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Unformatted text preview: This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Given: A uniform flexible chain whose mass is 7 . 5 kg and length is 3 m. A table whose top is frictionless. Initially you are holding the chain at rest and onehalf of the length of the chain is hung over the edge of the table. When you let loose of the chain it falls downward. The acceleration of gravity is 9 . 8 m / s 2 . a 3m 2 . 37m Radius of table is negligible compared to the length of chain Mass of chain is 7 . 5 kg . Find the acceleration a of the chain when the length of the chain hanging vertically is 2 . 37 m . Correct answer: 7 . 742 m / s 2 . Explanation: Note: The initial condition does not enter into the consideration for the acceleration. Let : g = 9 . 8 m / s 2 , L = 3 m , ℓ = 2 . 37 m , and m = 7 . 5 kg . The linear density of the chain is λ = m L = 7 . 5 kg 3 m = 2 . 5 kg / m . F = ℓ λg cm The free body diagram in the vertical di rection gives summationdisplay F y = ℓ λg = Lλa. Therefore a = ℓ L g (1) = 2 . 37 m 3 m (9 . 8 m / s 2 ) = 7 . 742 m / s 2 . 002 (part 2 of 2) 10.0 points Find the magnitude of the velocity of the of the chain when 2 . 37 m of the chain is hanging vertically. Correct answer: 3 . 3164 m / s. Explanation: The change in kinetic energy is Δ K = 1 2 mv 2 = 1 2 λLv 2 . (2) Let ℓ i = L 2 and ℓ f = ℓ . Using the table top as the origin of the ycoordinate and down as the positive y di rection y cm = m on table parenleftBig parenrightBig + m hanging parenleftbigg ℓ 2 parenrightbigg m on table + m hanging oldhomework 23 – Turner – (58185) 1 hinojosa (jlh3938) – oldhomework 23 – Turner – (58185) 2 y cm i = parenleftbigg L L 2 parenrightbigg λ 0 + L 2 λ parenleftbigg L 4 parenrightbigg λL y cm f = ( L ℓ ) λ 0 + ℓ λ parenleftbigg ℓ 2 parenrightbigg λL The vertical center of mass difference Δ y cm is Δ y cm = y cm f y cm i = λℓ ℓ 2 λ L 2 L 4 λL = 1 8 L [4 ℓ 2 L 2 ] . (3) The change in potential energy is Δ U = λLg Δ y cm = 1 8 λg [4 ℓ 2 L 2 ] . (4) From conservation of energy Δ K = Δ U , Eq. 2 and Eq. 4, we have 1 2 λLv 2 = 1 8 λg (4 ℓ 2 L 2 ) v 2 = g 4 L [4 ℓ 2 L 2 ] . (5) Therefore v = radicalbigg g 4 L [4 ℓ 2 L 2 ] (6) = radicalBigg (9 . 8 m / s 2 ) 4 (3 m) [4 (2 . 37 m) 2 (3 m) 2 ] = 3 . 3164 m / s . Alternative: You can use the kinematic expression and remember that the center of mass accelerates at g [not Eq. 1], since the acceleration of gravity is not a function of mass. v 2 = 2 a Δ y cm = 2 g 1 8 L [4 ℓ 2 L 2 ] = g 4 L [4 ℓ 2 L 2 ] , (7) where Δ y cm is obtained from Eq. 3....
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 Spring '08
 Turner
 Energy, Kinetic Energy, Mass, Momentum, Work, Hinojosa

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