10_InstSolManual_PC - ROTATIONAL MOTION E XERCISES Section 10.1 Angular Velocity and Acceleration 14 The angular speed is t ϖ θ = ∆ ∆(a 5 1 E

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Unformatted text preview: 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 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 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: 2 1 2 t t t ϖ ϖ α θ θ ϖ α = + = + + E VALUATE (a) The initial and final angular velocities are ϖ = and 3600 rev 2 (3600) rad 3600 rpm 377 rad/s 1 min...
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This note was uploaded on 09/12/2009 for the course PH PH101 taught by Professor Prof.kim during the Spring '09 term at Korea Advanced Institute of Science and Technology.

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10_InstSolManual_PC - ROTATIONAL MOTION E XERCISES Section 10.1 Angular Velocity and Acceleration 14 The angular speed is t ϖ θ = ∆ ∆(a 5 1 E

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