Physics Solution Manual for 1100 and 2101

This time is tarrow lv the time it takes for the edge

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Unformatted text preview: on 8.4 (with t0 = 0 s) as ω = ω0 + α t If α has the same sign as ω0, then the angular speed, which is the magnitude of the angular velocity ω, is increasing. On the other hand, If α and ω0 have opposite signs, then the angular speed is decreasing. SOLUTION According to Equation 8.4, we know that ω = ω0 + α t. Therefore, we find: ( ) ( 2.0 s ) = + 18 rad /s . The angular speed is 18 rad/s . ( ) ( ) ( ) ( 2.0 s ) = − 18 rad /s. The angular speed is 18 rad/s . (a) ω = + 12 rad /s + +3.0 rad /s 2 (b) ω = + 12 rad /s + −3.0 rad /s 2 ( 2.0 s ) = + 6.0 rad /s. The angular speed is 6.0 rad/s . (c) ω = − 12 rad /s + +3.0 rad /s 2 ( 2.0 s ) = − 6.0 rad /s. The angular speed is 6.0 rad /s . (d) ω = − 12 rad /s + −3.0 rad /s 2 Chapter 8 Problems 9. 397 SSM REASONING Equation 8.4 α = (ω − ω 0 ) / t indicates that the average angular acceleration is equal to the change in the angular velocity divided by the elapsed time. Since the wheel starts from rest, its initial angular velocity is ω0 = 0 rad/s. Its final angular velocity is given as ω = 0.24 rad/s. Since the average angular acceleration is given as α = 0.030 rad/s 2 , Equation 8.4 can be solved to determine the elapsed time t. SOLUTION Solving Equation 8.4 for the elapsed time gives t= ω − ω 0 0.24 rad/s − 0 rad/s = = 8.0 s α 0.030 rad/s 2 10. REASONING AND SOLUTION Using Equation 8.4 and the appropriate conversion factors, the average angular acceleration of the CD in rad/s2 is α= 2 ∆ω 210 rev/ min − 480 rev/ min 2π rad 1 min −3 2 = = – 6.4 × 10 rad/s ∆t 74 min 1 rev 60 s The magnitude of the average angular acceleration is 6.4 × 10−3 rad/s2 . 11. REASONING The average angular velocity ω is defined as the angular displacement ∆θ divided by the elapsed time ∆t during which the displacement occurs: ω = ∆θ / ∆t (Equation 8.2). Solving for the elapsed time gives ∆t = ∆θ / ω . We are given ∆θ and can calculate ω from the fact that the earth rotates on its axis once every 24.0 hours. −3 SOLUTION The sun itself subtends an angle of 9.28 × 10 rad. When the sun moves a −3 distance equal to its diameter, it moves through an angle that is also 9.28 × 10 rad; thus, −3 ∆θ = 9.28 × 10 rad. The average angular velocity ω at which the sun appears to move across the sky is the same as that of the earth rotating on its axis, ωearth , so ω = ωearth . Since the earth makes one revolution (2π rad) every 24.0 h, its average angular velocity is ωearth = ∆θ earth ∆tearth = 2π rad = 24.0 h 2π rad = 7.27 ×10−5 rad/s 3600 s ( 24.0 h ) 1h The time it takes for the sun to move a distance equal to its diameter is ∆t = ∆θ ωearth = 9.28 ×10−3 rad = 128 s (a little over 2 minutes) 7.27 ×10−5 rad/s 398 ROTATIONAL KINEMATICS ____________________________________________________________________________________________ 12. REASONING AND SOLUTION The angular displacements of the astronauts are equal. For A θ = sA/rA For B θ = sB/rB (8.1) Equating these two equations for θ and solving for sB gives sB = (rB/rA)sA = [(1.10 × 103 m)/(3.20 × 102 m)](2.40 × 102 m) = 825 m 13. REASONING AND SOLUTION The people meet at time t. At this time the magnitudes of their angular displacements must total 2π rad. θ1 + θ2 = 2π rad Then ω1t + ω2t = 2π rad t= 2π rad 2π rad = = 1200 s −3 ω1 + ω 2 1.7 × 10 rad/s + 3.4 × 10−3 rad/s 14. 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 ω increases with increasing arrow speed v. A longer arrow traveling at a given speed means that some part of the arrow is in the open space for a longer time. To avoid hitting the arrow, then, the blades must rotate more slowly. Thus, the maximum value of the angular speed ω decreases with increasing arrow length L. Chapter 8 Problems 399 The time during which some part of the arrow remains in the open angular space is tArrow. The time it takes for the edge of a propeller blade to rotate through the open angular space between the blades is tBlade. The maximum angular speed is the angular speed such that these two times...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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