oldhw 20 - alexander (jra2623) oldhomework 20 Turner...

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Unformatted text preview: alexander (jra2623) oldhomework 20 Turner (92510) 1 This print-out should have 14 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 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 x-axis is horizontal and to the right. The y-axis is vertically upward. Given: In the figure, B = 12 cm, C = 3 . 7 cm, D = 5 cm, E = 20 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 x-coordinate of the center of gravity? Correct answer: 3 . 73431 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 (5 cm 20 cm) rectangle are parenleftbigg D 2 , E 2 parenrightbigg , or (2 . 5 cm , 10 cm), and the ( x cm ,y cm ) coordinates of the bottom (3 . 7 cm 7 cm) rectangle are parenleftbigg B + D 2 , C 2 parenrightbigg , or (8 . 5 cm , 1 . 85 cm), since B- D 2 + D = B + D 2 . Thus, A top = D E = (5 cm) (20 cm) = 100 cm 2 A bottom = C W = (3 . 7 cm) (7 cm) = 25 . 9 cm 2 x cm = bracketleftbigg A top D 2 + A bottom B + D 2 bracketrightbigg ( A top + A bottom ) = (100 cm 2 ) (2 . 5 cm) (100 cm 2 ) + (25 . 9 cm 2 ) + (25 . 9 cm 2 ) (8 . 5 cm) (100 cm 2 ) + (25 . 9 cm 2 ) = 470 . 15 cm 3 125 . 9 cm 2 = 3 . 73431 cm . 002 (part 2 of 2) 10.0 points What is the y-coordinate of the center of grav- ity?...
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This note was uploaded on 10/20/2011 for the course PHY 302K taught by Professor Kaplunovsky during the Summer '08 term at University of Texas at Austin.

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oldhw 20 - alexander (jra2623) oldhomework 20 Turner...

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