phsx205_hw05_solutions

phsx205_hw05_solutions - Phys 205 Chapter 5 Solutions 5.3...

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Phys 205 Chapter 5 Solutions 5.3 The Earth has a radius of 6 . 4 × 10 6 m ad completes one revolution about its axis in 24 h. (a) Find the speed of a point on the equator. (b) Find the speed of New York City. This is a kinematics of uniform circular motion problem. Given: Radius of the Earth: r E = 6 . 4 × 10 6 m Period of rotation: T = 24 h = 86 , 400 s (a) We want to find the tangential speed of a point at the equator. Use Equation 5.1. v Eq = 2 πr E T v Eq = 2 π ( 6 . 4 × 10 6 m ) 86 , 400 s = 465 m / s (b) New York City has a latitude of 40.7 so the radius of its orbit will not be the full radius of the Earth (see the figure for Problem 5.9 below). The radius of its orbit will be: r = r E cos 40 . 7 = 4 . 85 × 10 6 m. v NY = 2 πr T v NY = 2 π ( 4 . 85 × 10 6 m ) 86 , 400 s = 353 m / s 5.6 Consider the motion of the hand of a mechanical clock. If the minute hand of the clock has a length of 6.0 cm, what is the centripetal acceleration of a point at the end of the hand? This is a kinematics of uniform circular motion problem. Given: Period T = 60 min = 3600 s Radius: r = 0 . 060 m We want to find the centripetal acceleration, a c . Use Equations 5.5 and 5.1. a c = v 2 r = 1 r ± 2 πr T ² 2 a c = 4 π 2 r T 2 This last equation is useful to use for problems like this one when you are given the period & radius and want the centripetal acceleration. a c = 4 π 2 (0 . 060 m) (3600 s) 2 = 1 . 8 × 10 - 7 m / s 2 5.9 Consider points on the Earth’s surface as sketched in the figure. Because of the Earth’s rotation, these points undergo uniform circular motion. Compute the centripetal acceler- ation of (a) a point at the equator, and (b) at latitude of 30 . This is a kinematics of uniform circular motion problem. Given: Radius of the Earth: r E = 6 . 4 × 10 6 m Period of rotation: T = 24 h = 86 , 400 s (a) We want to find the centripetal acceleration, a c , of a point at the equator. Use the equation from the previous problem. !"# a c = 4 π 2 r T 2 a c = 4 π 2 ( 6 . 4 × 10 6 m ) (86 , 400 s) 2 = 0 . 034 m / s 2 (b) At a latitude of 30 the radius of orbit will be: r = r E cos 30 = 5 . 54 × 10 6 m. a c = 4 π 2 r T 2 a c = 4 π 2 ( 5 . 54 × 10 6 m ) (86 , 400 s) 2 = 0 . 029 m / s 2 1
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5.12 When a fighter pilot makes a quick turn, he experiences a centripetal acceleration. When this acceleration is greater than about 8 × g , the pilot will usually lose consciousness (“black out”). consider a pilot flying at a speed of 900 m/s who wants to make a sharp turn. What is the minimum radius of curvature she can take without blacking out? This is a kinematics of uniform circular motion problem. (a) Given: Maximum centripetal acceleration: a c = 8 × g = 78 . 4 m / s 2 Tangential speed: v = 900 m / s We want to find the minimum radius of curvature which corresponds to the maximum centripetal acceleration. Use Equation 5.5. a c = v 2 r r = v 2 a c r = (900 m / s) 2 78 . 4 m / s 2 = 1 . 0 × 10 4 m The minimum radius is roughly 10 km. 5.14 The Daytona 500 stock car race is held on a track that is approximately 2.5 mi long, and the turns are banked at an angle of 31 . It is currently possible for cars to travel through the turns at a speed of 180 mi/h. Assuming these cars are on the verge of slipping into the outer wall of the racetrack (because they are racing!), find the
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phsx205_hw05_solutions - Phys 205 Chapter 5 Solutions 5.3...

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