20 (10) - hernandez (ejh742) oldhomework 20 Turner (58120)...

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Unformatted text preview: hernandez (ejh742) oldhomework 20 Turner (58120) 1 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 5 . 4 kg and length is 7 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 7m 4 . 27m Radius of table is negligible compared to the length of chain Mass of chain is 5 . 4 kg . Find the acceleration a of the chain when the length of the chain hanging vertically is 4 . 27 m . Correct answer: 5 . 978 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 = 7 m , = 4 . 27 m , and m = 5 . 4 kg . The linear density of the chain is = m L = 5 . 4 kg 7 m = 0 . 771429 kg / m . F = g cm The free body diagram in the vertical di- rection gives summationdisplay F y = g = La. Therefore a = L g (1) = 4 . 27 m 7 m (9 . 8 m / s 2 ) = 5 . 978 m / s 2 . 002 (part 2 of 2) 10.0 points Find the magnitude of the velocity of the of the chain when 4 . 27 m of the chain is hanging vertically. Correct answer: 2 . 89414 m / s. Explanation: The change in kinetic energy is K = 1 2 mv 2 = 1 2 Lv 2 . (2) Let i = L 2 and f = . 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 2 parenrightbigg m on table + m hanging hernandez (ejh742) oldhomework 20 Turner (58120) 2 y cm i = parenleftbigg L- L 2 parenrightbigg 0 + L 2 parenleftbigg L 4 parenrightbigg L y cm f = ( L- ) 0 + parenleftbigg 2 parenrightbigg L The vertical center of mass difference y cm is y cm = y cm f- y cm i = 2- L 2 L 4 L = 1 8 L [4 2- L 2 ] . (3) The change in potential energy is U = Lg y cm = 1 8 g [4 2- L 2 ] . (4) From conservation of energy K = U , Eq. 2 and Eq. 4, we have 1 2 Lv 2 = 1 8 g (4 2- L 2 ) v 2 = g 4 L [4 2- L 2 ] . (5) Therefore v = radicalbigg g 4 L [4 2- L 2 ] (6) = radicalBigg (9 . 8 m / s 2 ) 4 (7 m) [4 (4 . 27 m) 2- (7 m) 2 ] = 2 . 89414 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 y cm = 2 g 1 8 L [4 2- L 2 ] = g 4 L [4 2- L 2 ] , (7) where y cm is obtained from Eq. 3....
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This note was uploaded on 06/14/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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20 (10) - hernandez (ejh742) oldhomework 20 Turner (58120)...

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