281_PartUniversity Physics Solution

281_PartUniversity Physics Solution - Rotation of Rigid...

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Rotation of Rigid Bodies 9-3 (c) The average angular velocity is 6.00 rad s 8.00 rad s 1.00 rad s 2 −+ = . 0a v - z t θθ ω −= then leads to displacement of 7.00 rad after 7.00 s. EVALUATE: When z α and z ω have the same sign, the angular speed is increasing; this is the case for 3.00 s t = to 7.00 s t = . When z and z have opposite signs, the angular speed is decreasing; this is the case between 0 t = and 3.00 s t = . 9.9. IDENTIFY: Apply the constant angular acceleration equations. SET UP: Let the direction the wheel is rotating be positive. EXECUTE: (a) 2 0 1.50 rad s (0.300 rad s )(2.50 s) 2.25 rad s. zz z t ωω α =+= + = (b) 22 2 11 00 (1.50 rad/s)(2.50 s) (0.300 rad/s )(2.50 s) 4.69 rad tt −= + = + = . EVALUATE: 0 0 1.50 rad/s 2.25 rad/s (2.50 s) 4.69 rad t ωω θθ ++ ⎛⎞ = = ⎜⎟ ⎝⎠ , the same as calculated with another equation in part (b). 9.10. IDENTIFY: Apply the constant angular acceleration equations to the motion of the fan. (a) SET UP: 0 (500 rev/min)(1 min/60 s) 8.333 rev/s, z == (200 rev/min)(1 min/60 s) 3.333 rev/s, z 4.00 s, t = ? z = 0 z t ωα =+ EXECUTE: 2 0 3.333 rev/s 8.333 rev/s 1.25 rev/s 4.00 s z t −− = 0 ? 2 (8.333 rev/s)(4.00 s) ( 1.25 rev/s )(4.00 s) 23.3 rev = +− = (b) SET UP: 0 z = (comes to rest); 0 3.333 rev/s; z = 2 1.25 rev/s ; z =− ? t = 0 z t EXECUTE: 0 2 0 3.333 rev/s 2.67 s 1.25 rev/s z t = EVALUATE: The angular acceleration is negative because the angular velocity is decreasing. The average angular velocity during the 4.00 s time interval is 350 rev/min and v - z t gives 0 23.3 rev, which checks. 9.11. IDENTIFY: Apply the constant angular acceleration equations to the motion. The target variables are t and 0 . θ SET UP: (a) 2 1.50 rad/s ; z = 0 0 z = (starts from rest); 36.0 rad/s; z = ? t = 0 z t EXECUTE: 0 2 36.0 rad/s 0 24.0 s 1.50 rad/s z t = (b) 0 ? 2 0 (1.50 rad/s )(2.40 s) 432 rad = 0 432 rad(1 rev/2 rad) 68.8 rev π = EVALUATE: We could use 1 2 () t ω ω + to calculate 1 0 2 (0 36.0 rad/s)(24.0 s) 432 rad, = which checks. 9.12. IDENTIFY: In part (b) apply the equation derived in part (a). SET UP: Let the direction the propeller is rotating be positive. EXECUTE: (a) Solving Eq. (9.7) for t gives 0 z z z t = . Rewriting Eq. (9.11) as 1 2 z z θωω + and substituting for t gives 0 0 0 1 , 2 z z z z αα + = = which when rearranged gives Eq. (9.12). (b) 2 0 0 16.0 rad s 12.0 rad s 8.00 rad/s 7.00 rad z αω =
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9-4 Chapter 9 EVALUATE: We could also use 0 0 2 zz t ωω θθ + ⎛⎞ −= ⎜⎟ ⎝⎠ to calculate 0.500 s t = . Then 0 z t ω ωα =+ gives 2 8.00 rad/s z α = , which agrees with our results in part (b). 9.13. IDENTIFY: Use a constant angular acceleration equation and solve for 0 . z SET UP: Let the direction of rotation of the flywheel be positive. EXECUTE: 2 1 00 2 z z t θ θω −= + gives 2 0 11 0 22 60.0 rad (2.25 rad/s )(4.00 s) 10.5 rad/s 4.00 s t =− = = .
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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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281_PartUniversity Physics Solution - Rotation of Rigid...

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