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Unformatted text preview: homework 22 – BAUTISTA, ALDO – Due: Mar 24 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 1. 10 points. A uniform flat plate of metal is situated in the reference frame shown in the figure below. 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 x y Calculate the x coordinate of the center of mass of the metal plate. Correct answer: 7 . 33333 (tolerance ± 1 %). Explanation: Basic Concept: The center of mass coor dinate is x ≡ Z x dm m , where m ≡ Z dm , and dm = σ y dx , where σ is the areal density mass area of the plate. Solution: Let ( x 1 , y 1 ) = (6 , 0) ( x 2 , y 2 ) = (8 , 0) ( x 3 , y 3 ) = (8 , 4) . The equation for the hypotenuse is y y 1 x x 1 = y 3 y 1 x 3 x 1 . The slope of the hypotenuse is s = y 3 y 1 x 3 x 1 = 4 8 6 = +2 . Rewriting the equation, we have y = s ( x x 1 ) + y 1 = (+2) ( x 6) + 0 . The xcoordinate of the center of mass is x = σ Z x 2 x 1 x y dx σ Z x 2 x 1 y dx = Z x 2 x 1 x s ( x x 1 ) dx Z x 2 x 1 s ( x x 1 ) dx = Z x 2 x 1 x ( x x 1 ) dx Z x 2 x 1 ( x x 1 ) dx = 1 3 x 3 1 2 ( x 1 ) x 2 1 2 x 2 ( x 1 ) x x 2 x 1 = 1 3 ( x 3 2 x 3 1 ) 1 2 ( x 1 ) ( x 2 2 x 2 1 ) 1 2 ( x 2 2 x 2 1 ) ( x 1 ) ( x 2 x 1 ) = 3 x 1 ( x 2 2 2 x 1 x 2 + x 2 1 ) 3 ( x 2 2 2 x 1 x 2 + x 2 1 ) + 2 ( x 2 x 1 ) ( x 2 2 2 x 1 x 2 + x 2 1 ) 3 ( x 2 2 2 x 1 x 2 + x 2 1 ) = x 1 + 2 3 ( x 2 x 1 ) (1) = 6 + 2 3 (8 6) = 7 . 33333 . Alternate solution: The center of mass of a right triangle is 1 3 of the height or base of the triangle measured from its right angle. Therefore Eq. 1 is the xcoordinate of the center of mass of the metal plate. The ycoordinate of the center of mass of the metal plate is y = y 2 + 1 3 ( y 3 y 2 ) = 0 + 1 3 (4 0) = 1 . 33333 . homework 22 – BAUTISTA, ALDO – Due: Mar 24 2006, 4:00 am 2 Note: This problem has a different triangle for each student....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
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