Moment of Inertia
Sadie D. Hood Lab 8: Moment of Inertia Partner: Florence Doval Due 16 November 2011 Aims: To use a centripetal force apparatus to calculate the moment of inertia of rotating weights, using theories derived from ideas of energy transfer (Im = MR2 (g/2h)(t2-t02)) and point mass appoximation (m1r12 + m2r22). Set Up
Procedure First we measured the weights of two masses and wingnuts that secure them. Then we placed one of the masses on the very end of a horizontal rod on the centripetal force apparatus, 0.162 m away from the centre of the rod, and the other mass 0.115 m away from the centre of the rod. Then we attached a 0.2 kg mass to the bottom of a string and wound the string around the vertical shaft of the apparatus, so that the bottom of the weight rose to the bottom edge of the tabletop the apparatus was on. We measured the distance from the bottom of the weight to the floor, and then let the weight fall to the floor, and measured the time it took to do so. We repeated this measurement, using the same initial height, four more times. Then we took the masses and the wingnuts off the horizontal rod and let the 0.2 kg mass fall in the same way as before, five times. Then we replaced the masses and the wingnuts, but put them both on the edges of the horizontal rod, and repeated the same falling mass measurements five times. We moved the masses in towards the centre of the rod and continued to repeat the falling mass measurements. We moved the masses in toward the centre four more times, each time taking five falling mass measurements. Data First Pass: m1 = 0.06158 kg; r1 = 0.162 m; m2 = 0.06158 kg; r2 = 0.115 m h = 0.876 m; R = 0.0064 m; M = 0.2 kg tavg(s) Δtavg(s) t0avg (s) Δt0avg(s) Im(energy transfer) (Js2) ΔIm (e.t.) (Js2) Im (point slope) ΔIm (p.m.)(Js2) (Js2) 8.436 0.34 4.236 0.106 0.0024 3.15 x 10-4 0.0024 3.41 x 10-6 -4 [(2x x 10-6/6.4 x 10-4) + (1 x 10ΔIm (e.t.) = Im [(2ΔR/R) + (Δh/h) + (2Δt/t) + (2Δt0/t0)] = 2.44 x 10 1 4/.0876) + (2x x 0.34/8.436) + (2x
Procedure First we measured the weights of two masses and wingnuts that secure them. Then we placed one of the masses on the very end of a horizontal rod on the centripetal force apparatus, 0.162 m away from the centre of the rod, and the other mass 0.115 m away from the centre of the rod. Then we attached a 0.2 kg mass to the bottom of a string and wound the string around the vertical shaft of the apparatus, so that the bottom of the weight rose to the bottom edge of the tabletop the apparatus was on. We measured the distance from the bottom of the weight to the floor, and then let the weight fall to the floor, and measured the time it took to do so. We repeated this measurement, using the same initial height, four more times. Then we took the masses and the wingnuts off the horizontal rod and let the 0.2 kg mass fall in the same way as before, five times. Then we replaced the masses and the wingnuts, but put them both on the edges of the horizontal rod, and repeated the same falling mass measurements five times. We moved the masses in towards the centre of the rod and continued to repeat the falling mass measurements. We moved the masses in toward the centre four more times, each time taking five falling mass measurements. Data First Pass: m1 = 0.06158 kg; r1 = 0.162 m; m2 = 0.06158 kg; r2 = 0.115 m h = 0.876 m; R = 0.0064 m; M = 0.2 kg tavg(s) Δtavg(s) t0avg (s) Δt0avg(s) Im(energy transfer) (Js2) ΔIm (e.t.) (Js2) Im (point slope) ΔIm (p.m.)(Js2) (Js2) 8.436 0.34 4.236 0.106 0.0024 3.15 x 10-4 0.0024 3.41 x 10-6 -4 [(2x x 10-6/6.4 x 10-4) + (1 x 10ΔIm (e.t.) = Im [(2ΔR/R) + (Δh/h) + (2Δt/t) + (2Δt0/t0)] = 2.44 x 10 1 4/.0876) + (2x x 0.34/8.436) + (2x