finalsol

finalsol - Massachusetts Institute of Technology Physics...

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Unformatted text preview: Massachusetts Institute of Technology - Physics Department Final SOLUTIONS Physics - 8.01 Problem 1 16 points v sin 4 pts a) vy = v sin ; g t = 0 !t = 0 g 4 pts b) y = v + (v sin )t ; 1 g t2 2 v sin y = 0 + (v sin ) g ; 1 g v sgin 2 2 sin2 1 y = 2 v0 g 4 pts c) v0 t = 2 v sgin 4 pts d) x = v + (v cos )t 2 x = 2v0 sing cos 0 0 0 0 0 0 0 0 0 1 2 Fall 1999 Problem 2 15 points p2gl 2 5 pts a) mgl = 1 mvA !vA = 2 v 5 pts b) Net force up must be macen = m lA = 2mg T ; mg = 2mg !T = 3mg 5 pts c) gravity: mgl tension: 0 2 2 Problem 3 24 points 6 pts a) The diagrams: Nx Ny T2 mg 2 6 pts 6 pts T1 N N2 =m2g T2 m2 2 T1 m1g g The contact force N on the pulley must have components Nx = T2 and Ny = m2 g + T1 for the pulley to stay in place. 2 b) T2 = m2a = m2 g 2 a c) R(T1 ; T2) = I I = 1 m2 R2 = m24R2 =R 22 g a R(T1 ; m2 g ) = I R = 2IR = m28gR 2 ) T1 = m82g + m22g = 5 m2g 8 d) m1g ; T1 = m1a = m1 g 2 g = T = 5m g m1 2 1 8 2 6 pts 5 ) m1 = 4 m2 3 Problem 4 20 points 4 pts a) p = jrp F j = ;bM g sin ~ 1 4 pts b) Ip = Ic + M b2 = 2 M R2 + M b2 P 4 pts c) p = Ip 1 ;bM g sin = ( 2 M R2 + M b2) ) + 1 Rb2g+b2 sin = 0 2 4 pts d) Under the small angle approximation, sin , and the equation of motion is given by + 12 Rb2g+b2 = 0 which is simple harmonic with r an angular frequency given by ! = 12 Rb2g+b2 . The period is given by 1 2 2 4 pts s + T = ! = 2 2 Rbg b . e) There must be a force at P or the CM would accelerate straight downwards. 2 4 Problem 5 25 points 4 pts a) The diagram: T α ΣF mg 4 pts 4 pts 4 pts 9 pts b) ! = 2 v = !R !~ = 2 R ^ v z c) a = vR !~ = 4 22R x a ^ ~ d) F = m~ !F = m4 22R x a~ ^ e) T sin = m4 2R T cos = mg 2 2 !tan = 4 g2R 2 5 Problem 6 25 points 3 pts a) p = m1 v1x ~ ^ 4 pts b) p = m1 v1 x ~ ^ 1 2 4 pts c) 2 m1 v1 v 6 pts d) m1 v1 sin 1 = m2 v2 sin 2 ! v2 = m1 sin m2 sin 1 8 pts e) m1 v1 cos 1 + m2 v2 cos 2 = m1 v1 0 0 0 0 0 0 0 in m1v1 cos 1 + m1v1 ssin cos 2 = m1v1 in v1 cos 1 + ssin cos 2 = v1 0 0 1 2 0 !v1 = v1 cos 0 1 2 1 sin sin 2 2 +sin 1 cos 2 6 1 2 Problem 7 25 points r 1 mM G !v = 2MG 2 4 pts a) 2 mv = R R v 15 4 pts b) mv R sin 30 = mV (15R) ! V0 = sin 30 = 30 1 2 4 pts c) 2 mv0 ; mM G R 2 1 mM G 1 v0 mM G 5 pts d) same as in c. or 2 mV 2 ; 15R = 2 m 900 ; 15R v 1 8 pts e) 2 mv 2 ; mM G = 1 m 900 ; mM G R 2 15R 2 12 M G = 1 v0 ; M G 2 900 2 v0 ; R 15R � 0 � 0 2 0 7 Problem 8 18 points 4 pts a) M = V 6 pts b) W = W ; M g + Fbuoy Fbuoy = M g W = W ; M g + M g !W = W 8 pts c) mg + T = Fbuoy = M g !T = M g ; mg W = W ; M g + Fbuoy ; T = W ; T 0 0 0 0 8 Problem 9 15 points dP = ; g dy = vmass = N m = kP m olume V T dP = ; Pm g ! dP = ; mg dy dy kT P kT dP = Z h ; mg ! dy 0 P kT P log P = ; mgh kT ZP P0 0 P =Pe 0 ; mgh kT 9 Problem 10 17 points m R a2 F fr a1 X F = ma Ffr = ma2 X =I no slipping : RFfr = 2 mR2 5 a1 ; a2 = R Ffr = 2 mR 5 2 ma2 = 5 m(a1 ; a2) 7a = 2a 52 51 a2 = 2 a1 7 10 ...
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This note was uploaded on 11/20/2011 for the course PHY 203 taught by Professor Staff during the Fall '09 term at Rutgers.

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