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Unformatted text preview: Thevalingam, Donald Homework 18 Due: Feb 7 2007, midnight Inst: Eslami 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 6) 10 points A heavy ball swings on a string in a circular arc of radius 1 m. Q 1 m Q P 1 8 The two highest points of the balls trajec tory are Q and Q ; at these points the string is 18 from the vertical. Point P is the lowest point of the balls trajectory where the string hangs vertically down. The acceleration of gravity is 9 . 8 m / s 2 . What is the balls speed at the point P ? Ne glect air resistance and other frictional forces. Correct answer: 0 . 979435 m / s. Explanation: Let : g = 9 . 8 m / s 2 , m = 5 kg , = 1 m , and = 18 . The diagram for the velocity and accelera tion of the ball at point P is as follows, x y v a The difference of gravitational potential en ergy of the ball between point Q and point P is U = U Q U P = mg (1 cos ). The velocity of the ball at point Q is zero, so K P + U P = K Q + U Q K P + U P = 0 + U Q . Therefore 1 2 mv 2 P = K P = U Q U P = mg (1 cos ) v P = p 2 g (1 cos ) = q 2 (9 . 8 m / s 2 ) (1 m) (1 cos 18 ) = . 979435 m / s . 002 (part 2 of 6) 0 points What is the magnitude of the balls accelera tion at the point P ? Correct answer: 0 . 959292 m / s 2 . Explanation: At point P , the centripetal acceleration is the total acceleration: a P = v 2 P = (0 . 979435 m / s) 2 1 m = . 959292 m / s 2 . 003 (part 3 of 6) 10 points What is the balls speed at the point Q ? 1. 0 m / s < k v Q k < k v p k 2. k v Q k = 0 m/s correct 3. k v Q k < 0 m/s 4. k v Q k > k v p k 5. k v Q k = k v p k Explanation: Since Q is one of the highest points, the velocity at Q is zero . Thevalingam, Donald Homework 18 Due: Feb 7 2007, midnight Inst: Eslami 2 004 (part 4 of 6) 0 points What is the magnitude of the balls accelera tion at the point Q ? Correct answer: 3 . 02837 m / s 2 . Explanation: The diagram for the velocity and accelera tion of the ball at point Q is as follows, x y a v = 0 Since the velocity at point Q is zero, the centripetal acceleration (which should point along the string toward the point where the string is attached) is also zero. The tangential acceleration is thus the to tal acceleration. Since it is perpendicular to the string, the tension of the string does not contribute to it. It is provided by the tangen tial component of the gravitational force on the ball. a Q = mg sin m = g sin = (9 . 8 m / s 2 ) sin 18 = 3 . 02837 m / s 2 . 005 (part 5 of 6) 10 points After several swings, the string breaks. The mass of the string and air resistance are neg ligible....
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 Spring '08
 Turner
 Physics, Work

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