296_PartUniversity Physics Solution

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

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9-18 Chapter 9 (b) M M drum mass mg h K K =+ . 2 1 MM d r u m 2 mv mg h K =− and 2 drum 22 ( 2 5 0 . 0 J ) 2 2(3.71 m/s )(13.2 m) 8.04 m/s 15.0 kg K vg h m = = EVALUATE: We did the calculations without knowing the moment of inertia I of the drum, or the mass and radius of the drum. 9.71. IDENTIFY and SET UP: 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 . vr ω = EXECUTE: Figure 9.71 (a) 11 1 = 1 (60.0 rev/s)(2 rad/1 rev) 377 rad/s ωπ == 2 1 (0.45 10 m)(377 rad/s) 1.70 m/s × = (b) 12 vv = 2 2 rr = 21 2 1 ( / ) (0.45 cm/2.00 cm)(377 rad/s) 84.8 rad/s ωω = EVALUATE: 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. IDENTIFY: The speed of all points on the belt is the same, so = applies to the two pulleys. SET UP: 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 = . EXECUTE: (a) 2 rad s 0.208 m (2(3450 rev min)) 75.1 m s. 30 rev min 2 v ⎛⎞ ⎜⎟ ⎝⎠ (b) 2 2 42 rad rad s 0.208 m 2(3450 rev min) 5.43 10 m s , 30 rev min 2 ar so the force holding sawdust on the blade would have to be about 5500 times as strong as gravity. EVALUATE: In = and 2 rad = , must be in rad/s. 9.73. IDENTIFY and SET UP: Use Eq.(9.15) to relate rad a to and then use a constant acceleration equation to replace . EXECUTE: (a) 2 rad , = 2 rad,1 1 , = 2 rad,2 2 = rad rad,2 rad,1 2 1 () aa a r Δ= − = − One of the constant acceleration equations can be written 2 1 2( ) , zz ωα θ or 2 1 ) z −= Thus rad 2 1 2 1 )2 ( ) , r α θθ αθθ Δ= as was to be shown. (b) 2 rad 85.0 m/s 25.0 m/s 8.00 rad/s 2 ( ) 2(0.250 m)(15.0 rad) z a r Δ− = Then tan (0.250 m)(8.00 rad/s ) 2.00 m/s = EVALUATE: 2 is proportional to z and 0 so rad a is also proportional to these quantities. rad a increases while r stays fixed, z increases, and z is positive. IDENTIFY and SET UP: Use Eq.(9.17) to relate K and and then use a constant acceleration equation to replace . EXECUTE: (c) 2 1 2 ; K I = 2 1 2 , K I = 2 1 2 K I = 2 1 ( 2 ( ) ) ( ) , KK K I I I Δ= − = − = as was to be shown. (d) 2 2 45.0 J 20.0 J 0.208 kg m ( ) (8.00 rad/s )(15.0 rad) z K I αθ θ = EVALUATE: z is positive, increases, and K increases.
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Rotation of Rigid Bodies 9-19 9.74. IDENTIFY: wood lead . II I =+ mV ρ = , where is the volume density and mA σ = , where is the area density. SET UP: For a solid sphere, 2 2 5 Im R = . For the hollow sphere (foil), 2 2 3 R = . For a sphere, 3 4 3 VR π = and 2 4 AR = . 3 ww 4 3 R ρρ == . 2 LL 4 R σπ . EXECUTE: 22 3 2 2 2 4 w wL w L L 222 4 2 8 (4 ) 535 3 3 3 5 R R m R R R R R R πσ ⎛⎞ =+= + = + ⎜⎟ ⎝⎠ . 3 42 2 8 (800 kg m )(0.20 m) (0.20 m) 20 kg m 0.70 kg m 35 I ⎡⎤ = ⎢⎥ ⎣⎦ . EVALUATE: W 26.8 kg m = and 2 W 0.429 kg m I =⋅ . L 10.1 kg m = and 2 L 0.268 kg m I = . Even though the foil is only 27% of the total mass its contribution to I is about 38% of the total.
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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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296_PartUniversity Physics Solution - 9-18 Chapter 9 (b) mg...

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