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296_PartUniversity Physics Solution

296_PartUniversity Physics Solution - 9-18 Chapter 9(b mg M...

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9-18 Chapter 9 (b) M M drum mass mg h K K = + . 2 1 M M drum 2 mv mg h K = and 2 drum M M 2 2(250.0 J) 2 2(3.71 m/s )(13.2 m) 8.04 m/s 15.0 kg K v g h m = = = E VALUATE : We did the calculations without knowing the moment of inertia I of the drum, or the mass and radius of the drum. 9.71. I DENTIFY and S ET U P : All points on the belt move with the same speed. Since the belt doesn°t slip, the speed of the belt is the same as the speed of a point on the rim of the shaft and on the rim of the wheel, and these speeds are related to the angular speed of each circular object by . v r ω = E XECUTE : Figure 9.71 (a) 1 1 1 v r ω = 1 (60.0 rev/s)(2 rad/1 rev) 377 rad/s ω π = = 2 1 1 1 (0.45 10 m)(377 rad/s) 1.70 m/s v r ω = = × = (b) 1 2 v v = 1 1 2 2 r r ω ω = 2 1 2 1 ( / ) (0.45 cm/2.00 cm)(377 rad/s) 84.8 rad/s r r ω ω = = = E VALUATE : The wheel has a larger radius than the shaft so turns slower to have the same tangential speed for points on the rim. 9.72. I DENTIFY : The speed of all points on the belt is the same, so 1 1 2 2 r r ω ω = applies to the two pulleys. S ET U P : The second pulley, with half the diameter of the first, must have twice the angular velocity, and this is the angular velocity of the saw blade. rad/s 30 rev/min π = . E XECUTE : (a) 2 rad s 0.208 m (2(3450 rev min)) 75.1 m s. 30 rev min 2 v π = = ⎟⎜ (b) 2 2 4 2 rad rad s 0.208 m 2(3450 rev min) 5.43 10 m s , 30 rev min 2 a r π ω = = = × so the force holding sawdust on the blade would have to be about 5500 times as strong as gravity. E VALUATE : In v r ω = and 2 rad a r ω = , ω must be in rad/s. 9.73. I DENTIFY and S ET U P : Use Eq.(9.15) to relate rad a to ω and then use a constant acceleration equation to replace . ω E XECUTE : (a) 2 rad , a r ω = 2 rad,1 1 , a r ω = 2 rad,2 2 a r ω = 2 2 rad rad,2 rad,1 2 1 ( ) a a a r ω ω Δ = = One of the constant acceleration equations can be written 2 2 2 1 2 1 2 ( ), z z ω ω α θ θ = + or 2 2 2 1 2 1 2 ( ) z z z ω ω α θ θ = Thus rad 2 1 2 1 2 ( ) 2 ( ), z z a r r α θ θ α θ θ Δ = = as was to be shown. (b) 2 2 2 rad 2 1 85.0 m/s 25.0 m/s 8.00 rad/s 2 ( ) 2(0.250 m)(15.0 rad) z a r α θ θ Δ = = = Then 2 2 tan (0.250 m)(8.00 rad/s ) 2.00 m/s a r α = = = E VALUATE : 2 ω is proportional to z α and 0 ( ) θ θ so rad a is also proportional to these quantities. rad a increases while r stays fixed, z ω increases, and z α is positive. I DENTIFY and S ET U P : Use Eq.(9.17) to relate K and ω and then use a constant acceleration equation to replace . ω E XECUTE : (c) 2 1 2 ; K I ω = 2 1 2 2 2 , K I ω = 2 1 1 1 2 K I ω = 2 2 1 1 2 1 2 1 2 1 2 1 2 2 ( ) (2 ( )) ( ), z z K K K I I I ω ω α θ θ α θ θ Δ = = = = as was to be shown. (d) 2 2 2 1 45.0 J 20.0 J 0.208 kg m ( ) (8.00 rad/s )(15.0 rad) z K I α θ θ Δ = = = E VALUATE : z α is positive, ω increases, and K increases.
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Rotation of Rigid Bodies 9-19 9.74. I DENTIFY : wood lead . I I I = + m V ρ = , where ρ is the volume density and m A σ = , where σ is the area density.
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