SOLUTION From Equation 81 we have s r Using Equation 82 we can write the

Solution from equation 81 we have s r using equation

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SOLUTION From Equation 8.1 we have s r Using Equation 8.2, we can write the displacement as θ = t . With this substitution Equation 8.1 becomes s r t 0.15 m 1.4 rev /s 2 rad 1 rev Converts rev/s into rad/s       45 s 59 m The radian, being a quantity without units, is dropped from the final result, leaving the answer in meters. 11. REASONING     The average angular velocity     is defined as the angular displacement   divided by the elapsed time   t   during which the displacement occurs:   / t   (Equation 8.2). Solving for the elapsed time gives  / t    . We are given    and can calculate   from the fact that the earth rotates on its axis once every 24.0 hours. SOLUTION    The sun itself subtends an angle of 9.28   10 3  rad. When the sun moves a distance equal to its diameter, it moves through an angle that is also 9.28   10 3  rad; thus,  = 9.28   10 3  rad. The average angular velocity   at which the sun appears to move across the sky is the same as that of the earth rotating on its axis,  earth , so  earth . Since the earth makes one revolution (2   rad) every 24.0 h, its average angular velocity is  earth earth earth 2 rad 2 rad 24.0 h 24.0 h t 3600 s 1 h 5 7.27 10 rad/s The time it takes for the sun to move a distance equal to its diameter is 3 5 earth 9.28 10 rad 128 s (a little over 2 minutes) 7.27 10 rad/s t   ____________________________________________________________________________________________
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129 ROTATIONAL KINEMATICS 12. REASONING It does not matter whether the arrow is aimed closer to or farther away from the axis. The blade edge sweeps through the open angular space as a rigid unit. This means that a point closer to the axis has a smaller distance to travel along the circular arc in order to bridge the angular opening and correspondingly has a smaller tangential speed. A point farther from the axis has a greater distance to travel along the circular arc but correspondingly has a greater tangential speed. These speeds have just the right values so that all points on the blade edge bridge the angular opening in the same time interval. The rotational speed of the blades must not be so fast that one blade rotates into the open angular space while part of the arrow is still there. A faster arrow speed means that the arrow spends less time in the open space. Thus, the blades can rotate more quickly into the open space without hitting the arrow, so the maximum value of the angular speed increases with increasing arrow speed v .
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  • Winter '16
  • Angular velocity, Velocity, angular displacement

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