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solution_pdf6 - billing(cab4763 HW 6 opyrchal(11109 This...

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billing (cab4763) – HW 6 – opyrchal – (11109) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A small ball of mass 38 g is suspended from a string of length 84 cm and whirled in a circle lying in the horizontal plane. v r 9 . 8 m / s 2 84 cm 38 g 20 If the string makes an angle of 20 with the vertical, find the centripetal force experienced by the ball. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 0 . 135543 N. Explanation: T m g θ The centripetal force is supplied by the horizontal component of the tension. From Newton’s second law applied in the x and y directions, T cos θ = m g and T sin θ = F c . Dividing the second equation by the first, tan θ = F c m g F c = m g tan θ = (38 g)(9 . 8 m / s 2 ) tan 20 = 0 . 135543 N 002 10.0 points A curve of radius 52 . 9 m is banked so that a car traveling with uniform speed 53 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 . 7 Mg μ 0 θ What is θ ? Correct answer: 22 . 6891 . Explanation: Let : m = 1700 kg , v = 53 km / hr , r = 52 . 9 m , and μ 0 . Basic Concepts: Consider the free body diagram for the car. The forces acting on the car are the normal force, the force due to gravity, and possibly friction.
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billing (cab4763) – HW 6 – opyrchal – (11109) 2 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 g r (5) Solution: Solution in an Inertial Frame: Watching from the Point of View of Some- one Standing on the Ground. The car is performing circular motion with a constant speed, so its acceleration is just the centripetal acceleration, a c = v 2 r . The net force on the car is F net = m a c = m v 2 r The component of this force parallel to the
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