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Unformatted text preview: tube. Table II Energy (J) KE t 0.007 J KE s 0.009 J Calculations: KE t = (1/2)mv 2 KE t = (1/2)(0.00832 kg)(1.26 m/s) 2 = 0.006604 KE s = (7/5)E t KE s = (7/5)(0.007 J) = 0.009246 J Table III shows percent contribution of translational kinetic energy to the total kinetic energy, and the moment of inertia of the sphere compared with theory. Table III Results Expt. Theory % Error KE t /KE s 0.78 0.71 9.9% I(kgm 2 ) 6.30e8 kgm 2 8.32e8 kgm 2 24.3% Calculations: KE t /KE s = (0.007 J)/(0.009 J) = 0.78 Actual – Experimental/Actual x 100 = % Error 0.71 – 0.78/0.71 x 100 = 9.9% Theory: I = (2/5)mr 2 I = (2/5)(0.00832 kg)(0.005 m) 2 = 8.32e8 kgm 2 Expt: KE s = KE t + KE r 0.009 J = 0.007 J + KE r KE r = 0.002 J KE r = (1/2)Iω 2 , ω = v/r 0.002 J = (1/2)I[(1.26 m/s)/(0.005 m)] 2 I = 6.30e8 % Error = 8.32e8 6.30e8 /8.32e8 x 100 = 24.3%...
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 Spring '06
 Dr.Orehotsky
 Physics, Energy, Inertia, Kinetic Energy, Moment Of Inertia, 0.6 m, translational kinetic energy, 0.009246 J

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