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 + = . 0 av- z t θ θ ω = then leads to displacement of 7.00 rad after 7.00 s. E VALUATE : 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 α and z ω have opposite signs, the angular speed is decreasing; this is the case between 0 t = and 3.00 s t = . 9.9. I DENTIFY : Apply the constant angular acceleration equations. S ET U P : Let the direction the wheel is rotating be positive. E XECUTE : (a) 2 0 1.50 rad s (0.300 rad s )(2.50 s) 2.25 rad s. z z z t ω ω α = + = + = (b) 2 2 2 1 1 0 0 2 2 (1.50 rad/s)(2.50 s) (0.300 rad/s )(2.50 s) 4.69 rad z z t t θ θ ω α = + = + = . E VALUATE : 0 0 1.50 rad/s 2.25 rad/s (2.50 s) 4.69 rad 2 2 z z t ω ω θ θ + + = = = , the same as calculated with another equation in part (b). 9.10. I DENTIFY : Apply the constant angular acceleration equations to the motion of the fan. (a) S ET U P : 0 (500 rev/min)(1 min/60 s) 8.333 rev/s, z ω = = (200 rev/min)(1 min/60 s) 3.333 rev/s, z ω = = 4.00 s, t = ? z α = 0 z z z t ω ω α = + E XECUTE : 2 0 3.333 rev/s 8.333 rev/s 1.25 rev/s 4.00 s z z z t ω ω α = = = − 0 ? θ θ = 2 2 2 1 1 0 0 2 2 (8.333 rev/s)(4.00 s) ( 1.25 rev/s )(4.00 s) 23.3 rev z z t t θ θ ω α = + = + = (b) S ET U P : 0 z ω = (comes to rest); 0 3.333 rev/s; z ω = 2 1.25 rev/s ; z α = − ? t = 0 z z z t ω ω α = + E XECUTE : 0 2 0 3.333 rev/s 2.67 s 1.25 rev/s z z z t ω ω α = = = E VALUATE : 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 0 av- z t θ θ ω = gives 0 23.3 rev, θ θ = which checks. 9.11. I DENTIFY : Apply the constant angular acceleration equations to the motion. The target variables are t and 0 . θ θ S ET U P : (a) 2 1.50 rad/s ; z α = 0 0 z ω = (starts from rest); 36.0 rad/s; z ω = ? t = 0 z z z t ω ω α = + E XECUTE : 0 2 36.0 rad/s 0 24.0 s 1.50 rad/s z z z t ω ω α = = = (b) 0 ? θ θ = 2 2 2 1 1 0 0 2 2 0 (1.50 rad/s )(2.40 s) 432 rad z z t t θ θ ω α = + = + = 0 432 rad(1 rev/2 rad) 68.8 rev θ θ π = = E VALUATE : We could use 1 0 0 2 ( ) z z t θ θ ω ω = + to calculate 1 0 2 (0 36.0 rad/s)(24.0 s) 432 rad, θ θ = + = which checks. 9.12. I DENTIFY : In part (b) apply the equation derived in part (a). S ET U P : Let the direction the propeller is rotating be positive. E XECUTE : (a) Solving Eq. (9.7) for t gives 0 z z z t ω ω α = . Rewriting Eq. (9.11) as 1 0 0 2 ( ) z z t t θ θ ω ω = + and substituting for t gives 2 2 0 0 0 0 0 0 0 1 1 1 ( ) ( ) ( ), 2 2 2 z z z z z z z z z z z z z z ω ω ω ω θ θ ω ω ω ω ω ω ω α α α + = + = = ⎟⎜ which when rearranged gives Eq. (9.12). (b) ( ) ( ) ( ) ( ) 2 2 2 2 2 1 1 0 2 2 0 1 1 16.0 rad s 12.0 rad s 8.00 rad/s 7.00 rad z z z α ω ω θ θ = = =
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9-4 Chapter 9 E VALUATE : We could also use 0 0 2 z z t ω ω θ θ + = to calculate 0.500 s t = . Then 0 z z z t ω ω α = + gives 2 8.00 rad/s z α = , which agrees with our results in part (b).
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