42O0kg)(25Ky = 30,000N 25In

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Unit 4: Uniform Circular Motion Name_______Key_______________________ WS-3 Vertical Motion and Banked Turns Date_______________ Block___________ 1. A car travels through a valley at constant speed, though not at constant velocity. a. Explain how this is possible. thereto - its direction e The car is constantly changing it is accelerating b. Is the car accelerating? What direction is the car's acceleration? (Explain how you know.) towards the center of the circle yes , c. Construct a qualitative force diagram for the car at the moment it is at the bottom of the valley. Are the forces balanced? Justify the relative sizes of the forces. ^ F w F T ÷ 8 " mg d. If the car's speed is 25 m/s, its mass is 1200 kg and the radius of valley (r) is 25 meters, determine the magnitude of the centripetal force acting on the car. t.my?42o0kg)(25kY = 30,000N 25in e. Construct a quantitative force diagram for the car at the bottom of the valley. - a N = 42,000 n mg 30,000N = (10%2) ÷ mg 1200kg u 12,000N 2. A car travels over a hill at constant speed. ftw • Img a. Is the car accelerating? What direction is the car's acceleration? (Explain how you know.) The car is accelerating. At the top the acceleration is down. The acceleration is always perpendicular to the tangent at the location b. If the speed of the car is 43 km/h, its mass is 1200 kg and the radius of the hill (r) is 25m, determine the magnitude of the centripetal force acting on the car. - V - 43€ III. 94% 4200197/1194%12 her E=m= - 25in m=l200kg E= 6800N - r -25in c. Construct a quantitative force diagram for the car at the moment it is at the top of the hill. Are the forces balanced? Justify the relative sizes of the forces. n 200N - - -_ EF - Fw 6800N my • • 12,000 - 6800 -_ Fw umgME v12 Fw= 5,200N 5. ,000N 3. A car goes around a 20 degree banked turn at a radius of 50 meters. The coefficient of static friction between the tires and the road is 0.80. What is the maximum speed the car can be going to make it around the corner? -_ - - O EE mac r Etty ? Fucose - -0 - =O Fns.net/FosO---mr-v Isar mg ' - - - - - - O cost me Euro Fucose ufus.no Fusaro -1mF mg r ° '¥ " ? - !I .no Fnfcoso us )=mg Fnfsiroteecoso)=m¥ ) , r- - E=mg_- Cosa - =m¥ Masino C osf Usnrf I rmlgfsihottrcoso=rg(sirO+µcosO# (4) (b) ( Suh 200+0.86520 ) EM.no#zoo-o.ss.uzoo)v--28.7Ys✓.

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Circular Motion Dynamics
Problem Solving Circular Motion Dynamics Problem 1: Double Star System Consider a double star system under the influence of gravitational force between the stars. Star 1 has mass m1 and star 2 has mass m2 . Assume that each star undergoes uniform circular motion about the center of mass of the system. If the stars are always a fixed distance s apart, what is the period of the orbit? Solution: Choose radial coordinates for each star with origin at center of mass. Let rˆ1 be a unit vector at Star 1 pointing radially away from the center of mass. Let rˆ2 be a unit vector at Star 2 pointing radially away from the center of mass. The force diagrams on the two stars are shown in the figure below. ! ! From Newton’s Second Law, F1 = m 1a 1 , for Star 1 in the radial direction is m m ˆ 1 2 2 r1 : "G 2 = "m1 r 1 ! . s We can solve this for r1 , m r = G 2 . 1 ! 2s 2 ! ! Newton’s Second Law, F2 = m 2a 2 , for Star 2 in the radial direction is m m ˆ 1 2 2 r2 : "G 2 = "m2 r 2 ! . s We can solve this for r2 , m r = G 1 . 2 ! 2s 2 Since s , the distance between the stars, is constant m2 m1 (m2 + m1 ) s = r + r = G + G = G 1 2 ! 2s 2! 2s 2 ! 2s 2 . Thus the angular velocity is 1 2 " (m2 + m1 ) # ! = $G 3 % & s ' and the period is then 1 2 2! # 4! 2s 3 $ T = = % & .
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Assignment 2 Centripetal Force & Univ
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April 2018 the Privileged View Steve Beste, President
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FAA-H-8083-3A, Airplane Flying Handbook -- 3 of 7 Files
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