296_PartUniversity Physics Solution

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

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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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