Chapter 07

# Chapter 07 - CHAPTER 7 Quick Quizzes 1(c For a rotation of...

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C HAPTER 7 Quick Quizzes 1. (c). For a rotation of more than 180°, the angular displacement must be larger than 3.14 rad π = . The angular displacements in the three choices are (a) , (b) 6 rad 3 rad 3 rad −= ( ) 1 rad 2 rad −− = , (c) 5 rad 1 rad 4 rad = . 2. (b). Because all angular displacements occurred in the same time interval, the displacement with the lowest value will be associated with the lowest average angular speed. 3. (b) and (d). Points on the spokes which are closer to the axis of rotation have smaller values of tangential speed and acceleration than do points on the rim of the wheel. 4. (e), (a), (b). 5. (c). Both change in direction, but have constant magnitudes. 6. (b) and (c). According to Newton’s law of universal gravitation, the force between the ball and the Earth depends on the product of their masses, so both forces, that of the ball on the Earth, and that of the Earth on the ball, are equal in magnitude. This follows also, of course, from Newton’s third law. The ball has large motion compared to the Earth because according to Newton’s second law, the force gives a much greater acceleration to the small mass of the ball. 7. (e). 8. (a). From Kepler’s third law and the given period, the major axis of the comet can be calculated. It is found to be . Because this is smaller than the Earth-Sun distance, the comet cannot possibly collide with the Earth. 11 1.2 10 m × 209

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CHAPTER 7 Problem Solutions 7.1 (a) 60 000 mi 5280 ft 1.0 ft 1 mi s r θ ⎛⎞ == = ⎜⎟ ⎝⎠ 8 3.2 10 rad × . (b) 8 1 rev 3.2 2 r a d π = 7 5.0 10 rev × . 7.2 The distance traveled is s r = , where is in radians. For 30°, () rad 4.1 m 30 180 sr ° = ° 2.1 m . For 30 radians, ( )( ) 4.1 m 30 rad = 2 1.2 10 m × . For 30 revolutions, r a d 4.1 m 30 rev 1 rev = 2 7.7 × . 7.3 The Earth moves through 2 π rad in one year ( ) 7 3.156 10 s × . Thus, 7 r a d 3.156 ω × 7 1.99 10 rad s × . Alternatively, the Earth moves through 360° in one year (365.242 days). Thus, 360 365.2 days ° 0.986 deg day . 7.4 We use f t i α = and find 0.20 rev s 0 2 rad 30 s 1 rev -2 2 4.2 10 rad s × . 7.5 (a) ( ) 4 2.51 10 rev min 0 r a d 1 min 3.20 s 1 rev 60.0 s ×− 2 821 rad s 210
CHAPTER 7 (b) () 2 2 2 11 r a d 0 821 3.20 s 22 s i tt θω α ⎛⎞ =+ = + = 3 4.21 10 rad × ⎜⎟ ⎝⎠ 7.6 2 r a d rev 1 min 100 10.47 rad s min 1 rev 60.0 s i π ω == (a) From fi t , 2 01 0 . 4 7 r a d s 2.00 rad s t = 5.24 s . (b) 0 10.47 rad s 5.24 s + + = = 27.4 rad . 7.7 The final linear speed of the car is ( ) ( ) 2 17.0 m s 2.00 m s 5.00 s 27.0 m s vva t =+= + = , and the distance traveled in this time interval is 27.0 17.0 m s 5.00 s 110 m vv sv t t + + = = . The angular displacement of the wheel is 110 m 1 rev = 0.48 m 2 rad s r θ 36.5 rev . 7.8 The initial angular velocity is 24.0 m s 50.0 rad s 0.480 m i i v r = . Then, from ( ) 2 ωα =+ ∆ , the angular displacement as the wheel stops is ( ) ( ) 2 2 05 0 . 0 r a d s 1 rev 2 -1.35 rad s ωω απ ∆= = = r a d 147 rev . 211

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CHAPTER 7 7.9 Main Rotor: () 2 r a d rev 1 min 3.80 m 450 min 1 rev 60 s vr π ω ⎛⎞ ⎛⎞ ⎛ = == ⎜⎟ ⎜ ⎜⎟ ⎝⎠ ⎝ ⎝⎠ 179 m s m = 179 = s3 4 3 m s sound v v 0.522 sound v Tail Rotor: r a d 1 min 0.510 m 4138 60 s ⎛⎞⎛ = ⎜⎟⎜ ⎝⎠⎝ 221 m s m = 221 = 4 3 m s sound v v 0.644 sound v 7.10
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Chapter 07 - CHAPTER 7 Quick Quizzes 1(c For a rotation of...

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