This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: oldhomewk 23 – PAPAGEORGE, MATT – 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 6 . 4 kg and length is 5 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 5m 3 . 9m Radius of table is negligible compared to the length of chain Mass of chain is 6 . 4 kg . Find the acceleration a of the chain when the length of the chain hanging vertically is 3 . 9 m . Correct answer: 7 . 644 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 = 5 m , ℓ = 3 . 9 m , and m = 6 . 4 kg . The linear density of the chain is λ = m L = 6 . 4 kg 5 m = 1 . 28 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 . 9 m 5 m (9 . 8 m / s 2 ) = 7 . 644 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 . 9 m of the chain is hanging vertically. Correct answer: 4 . 19066 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 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 oldhomewk 23 – PAPAGEORGE, MATT – 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 (5 m) [4 (3 . 9 m) 2 (5 m) 2 ] = 4 . 19066 m / s ....
View
Full
Document
This note was uploaded on 03/02/2009 for the course PHY 58235 taught by Professor Kleinman during the Spring '09 term at University of Texas.
 Spring '09
 KLEINMAN
 Physics, Friction, Mass

Click to edit the document details