c The following drawing shows the relevant geometry r s celestial body person

C the following drawing shows the relevant geometry r

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c. The following drawing shows the relevant geometry: r s celestial body person on earth pe rson on e arth moon r moon r sun s sun R sun sun moon s b R b
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388 ROTATIONAL KINEMATICS The apparent circular area of the sun as measured by a person standing on the earth is given by: 2 sun sun A R , where R sun is the radius of the sun. The apparent circular area of the sun that is blocked by the moon is 2 blocked b A R , where R b is shown in the figure above. Also from the figure above, it follows that R sun = (1/2) s sun and R b = (1/2) s b Therefore, the fraction of the apparent circular area of the sun that is blocked by the moon is 2 2 2 2 blocked b b b moon sun 2 2 sun sun sun sun sun sun 2 2 3 moon 3 sun ( / 2) ( / 2) 9.04 10 rad 0.951 9.27 10 rad A R s s r A s r R s The moon blocks out 95.1 percent of the apparent circular area of the sun. 19. REASONING The golf ball must travel a distance equal to its diameter in a maximum time equal to the time required for one blade to move into the position of the previous blade. SOLUTION The time required for the golf ball to pass through the opening between two blades is given by / t      , with 1.25 rad/s and (2 rad)/16 = 0.393 rad . Therefore, the ball must pass between two blades in a maximum time of t 0.393 rad 1.25 rad/s 0.314 s The minimum linear speed of the ball is v x t 4.50 10 –2 m 0.314 s 1.43 10 –1 m/s 20. REASONING We know that the skater’s final angular velocity is 0.0 rad /s , since she comes to a stop. We also that her initial angular velocity is 0 15 rad / s   and that her angular displacement while coming to a stop is 5.1 rad   . With these three values, we can use 2 2 0 2  (Equation 8.8) to calculate her angular acceleration α . Then, knowing α , we can use 0 t (Equation 8.4) to determine the time t during which she comes to a halt. SOLUTION a. With 0.0 rad /s for the skater’s final angular velocity, Equation 8.8 becomes 2 0 0 2  . Solving for , we obtain
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Chapter 8 Problems 389 2 2 2 0 (15 rad/s) –22 rad/s 2 2(5.1 rad)     b. With 0.0 rad /s for the skater’s final angular velocity, Equation 8.4 becomes 0 0 t . Solving for t , we obtain 0 2 15 rad/s 0.68 s –22 rad/s t     ____________________________________________________________________________________________ 21. SSM REASONING The angular displacement is given as θ = 0.500 rev, while the initial angular velocity is given as 0 = 3.00 rev/s and the final angular velocity as = 5.00 rev/s. Since we seek the time t , we can use Equation 8.6 1 2 0 t from the equations of rotational kinematics to obtain it.
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