Massachusetts Institute of Technology Department of Physics Physics 8.01L SAMPLE FINAL EXAM SOLUTIONS Problem 1 a) V = N kT P , VN EW = N k T3 P 10 = 10 N kT 3 P , ⇒ V goes up b) (i) F = mg = 8(10) = 80N (ii)FBuoy = (1000)(.002)(10) = 20N, ∴ F = 80 − 20 = 60N c) Velocity goes down, pressure goes up. Problem 2 a) Equilibrium, so FBuoyant − w = 0 ⇒ FB = w. FB = Vf ρf g = (πr2 d)ρg = w, d = πrw2 ρg 2 b) FT OT = M a = FB − w, FB = πr2 ( d2 )ρg = d( πr2ρg ) 2 From (a), d = πrw2 ρg ⇒ FB = πrw2 ρg ( πr2ρg ) = w2 Half as deep ⇒ half as large buoyancy force. −w FT OT = M a = w2 − w = −w 2 , a = 2m , but w = M g. −g ⇒ a = 2 , accelerating downward at g2 . Problem 3 a) P + ρgy = constant, P1 + ρg(0) = P2 + ρg(6500) P2 = P1 − ρg(y2 ) = 1.013 × 105 − (0.95)(9.8)(6500), P2 = 4.08 × 104 N/m2 = 0.40 atm b) N = const, V = const, P V = N kT ⇒ P1 T1 = P2 T2 , T1 = 293, P1 = 1.5 × 107 , T2 = 253, P2 = P1 T2 = 1.30 × 107 N/m2 T1 Problem 4 ii) f exerts torque around center of mass, so you fall over. 1 January 25, 2006 iii) Now N exerts torque which can balance torque due to friction. Problem 5 a) b) Take torques around toes: M gL cos(θ) − F ( 43L ) cos(θ) = 0, F = c)T + F = M g, T = 3 Mg 4 1 Mg . 4 Problem 6 a) 3M g + M g = 4M g. b) Take torques about left end: 4M gD − LM g = 0, D = L4 . Check torque around weight: 0 = D(3M g) − M g(L − D), D(4M g) = M gL, D = Problem 7 a) b) � c) d) � Fx = N2 + N1 sin(θ) − f1 cos(θ) = 0 Fy = f2 − M g + N1 cos(θ) + f1 sin(θ) = 0 � � τ = M g( L2 cos(θ)) − N1 L = 0. τ = N2 L sin(θ) + f2 L cos(θ) − M g L2 cos(θ) = 0. Problem 8 a) τ = Iα, take torques about hinge. 2 (M g)( L2 )(sin(90◦ )) = ( ML 3 )(α) ⇒ α = gL 2 L2 3 = 3g 2L , α= 2 3g 2L L 4. b) F = M acm , acm = α( L2 ), downward. All forces and acceleration are vertical ⇒ FH = 0 . 3g )( L2 ) = −3Mg FV − M g = −M a = −M α( L2 ) = −M ( 2L 4 Mg Mg 3Mg FV = M g − 4 ⇒ FV = , FT OT = , up . 4 4 c) Used ﬁxed pivot: KEI = 0, P EI = M gL, KEF = 12 Iend ω 2 , P EF = M g( L2 ), Work = 0. � 3g 3g MgL 2 L ML2 ω 2 1 ML2 2 or ω = = 2 , ω = . 2 ( 3 )ω + M g 2 = M gL, 6 L L Used center of mass: 2 + 21 ICM ω 2 , vCM = ω( L2 ) KEI = 0, KEF = 21 M vCM 1 L 2 2 1 ML2 1 3 1 KEF = 2 M ( 2 ) ω + 2 ( 12 )ω 2 = M L2 ω 2 ( 81 + 24 ) = M L2 ω 2 ( 24 + 24 ) 2 2 1 = M L ω ( 6 ) ⇒ Same answer. Problem 9 a) L is conserved: mvR = (I0 + mR2 )ωf mvR ωf = I0 + mR2 2 2 2 1 m v R 2 b) KEI = 12 mv 2 , KEF = 21 (I0 + mR2 )( I0mvR +mR2 ) = 2 ( I0 +mR2 ) mR2 KEF = KEI I0 + mR2 Problem 10 a) Left b) Yes, gravity. c) Out of the page; Counter-clockwise. d) Yes, pivot force. e) Out of the page; Counter-clockwise. f ) Out of the page. Problem 11 Take clockwise to be positive. Angular momentum is conserved: IωI − mvI d = Iωf + mvf d 0.30(ω) − 0.15(50)(0.8) = 0.3(0.35ω) + 0.15(40)(0.8) 0.20ω = 6 + 4.8 ⇒ ω = 54rad/s . Period = 0.12 sec. 3

8.01L Practice Final Exam - Solutions

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