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# hw8a - khounvivongsy(sk27799 – Homework 8 – Weathers...

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khounvivongsy (sk27799) – Homework 8 – Weathers – (17104) 1 This print-out should have 12 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 11 . 3 m in diameter from which 10 kg seats are suspended at the end of 3 . 65 m massless chains. When the system rotates, the chains make an angle of 38 . 1 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: 7 . 79242 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 = (3 . 65 m) sin 38 . 1 + 11 . 3 m 2 = 7 . 90218 m , we have v = radicalbig g r tan θ = radicalBig (9 . 8 m / s 2 ) (7 . 90218 m) tan 38 . 1 = 7 . 79242 m / s . 002 (part 2 of 2) 10.0 points If a child of mass 46 . 4 kg sits in a seat, what is the tension in the chain (for the same angle)? Correct answer: 702 . 371 N. Explanation: M = 46 . 4 kg From the first part we have T cos θ = ( m + M ) g T = ( m + M ) g cos θ = (10 kg + 46 . 4 kg) (9 . 8 m / s 2 ) cos 38 . 1 = 702 . 371 N . 003 10.0 points A curve of radius 53 . 6 m is banked so that a car traveling with uniform speed 48 km / hr can round the curve without relying on fric- tion to keep it from slipping to its left or right. The acceleration of gravity is 9 . 8 m / s 2 . 1 . 9 Mg μ 0 θ What is θ ? Correct answer: 18 . 6981 . Explanation: Let : m = 1900 kg , v = 48 km / hr , r = 53 . 6 m , and μ 0 . Basic Concepts: Consider the free body diagram for the car. The forces acting on

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khounvivongsy (sk27799) – Homework 8 – Weathers – (17104) 2 the car are the normal force, the force due to gravity, and possibly friction. μ N N N cos θ m g N sin θ x y To keep an object moving in a circle re- quires a force directed toward the center of the circle; the magnitude of the force is F c = m a c = m v 2 r . Also remember, vector F = summationdisplay i vector F i . Using the free-body diagram, we have summationdisplay i F x N sin θ - μ N cos θ = m v 2 r (1) summationdisplay i F y N cos θ + μ N sin θ = m g (2) ( m g ) bardbl = m g sin θ (3) m a bardbl = m v 2 r cos θ (4) and , if μ = 0 , we have tan θ = v
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