged out. Effective mass should not be confused with reduced mass, which is a concept from Newtonian mechanics. Effective mass can only be understood with quantum mechanics. 6 Regardless, since this question is asked in the classical mechanics portion of the exam, and every other classical mechanics reference I looked at refers positronium s reduced mass I assume effective mass is equivalent to the reduced mass. m Ps µ Ps For classical treatment m e m e + m e + m e + m e m e m Ps m e 4 b At what point in a bound Kepler orbit is the speed of the satellite at its minimum? Figure : A particle in bound Kepler Orbit about a more massive object. Point P is the periapsis and point A is the apsis. Short Answer: Kepler s Second Law of Planetary Motion states that the radius vector of an object caught in an inverse square potential sweeps out equal areas in equal times as in Figure. In order to adhere to this law, the object must travel fastest at the closest point of orbit, known as periapsis think

86 Jeff Kissel November 5, 6 Classical Mechanics perigee, perihelion, or periastron, and slowest at the farthest point, known as apsis apogee, aphelion, apasteron, etc.. In other words, point A in Figure is the point at which the satellite s speed in minimum. Thorough Answer: The Lagrangian for a particle in a Keplerian orbit in physics spherical-polar coordinates; θ is the angle from the ẑ axis, φ is that from the +ˆx axis is L mṙ + r θ + r sin θ φ + k r 5 Because Eq. 5 is explicitly independent of θ and φ we know p θ and p φ are conserved quantities. The later fact restricts motion to a plane, i.e. we can ignore the third term in the Lagrangian. The former yields Kepler s Second Law: d dt d L dt θ mr θ L mṙ + r θ + k r L θ r θ constant 6 The factor is inserted because r θ is the areal velocity the triangular area swept out by the radius vector per unit time. Figure : The area swept out but the radius vector in a time dt. The interpretation follows from Figure above, the differential area swept out in a time dt being da r r dθ 7 3

87 Jeff Kissel November 5, 6 Classical Mechanics da dθ r 8 dt dt which from Eq. 6 we know is constant. So, for a particle in Keplerian orbit to sweep out the same area per unit time, the object must travel fastest at the closest point of orbit, known as periapsis think perigee, perihelion, periastron, etc., and slowest at the farthest point, known as apsis apogee, aphelion, apasteron, etc.. In other words, point A in Figure is the point at which the satellite s speed in minimum. c If you have two spheres of the same mass and radius, describe a non-destructive test by which you could distinguish between the one that is solid and the one that is hollow. Short Answer: If one rolls both spheres down a ramp, the solid sphere will reach the bottom faster. The solid sphere has a smaller moment of rotational inertia, i.e. less energy is put into rotating the solid sphere so more energy can by used in linear velocity of the center of mass. Thus, the solid sphere will travel faster down the ramp. Thorough answer: The moments of inertia for a solid and hollow sphere abbreviated as SS and HS, respectively are I SS 5 MR I HS 3 MR 9 Remember v ωr, so each sphere s kinetic energy is T SS Mv CoM,SS + I SSω Mv CoM,SS + 5 M R vcom,ss R Mv CoM,SS + 5 Mv CoM,SS T SS 7 Mv CoM,SS T HS Mv CoM,HS + I HSω Mv CoM,HS + 3 M R vcom,hs R Mv CoM,HS + 3 Mv CoM,HS T HS 5 6 Mv CoM,HS Energy is conserved in this system, so the gravitational potential at the top of the ramp will equal the kinetic energy at the bottom, so V i T f 4

88 Jeff Kissel November 5, 6 Classical Mechanics Mgh 7 Mv CoM,SS Mgh 5 6 Mv CoM,HS vcom,ss 7 gh v CoM,HS 6 gh 5 which means the solid sphere will arrive first, since it is traveling at a faster linear velocity /7 > 6/5 down the ramp. d Which way is a particle moving from east to west in the southern hemisphere deflected by the Coriolis force arising from the Earth s rotation? Figure 3: A particle moves from East to West in the Southern hemisphere, along the vector r. The Coriolis Force is defined as F Cor m ω v r 3 Thus, as in Figure 3, if v r points west, ω is perpendicular to Earth, then m ω v r points South. There s not much to this problem, so I won t bother with a thorough answer.if you re wondering where the definition of the Coriolis force comes from, check out Goldstein pp e In a bound Keplerian orbit, what is the ratio of the average kinetic energy to the average potential energy? Short Answer: The only force experienced by a particle in a bound Keplerian orbit is the central, gravitational inverse square force a power law of order n. This gravitational force is derivable from a scalar is derivable from a scalar potential V. Any system satisfying these conditions follows the Virial Theorem: the average kinetic energy is twice that of the average potential, or T V 4 5