Homework Set 2 Physics 319 Classical Mechanics

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Dominick Gibbs
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1 Homewor Se Physics 19 Classical Mechanics Problem.7 a) The roce velociy equaion (no graviy) is m v v ln m Afer wo minues he velociy is m/sec ln = 79 m/sec. b) The rae a which mass is ejeced is ( )g/1 sec. This wors ou o (8+1/) 1 g/sec. To ge he hrus muliply by m/sec, or 5 MN. The iniial weigh is simply mg or 19.6 MN. (M for mega is he inernaionally agreed symbol for million (1 6 )). The hrus ceeds he weigh, necessary for lif off! Problem.1 The modified (from he lecure noes) roce velociy equaion including graviy is mv mv mg dln m v v g d m m v v ln g Assuming he velociy and posiion are zero a =, he heigh is m y v ln gd m v lnm v ln m d g v v ln m m ydy g ln v v ln m m yln m y m y g v v ln m m ln m m m ln m m g m vm m v v ln ln g m m v m v m ln g m

2 Using he numbers given a wo minues he shule will be a 1 m 1 6 ln/((8+1/) 1 ) m / m = ( ) m = 9.9 m. Problem. This problem is a fairly sraighforward unwinding of he definiions. By he definiions of he CM for he individual sysems m R1 M1R1 m r m R M R m r where he sum is over he masses in sysem 1 and he sum is over he masses in sysem. By he definiion of he combined CM m m R M M R m r m r M R M R M R R M M R M Problem.7 a) For a plane moving around he sun he angular momenum normal o he plane of revoluion is (Taylor uses where I use ) ˆ l r mv rrˆ m rrˆ r mr rˆ ˆ l l mr b) From he lecures he area of he riangle formed by r d, dr r d r is r, and 1 1 da r dr r r dˆ. In he limi d, he rae area is swep ou is da 1 l r d m Because angular momenum is conserved (a consan) so is he rae a which area is swep ou (Kepler s Second Law).

3 Problem. For a uniform sphere he densiy M / 4 R /. Now I dmr axis where r axis is he disance o he roaion axis. Using Figure 1, and he volume elemen for spherical coordinaes o ge he mass incremen R 1 I r sin r drd cos d 1 R 1 4 r 1cos drd cos d 1 5 M R / 4 R 5 5 MR θ Figure 1 Problem 4.8 Choose a polar coordinae sysem cenered on he cener of he sphere and whose angle is measured from verical. By energy conservaion mgr mgr cos m v / By he equaions of moion in polar coordinaes R cons R g cos N r d dr v rrˆ rˆ rˆ Rˆ d d where N r is he (radial) normal force ered by he sphere. The goal is o find he insan when his force vanishes. When N r vanishes gr R R / R / or 1cos 1/ or he heigh above he cener of he sphere is Rcos R/.

4 Problem 4.18 a) Le xs be any curve in he surface f x f r wih x r. The funcion f x s is a consan and so is derivaive is zero as a funcion of s. For he small displacemen dr dx / ds s, by 4.5 / d f x s f r f r dx ds s f r dx f x s lim lim f ds s s s s ds where he vecor dx / ds is angen o he curve (and surface) a s dx f. ds. Therefore Because his is so for all curves in he surface, i.e., for all angens o he surface, f is normal o he surface. b) For a small displacemen dr uˆ df f r uˆ f r f uˆ Problem 4. a) b) f r uˆ f r df lim f uˆ f cos d This quaniy is maximum when he uni vecor is aligned wih he gradien. xˆ yˆ zˆ xˆ yˆ zˆ x y z x y z Conservaive. The poenial is (use sraigh pahs parallel o axes) x xxˆ xxˆ yyˆ xxˆ yyˆ zzˆ U x F dr xdx ydy zdz xxˆ xxˆ yyˆ x z y U xxˆ yyˆ zzˆ xˆ yˆ zˆ xˆ yˆ 1 1 zˆ x y z y x Conservaive. The poenial is (use same pah as a))

5 c) xxˆ xxˆ yyˆ xxˆ yyˆ zzˆ U x dx x dy dz xxˆ xxˆ yyˆ ˆ ˆ xy U yx xy xˆ yˆ zˆ xˆ yˆ 1 1 zˆ zˆ x y z y x No conservaive. Problem 4.6 a) Referring o Fig 4.7 U MgH mgh b H l sin h bco b U Mgl Mg mgb co sin For equilibrium Mgb cos cos du / d mgb mgb sin sin M cos m M m This condiion is obviously correc by realizing ha he sring ension has he same magniude a eiher end of he sring (bu differen direcions!) b) For sabiliy ae he second derivaive d m M cos m M coscos sin d U / d M d sin sin sin M sin This will be posiive (indicaing sabiliy) for all possible /.

6 Problem 4.49 U r1 r r r 1 r r r r r r ˆ 1x x 1y y 1z z U U U 1 r ˆ ˆ ˆ 1x r x x r1 y r y y r1 z r z z F U xˆ yˆ zˆ 1 1 / x 1 y1 z1 r1 x r x r1 y r y r1 z r z r r r r F 1 1 U 1 r r r r F r r x r r yˆ r r 1 x 1x y 1y z r r r r r r 1x x 1y y 1z z These, of course, are he pressions consisen wih he formula for he graviaional force on page 19. 1z zˆ /

2.1: What is physics? Ch02: Motion along a straight line. 2.2: Motion. 2.3: Position, Displacement, Distance

2.1: What is physics? Ch02: Motion along a straight line. 2.2: Motion. 2.3: Position, Displacement, Distance Ch: Moion along a sraigh line Moion Posiion and Displacemen Average Velociy and Average Speed Insananeous Velociy and Speed Acceleraion Consan Acceleraion: A Special Case Anoher Look a Consan Acceleraion