Experiment 15 - tube Table II Energy(J KE t 0.007 J KE s...

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Experiment 15: Moment of Inertia – Rotational and Translational Kinetic Energy Question: Derive the above statement that E t /E s = 0.71 KE s = KE t + KE r = (1/2)mv 2 + (1/2)Iω 2 Using I = (2/5)mr 2 and ω = v/r, KE s = KE t + KE r = (1/2)mv 2 + (1/2)[(2/5)mr 2 ](v/r) 2 = (1/2)mv 2 + (1/2)[(2/5)mv 2 ] = (1/2)mv 2 + (2/10)mv 2 = (5/10)mv 2 + (2/10)mv 2 = (7/5)(1/2)mv 2 E s = (7/5)E t E t /E s = (5/7) = 0.71 Table I shows the x displacement range traveled by the ball in projectile motion, the average of the values, the measured value of y, the exit muzzle velocity, and the measured value of h. Table I Trial X displmt (m) 1 0.55 m 2 0.53 m 3 0.61 m 4 0.63 m 5 0.66 m 6 0.59 m xav 0.60 m y(m) 1.104 m velocity (m/s) 1.26 m/s h(m) 0.245 m Calculations: xav = (0.55 m + 0.53 m + 0.61 m + 0.63 m + 0.66 m + 0.59 m)/6 = 0.595 m
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v 0 = (4.9/y) 1/2 x v 0 = [4.9/(1.104 m)] 1/2 (0.6 m) = 1.26405 Table II shows the calculated value of the translational kinetic energy of the exiting sphere and the calculated value of the total kinetic energy of the sphere upon exiting the
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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 ) 6.30e-8 kgm 2 8.32e-8 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 = 8.32e-8 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 = 6.30e-8 % Error = |8.32e-8- 6.30e-8 |/8.32e-8 x 100 = 24.3%...
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