oldhw23_s

# oldhw23_s - oldhomewk 23 – YOO HEE – Due Apr 1 2008...

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Unformatted text preview: oldhomewk 23 – YOO, HEE – Due: Apr 1 2008, 4:00 am 1 Question 1, chap 10, sect 3. part 1 of 2 10 points Given: A uniform flexible chain whose mass is 8 . 4 kg and length is 4 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 4m 3m Radius of table is negligible compared to the length of chain Mass of chain is 8 . 4 kg . Find the acceleration a of the chain when the length of the chain hanging vertically is 3 m . Correct answer: 7 . 35 m / s 2 (tolerance ± 1 %). Explanation: Note: The initial condition does not enter into the consideration for the acceleration. Let : g = 9 . 8 m / s 2 , L = 4 m , ℓ = 3 m , and m = 8 . 4 kg . The linear density of the chain is λ = m L = 8 . 4 kg 4 m = 2 . 1 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) = 3 m 4 m (9 . 8 m / s 2 ) = 7 . 35 m / s 2 . Question 2, chap 10, sect 3. part 2 of 2 10 points Find the magnitude of the velocity of the of the chain when 3 m of the chain is hanging vertically. Correct answer: 3 . 5 m / s (tolerance ± 1 %). 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 oldhomewk 23 – YOO, HEE – Due: Apr 1 2008, 4:00 am 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 (4 m) [4 (3 m) 2- (4 m) 2 ] = 3 . 5 m / s ....
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## This note was uploaded on 05/04/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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oldhw23_s - oldhomewk 23 – YOO HEE – Due Apr 1 2008...

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