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# solution12_pdf - arellano(aa39398 Homework 8 weathers(17101...

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arellano (aa39398) – Homework 8 – weathers – (17101) 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 9 . 07 m in diameter from which 10 kg seats are suspended at the end of 1 . 61 m massless chains. When the system rotates, the chains make an angle of 16 . 9 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: 3 . 85958 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 = (1 . 61 m) sin 16 . 9 + 9 . 07 m 2 = 5 . 00303 m , we have v = radicalbig g r tan θ = radicalBig (9 . 8 m / s 2 ) (5 . 00303 m) tan 16 . 9 = 3 . 85958 m / s . 002 (part 2 of 2) 10.0 points If a child of mass 47 . 9 kg sits in a seat, what is the tension in the chain (for the same angle)? Correct answer: 593 . 031 N. Explanation: M = 47 . 9 kg From the first part we have T cos θ = ( m + M ) g T = ( m + M ) g cos θ = (10 kg + 47 . 9 kg) (9 . 8 m / s 2 ) cos 16 . 9 = 593 . 031 N . 003 10.0 points A curve of radius 50 . 8 m is banked so that a car traveling with uniform speed 51 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 . 2 . 1 Mg μ 0 θ What is θ ? Correct answer: 21 . 9559 . Explanation: Let : m = 2100 kg , v = 51 km / hr , r = 50 . 8 m , and μ 0 . Basic Concepts: Consider the free body diagram for the car. The forces acting on

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arellano (aa39398) – Homework 8 – weathers – (17101) 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 2
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