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# SOL18 - Thevalingam Donald Homework 18 Due Feb 7 2007...

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Thevalingam, Donald – Homework 18 – Due: Feb 7 2007, midnight – Inst: Eslami 1 This print-out should have 17 questions. Multiple-choice 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 0 P 18 The two highest points of the ball’s trajec- tory are Q and Q 0 ; at these points the string is ± 18 from the vertical. Point P is the lowest point of the ball’s trajectory where the string hangs vertically down. The acceleration of gravity is 9 . 8 m / s 2 . What is the ball’s 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 θ = 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 Δ U = U Q - U P = m g ‘ (1 - cos θ ). 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 m v 2 P = K P = U Q - U P = m g ‘ (1 - cos θ ) v P = p 2 g ‘ (1 - cos θ ) = q 2 (9 . 8 m / s 2 ) (1 m) (1 - cos 18 ) = 0 . 979435 m / s . 002 (part 2 of 6) 0 points What is the magnitude of the ball’s 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 = 0 . 959292 m / s 2 . 003 (part 3 of 6) 10 points What is the ball’s 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 .

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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 ball’s 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 θ 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 = m g sin θ m = g sin θ = (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. If the break occurs when the ball is at point P , describe the motion of the ball after the break. 1. It moves horizontally with a constant speed. 2. The ball falls vertically straight down.
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SOL18 - Thevalingam Donald Homework 18 Due Feb 7 2007...

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