# chap 10 - ROTATIONAL MOTION 10 EXERCISES Section 10.1...

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10.1 ROTATIONAL MOTION EXERCISES Section 10.1 Angular Velocity and Acceleration 14. The angular speed is ωθ =∆ ∆ /. t (a) ωπ E rev/ d s s == = × −− 1 1 2 86 400 7 27 10 51 /, . . (b) min rev/ h s s = × 1 1 2 3600 1 75 10 31 . (c) ωω hr rev/ h s = × 1 12 1 45 10 1 12 41 min .. (d) × = 300 300 2 60 31 4 1 rev s s /min / . . ( Note: Radians are a dimensionless angular measure, i.e., pure numbers; therefore angular speed can be expressed in units of inverse seconds.) 15. INTERPRET We are asked to compute the linear speed at some location on Earth. The problem involves the rotational motion of the Earth. DEVELOP We first calculate the angular speed of the Earth using Equation 10.1: ω θ π E d rad s s = = × t 1 1 2 86 400 727 10 5 , . 1 The linear speed can then be computed using Equation 10.3: vr = . EVALUATE (a) On the equator, vR ==× × = EE sm m / s (. ) ) 7 27 10 6 37 10 463 6 (b) At latitude θθ ,c o s rR = E so E () c o s . 463 m/s ASSESS The angle = 0 corresponds to the Equator. So the result found in (b) agrees with (a) . In addition, if we take 90 , then we are at the poles, and the linear speed is zero there. 16. (a) ( /min)( / )(min/ ) . 720 2 60 24 75 4 1 s s ππ s 1 . (b) (/ ) ( / ) ( ) . . 50 180 3600 2 42 10 °° = × h h/ s s (c) ) ( / ) . 1000 2 2000 6 28 10 13 1 rev s s s < . (d) ) / ( ) . 1 2 10 2 10 77 1 rev y s s < ×= × (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. INTERPRET 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. DEVELOP We first convert the angular speed to rad/s: = = t 3500 rev min s 1 2 3500 60 367 /s The linear speed can then be computed using Equation 10.3: = . EVALUATE The radius of the circular saw is r 12 5 0 125 cm m. Therefore, its linear speed is = ( . 367 0 1 rad/s 25 m) 45.8 m/s ASSESS The linear speed of the saw is more than1 mi/h! 00 18. From Equation 10.4 (before the limit is taken), αω av rpm/ s × = ∆∆ /( )( ) t 500 200 74 60 6 76 10 7 08 10 23 2 . × rpm/s s (Recall that 12 6 0 rpm s = ) 19. INTERPRET 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

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10.2 Chapter 10 DEVELOP 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: ωω α θθ ω =+ =+ + 0 00 2 1 2 t tt EVALUATE (a) The initial and final angular velocities are ω 0 0 = and π === = 3600 1 2 3600 60 377 rpm 3600 rev min rad s () rad/s Therefore, the amount of time it takes to reach this angular speed is t = = == ωω 0 377 0 052 725 12 1 rad/s rad/s s 2 . . (b) Using Equation 10.8, we find the number of turns made during this time interval to be = = = 22 2 1 2 1 2 1 2 0 52 725 t (. ) ( ) rad/s s 2 1 37 10 2 17 10 54 .. ×= × rev ASSESS 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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## This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at University of California, Berkeley.

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chap 10 - ROTATIONAL MOTION 10 EXERCISES Section 10.1...

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