From a etotal mgh at b e 1 m v 2 mgh 2 2 r r

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Unformatted text preview: Etop # Ttop $ Utop # 0 $ mgh EA # TA $ U A # 1 mv 2 $ 0 2 Etop # EA % v # 2 gh So N # mg $ m2 gh R 2h ' & N # mg ( 1 $ ) * R+ b) At B the forces are: N mg 45˚ N # mv 2 R $ mg cos 45, # mv 2 R $ mg 2 (1) Get v by conservation of energy. From a), Etotal # mgh . At B, E # 1 m v 2 $ mgh 2 2 R R 45˚ R R cos 45˚ = R 2 h′ R! R " h# 2 or 1% $ h# ! R ' 1 & ( ) 2* So Etotal ! TB " UB becomes: 1% 1 $ 2 mgh ! mgR ' 1 & ( " mv ) 2* 2 Solving for v 2 + 1 %, $ 2 2 - gh & gR ' 1 & ( !v ) 2*. / 0 Substituting into (1): + 2h $ 3 %, N ! mg - " ' & 2( . *0 /R ) 2 2 c) From b) vB ! 2 g + h & R " R / 2, 0 1 v ! +2 g h & R " R / 2 2, 0 12 d) This is a projectile motion problem 45˚ B 45˚ A Put the origin at A. The equations: x ! x0 " vx 0 t y ! y0 " v y 0 t & 12 gt 2 become x! R v " Bt 2 2 y ! h# " Solve (3) for t when y = 0 (ball lands). vB 1 t & gt 2 2 2 (2) (3) gt 2 ! 2 vBt ! 2h # " 0 t" 2 2 vB $ 2vB % 8 gh # 2g We discard the negative root since it gives a negative time. Substituting into (2): & 2 v $ 2v 2 % 8 g h # ' 71 B B ( ) 2g 2 2( ) * + The equilibrium point (where dU d! " 0 ) that we wish to look at is clearly ! = 0. At that point, Using the previous expressions for vB and h# yields we have d 2U d! 2 " mg # R % b 2$ , which is stable for R & b 2 and unstable for R ' b( 2 . We can 12 use the results of Problem 2-46"to obtainR % h % & hfor 3 R2case 2R " ' 2 , where we will find that b x 2 ! 1 stability 2 ! the % R2 ) ( 2 + the first non-trivial result is in fourth order and*is negative. We therefore have an equilibrium at ! = 0 which is stable for R & b 2 and unstable for R ) b 2 . e) U ( x) " mgy( x ) , with y(0) " h , so U ( x) has the shape of the track. NEWTONIAN MECHANICS—SINGLE PAR RTICLE vB x" % , - 2-43. F " % kx + kx 3 * 2 2-26. All of the kinetic energy of the block goes into compressing the spring, so that mv 2 2 " kx 2 2 , or x " v m k...
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