286_PartUniversity Physics Solution

286_PartUniversity Physics Solution - 9-8 9.26 Chapter 9...

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9-8 Chapter 9 9.26. IDENTIFY: Apply constant angular acceleration equations. vr ω = . A point on the rim has both tangential and radial components of acceleration. SET UP: tan ar α = and 2 rad = . EXECUTE: (a) 2 0 = 0.250 rev/s (0.900 rev/s )(0.200 s) 0.430 rev/s zz z t ωω α =+ + = (Note that since 0 z and z are given in terms of revolutions, it&s not necessary to convert to radians). (b) av- (0.340 rev s)(0.2 s) 0.068 rev z t Δ= = . (c) Here, the conversion to radians must be made to use Eq. (9.13), and () 0.750 m 0.430 rev/s 2 rad rev 1.01 m s. 2 ωπ ⎛⎞ == = ⎜⎟ ⎝⎠ (d) Combining equations (9.14) and (9.15), 22 2 2 2 rad tan ()( ) aaa r r ωα = + . ((0.430 rev/s)(2 rad/rev)) (0.375 m) (0.900 rev/s )(2 rad/rev)(0.375 m) a ππ ⎡⎤ ⎣⎦ . 2 3.46 m s a = . EVALUATE: If the angular acceleration is constant, tan a is constant but rad a increases as increases. 9.27. IDENTIFY: Use Eq.(9.15) and solve for r . SET UP: 2 rad = so 2 rad /, ra = where must be in rad/s EXECUTE: rad 3000 3000(9.80 m/s ) 29,400 m/s ag = 1 min 2 rad (5000 rev/min) 523.6 rad/s 60 s 1 rev π Then 2 rad 29,400 m/s 0.107 m. (523.6 rad/s) a r = EVALUATE: The diameter is then 0.214 m, which is larger than 0.127 m, so the claim is not realistic. 9.28. IDENTIFY: In part (b) apply the result derived in part (a). SET UP: 2 rad = and = ; combine to eliminate r . EXECUTE: (a) rad = . v v ωω (b) From the result of part (a), 2 rad 0.500 m s 0.250 rad s. 2.00 m s a v = EVALUATE: 2 rad = and = both require that be in rad/s, so in rad av = , is in rad/s. 9.29. IDENTIFY: = and rad / v r . SET UP: 1 rev = , so rad/s 30 rev/min = . EXECUTE: (a) ( ) 3 12.7 10 m rad/s (1250 rev min) 0.831 m s. 30 rev/min 2 r × (b) 2 3 (0.831 m s) 109 m s . (12.7 10 m) 2 v r × EVALUATE: In = , must be in rad/s. 9.30. IDENTIFY: tan = , = and 2 rad / r = . 0a v - z t θθ ω −= . SET UP: When z is constant, 0 av- 2 z z z + = . Let the direction the wheel is rotating be positive. EXECUTE: (a) 2 2 tan 10.0 m s 50.0 rad s 0.200 m a r = (b) At 3.00 s t = , 50.0 m s v = and 50.0 m s 250 rad s 0.200 m v r = and at 0, t = 2 50.0 m s ( 10.0 m s )(0 3.00 s) 80.0 m s v = , 400 rad s. = (c) av- (325 rad s)(3.00 s) 975 rad 155 rev z t = .
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Rotation of Rigid Bodies 9-9 (d) 2 rad (9.80 m/s )(0.200 m) 1.40 m/s. va r == = This speed will be reached at time 50.0 m/s 1.40 m/s 4.86 s 10.0 m/s = after 3.00 s t = , or at 7.86 s t = . (There are many equivalent ways to do this calculation.) EVALUATE: At 0 t = , 24 2 rad 3.20 10 m/s ar ω ==× . At 3.00 s t = , 42 rad 1.25 10 m/s a . For rad ag = the wheel must be rotating more slowly than at 3.00 s so it occurs some time after 3.00 s. 9.31. IDENTIFY and SET UP: Use Eq.(9.15) to relate to rad a and m = F a ! ! to relate rad a to rad . F Use Eq.(9.13) to relate and v , where v is the tangential speed. EXECUTE: (a) 2 rad = and 2 rad rad Fm am r 2 2 rad,2 2 rad,1 1 640 rev/min 2.29 423 rev/min F F ⎛⎞ = ⎜⎟ ⎝⎠ (b) vr = 22 11 640 rev/min 1.51 423 rev/min v v = (c) = 1 min 2 rad (640 rev/min) 67.0 rad/s 60 s 1 rev π Then (0.235 m)(67.0 rad/s) 15.7 m/s. = 2 rad (0.235 m)(67.0 rad/s) 1060 m/s = 2 rad 2 1060 m/s 108; 9.80 m/s a g 108 = EVALUATE: In parts (a) and (b), since a ratio is used the units cancel and there is no need to convert to rad/s.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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286_PartUniversity Physics Solution - 9-8 9.26 Chapter 9...

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