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10_InstSolManual_PC - ROTATIONAL MOTION 10 EXERCISES...

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ROTATIONAL MOTION E XERCISES Section 10.1 Angular Velocity and Acceleration 14. The angular speed is / . t ϖ θ = ∆ (a) 5 1 E 1 rev/1 d 2 /86,400 s 7.27 10 s . ϖ π - - = = = × (b) 3 1 min 1 rev/1 h 2 /3600 s 1.75 10 s . ϖ π - - = = = × (c) 4 1 1 hr min 12 1 rev/12 h 1.45 10 s . ϖ ϖ - - = = = × (d) 1 300 rev/min 300 2 /60 s 31.4 s . ϖ π - = = × = ( Note: Radians are a dimensionless angular measure, i.e., pure numbers; therefore angular speed can be expressed in units of inverse seconds.) 15. I NTERPRET We are asked to compute the linear speed at some location on Earth. The problem involves the rotational motion of the Earth. D EVELOP We first calculate the angular speed of the Earth using Equation 10.1: 5 1 E 1 rev 2 rad 7.27 10 s 1 d 86,400 s t θ π ϖ - - = = = = × The linear speed can then be computed using Equation 10.3: . v r ϖ = E VALUATE (a) On the equator, 5 1 6 E E (7.27 10 s )(6.37 10 m) 463 m/s v R ϖ - - = = × × = (b) At latitude E , cos r R θ θ = so (463 m/s)cos . E v r ϖ θ = = A SSESS The angle 0 θ = corresponds to the Equator. So the result found in (b) agrees with (a) . In addition, if we take 90 , θ = ° then we are at the poles, and the linear speed is zero there. 16. (a) 1 1 (720 rev/min)(2 /rev)(min/60s) 24 s 75.4 s . π π - - = = (b) 4 1 (50 / )( /180 )(h/3600 s) 2.42 10 s . h π - - ° ° = × (c) 1 3 1 (1000 rev/s)(2 /rev) 2000 s 6.28 10 s . π π - - = × < (d) 7 7 1 (1 rev/y) 2 /( 10 s) 2 10 s . π π - - × = × < (See note in solution to Exercise 14. The approximate value for 1 y used in part (d) is often handy for estimates, and is fairly accurate; see Chapter 1, Problem 20.) 17. I NTERPRET The problem asks about the linear speed of the straight wood saw, but it’s equivalent to finding the linear speed of the circular saw. D EVELOP We first convert the angular speed to rad/s: 3500 rev 2 (3500) rad 367 rad/s 1 min 60 s t θ π ϖ = = = = The linear speed can then be computed using Equation 10.3: . v r ϖ = E VALUATE The radius of the circular saw is 12.5 cm 0.125 m. r = = Therefore, its linear speed is (367 rad/s)(0.125 m) 45.8 m/s v r ϖ = = = A SSESS The linear speed of the saw is more than 100 mi/h! 18. From Equation 10.4 (before the limit is taken), av / (500 200) rpm/(74 60 s) t α ϖ = ∆ ∆ = - × = 2 3 2 6.76 10 rpm/s 7.08 10 s . - - - × = × (Recall that 1 rpm 2 /60 s.) π = 19. I NTERPRET In this problem we are asked to find the angular speed and number of revolutions of a turbine, given its angular acceleration. The key to this type of rotational problem is to identify the analogous situation for linear motion. The analogies are summarized in Table 10.1. 10.1 10
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10.2 Chapter 10 D EVELOP Given a constant angular acceleration α , the angular velocity and angular position at a later time t can be found using Equations 10.7 and 10.8: 0 2 0 0 1 2 t t t ϖ ϖ α θ θ ϖ α = + = + + E VALUATE (a) The initial and final angular velocities are 0 0 ϖ = and 3600 rev 2 (3600) rad 3600 rpm 377 rad/s 1 min 60 s π ϖ = = = = Therefore, the amount of time it takes to reach this angular speed is 0 2 377 rad/s 0 725 s 12.1 min 0.52 rad/s t ϖ ϖ α - - = = = = (b) Using Equation 10.8, we find the number of turns made during this time interval to be 2 2 2 2 5 4 0 0 1 1 1 (0.52 rad/s )(725 s) 1.37 10 rad 2.17 10 rev 2 2 2 t t t θ θ ϖ α α = + + = = = × = × A SSESS The turbine turns very fast. After 12.1 min, it has reached an angular speed of 377 rad/s, or 60 rev/s!
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