finalsol

finalsol - Massachusetts Institute of Technology Physics...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

This note was uploaded on 11/20/2011 for the course PHY 203 taught by Professor Staff during the Fall '09 term at Rutgers.

Ask a homework question - tutors are online