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**Unformatted text preview: **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 cos40 . 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 → 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 cos30 ◦ = 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 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....

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