This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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%...
View
Full
Document
This note was uploaded on 04/30/2008 for the course PHY 201 taught by Professor Dr.orehotsky during the Spring '06 term at Wilkes.
 Spring '06
 Dr.Orehotsky
 Physics, Energy, Inertia, Kinetic Energy

Click to edit the document details