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Unformatted text preview: ikram (ai2734) hw1020 Swinney (58785) 1 This printout should have 8 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points A carpenters square of uniform density has the shape of an L, as shown in the figure. Assume: A ( x,y ) coordinate frame with the origin at the lower left corner of the car penters square. The xaxis is horizontal and to the right. The yaxis is vertically upward. Given: In the figure, B = 14 cm, C = 4 . 8 cm, D = 4 . 2 cm, E = 19 cm. Because the square is uniform in thickness and has a small thickness, we can assume that the weight of each segment of the square is proportional to its area. D E B C What is the xcoordinate of the center of gravity? Correct answer: 4 . 69603 cm. Explanation: Basic Concepts: Center of mass: vectorr c = i m i vectorr i i m i . Solution: Take the origin to be at the outside corner of the L . Divide the carpenters square into two rectangles (simple shapes) and find the area and the center of mass of each piece. Each piece will behave as if it were a point particle with mass equal to the mass of the piece and located at the pieces center of mass. Thus, the center of mass of the whole square will be vectorr c = m top vectorr c top + m bot vectorr c bot m top + m bot , where m top and m bot are the masses of the two pieces and r c top and r c bot are the centers of mass of the two pieces. The mass of each piece is proportional to its area, so if we let represent the uniform mass per area of the pieces, m top = A top and m bot = A bot . For objects where the mass is uniformly distributed, ( i.e. , the density is uniform for thick objects or the mass per area is uni form for thin objects) the center of mass is at the same position as the objects geomet ric center. This means that the ( x cm ,y cm ) coordinates of the left/top (4 . 2 cm 19 cm) rectangle are parenleftbigg D 2 , E 2 parenrightbigg , or (2 . 1 cm , 9 . 5 cm), and the ( x cm ,y cm ) coordinates of the bot tom (4 . 8 cm 9 . 8 cm) rectangle are parenleftbigg B + D 2 , C 2 parenrightbigg , or (9 . 1 cm , 2 . 4 cm), since...
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This note was uploaded on 02/02/2010 for the course PHY 301 taught by Professor Swinney during the Spring '07 term at University of Texas at Austin.
 Spring '07
 Swinney
 mechanics

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