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Unformatted text preview: midterm 03 SAENZ, LORENZO Due: Apr 4 2007, 11:00 pm 1 Question 1 part 1 of 1 10 points At t 1 = 0, a wheel rotating about a fixed axis at a constant angular accelera tion = . 88 rad / s 2 has an angular ve locity = 3 . 7 rad / s and an angular position 1 = 6 . 6 rad. What is the angular position 2 of the wheel at time t 2 = 3 s? 1. 11 . 375 rad 2. 11 . 73 rad 3. 12 . 095 rad 4. 12 . 48 rad 5. 12 . 88 rad 6. 13 . 285 rad 7. 13 . 74 rad correct 8. 14 . 175 rad 9. 14 . 615 rad 10. 15 . 08 rad Explanation: In the usual notation s r = d dt = v r = d dt = a r , the only equation we need is the one for an gular position = t + 1 2 t 2 . For our problem it reads 2 = 1 + 1 t 2 + 1 2 t 2 2 = (6 . 6 rad) + (3 . 7 rad / s) (3 s) + 1 2 ( . 88 rad / s 2 ) (3 s) 2 = (6 . 6 rad) + (11 . 1 rad) + ( 3 . 96 rad) = 13 . 74 rad . Question 2 part 1 of 1 10 points A mobile consisting of four weights hanging on three rods of negligible mass. 34 N W 1 W 2 W 3 5 m 9 m 5 m 6 m 5 m 8 m Find the weight of W 3 . 1. 54 . 4762 N 2. 56 . 2731 N 3. 58 . 1778 N 4. 60 . 0247 N 5. 61 . 9259 N 6. 64 N 7. 66 N 8. 68 . 0952 N 9. 70 . 496 N 10. 72 . 7222 N correct Explanation: Let : W = 34 N , r 1 = 9 m , l 1 = 5 m , r 2 = 6 m , l 2 = 5 m , r 3 = 8 m , and l 3 = 5 m . We can apply the balance condition summationdisplay vector = 0 successively, starting with the lowest part of the mobile, to find the value of each unknown weight. Apply summationdisplay vector = 0 about an axis through the point of suspension of the lowest part of the mobile, we obtain l 1 W r 1 W 1 = 0 midterm 03 SAENZ, LORENZO Due: Apr 4 2007, 11:00 pm 2 W 1 = l 1 W r 1 = (5 m) (34 N) (9 m) = 18 . 8889 N . Apply summationdisplay vector = 0 about an axis through the point of suspension of the middle part of the mobile, we obtain l w W 2 r 2 [ W + W 1 ] = 0 W 2 = r 2 [ W + W 1 ] l w = (6 m) [(34 N) + (18 . 8889 N)] (5 m) = 63 . 4667 N . Apply summationdisplay vector = 0 about an axis through the point of suspension of the top part of the mobile, we obtain l e [ W + W 1 + W 2 ] r 3 W 3 = 0 W 3 = l 3 [ W + W 1 + W 2 ] r 3 = (5 m) (8 m) bracketleftBig (34 N) + (18 . 8889 N) + (63 . 4667 N) bracketrightBig = 72 . 7222 N . Question 3 part 1 of 3 10 points A beetle takes a joy ride on a pendulum. The string supporting the mass of the pendu lum is 161 cm long. If the beetle rides through a swing of 28 , how far has he traveled along the path of the pendulum? 1. 73 . 7926 cm 2. 76 . 236 cm 3. 78 . 6795 cm correct 4. 81 . 1578 cm 5. 83 . 7758 cm 6. 86 . 3938 cm 7. 89 . 0816 cm 8. 91 . 8742 cm 9. 94 . 7714 cm 10. 97 . 7385 cm Explanation: Arc length is defined as s = r = (161 cm)(0 . 488692 rad) = 78 . 6795 cm , where theta is in radians....
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This note was uploaded on 02/09/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Acceleration

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