# oldhw23 - This print-out should have 11 questions...

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Unformatted text preview: This print-out should have 11 questions. Multiple-choice 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 one-half 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 y-coordinate 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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oldhw23 - This print-out should have 11 questions...

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