# hw5 - patel(kp9583 HW5 Mackie(20211 This print-out should...

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patel (kp9583) – HW5 – Mackie – (20211) 1 This print-out should have 8 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 An amusement park ride consists of a rotating circular platform 8 . 69 m in diameter from which 10 kg seats are suspended at the end of 2 . 21 m massless chains. When the system rotates, the chains make an angle of 21 . 8 with the vertical. The acceleration of gravity is 9 . 8 m / s 2 . θ l d What is the speed of each seat? Correct answer: 4 . 4998 m / s. Explanation: In the vertical direction we have T cos θ = m g , where T is the tension in the chain. In the horizontal direction we have T sin θ = m v 2 r . Since r = sin θ + d 2 = (2 . 21 m) sin 21 . 8 + 8 . 69 m 2 = 5 . 16572 m , we have v = radicalbig g r tan θ = radicalBig (9 . 8 m / s 2 ) (5 . 16572 m) tan 21 . 8 = 4 . 4998 m / s . 002 (part 2 of 2) 10.0 points If a child of mass 38 . 5 kg sits in a seat, what is the tension in the chain (for the same angle)? Correct answer: 511 . 909 N. Explanation: M = 38 . 5 kg From the first part we have T cos θ = ( m + M ) g T = ( m + M ) g cos θ = (10 kg + 38 . 5 kg) (9 . 8 m / s 2 ) cos 21 . 8 = 511 . 909 N . 003 (part 1 of 2) 10.0 points A small metal ball is suspended from the ceil- ing by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread describes a cone. The acceleration of gravity is 9 . 8 m / s 2 . v 9 . 8 m / s 2 2 m 6 . 2 kg 24 What is the speed of the ball when it is in circular motion? Correct answer: 1 . 88398 m / s. Explanation: Let : = 2 m , θ = 24 , g = 9 . 8 m / s 2 , and m = 6 . 2 kg .

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patel (kp9583) – HW5 – Mackie – (20211) 2 Use the free body diagram below. T m g θ The tension on the string can be decom- posed into a vertical component which bal- ances the weight of the ball and a horizontal component which causes the centripetal ac- celeration, a centrip that keeps the ball on its horizontal circular path at radius r = sin θ . If T is the magnitude of the tension in the string, then T vertical = T cos θ = m g (1) and T horiz = m a centrip or T sin θ = m v 2 ball sin θ . (2) Solving (1) for T yields T = m g cos θ (3) and substituting (3) into (2) gives m g tan θ = m v 2 ball
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