Conservation of Angular Momentum (Recovered).docx

M 2 s 000503 kg m 2 s 000496 kg m 2 s 2 14

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m 2 s ( 0.00503 kg m 2 s + 0.00496 kg m 2 s )/ 2 =1.4% Calculating Change in Angular Velocity due to Friction Δω= Δt*α; 0.342s*-0.4842 rad/s 2 =-0.2034 rad/s Calculating Change in Angular Momentum due to Friction ΔL= Δω* I 1 ; -0.2034 rad/s*1.0456E-4 kg*m 2 =-2.127E-5 kg m 2 s Data Table: No data tables were used during this lab. Graphs: Graph 1: Graph 2: Graph 3:
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Graph 4: Graph 5: Analysis/Questions: 1. Moment of Inertia for a Disk: I= 1 2 M R 2 ; Mass and Radius of 1 st Aluminum Disk: 105.6g, 8.9cm o Moment of Inertia: I 1 = 0.0445 m ¿ ¿ 1 2 0.1056 kg ¿ = 1.0456E-4 kg*m 2 Mass and Radius of 2 nd Aluminum Disk: 106.5g, 8.9 cm
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o Moment of Inertia: I 2 = 1 2 0.1065 kg ∗( 0.0445 m ) 2 =1.054E-4 kg*m 2 Moment of Inertia for Cylindrical Tube: I= 1 2 M ( R 1 2 + R 2 2 ) Mass and Radii of Steel Disk: 265.1g, R 1 =4.35cm, R 2 =1.4cm o Moment of Inertia: I 3 = .0435 m 2 1 2 .2651 kg ¿ + .014 m 2 ¿ =2.768E-4 kg*m 2 2. The rate of change of angular velocity for the 1 st aluminum disk was -0.5113 rad/s 2 . The rate of change of angular velocity for the Steel Disk was -0.3255 rad/s 2 . The difference in the angular acceleration was due to the differing moments of inertia. Because the moment of inertia of the steel disk, 2.768E-4 kg*m 2 , was higher than the moment of inertia of the 1 st aluminum disk, 1.0456E-4 kg*m 2 , the angular acceleration had to be lower in order to compensate, following an inversely proportional relationship, T=I α. As a result, the angular acceleration of the steel disk was lower than the angular acceleration of the 1 st aluminum disk. Run 2: 3. ω 1 = 47.43 rad/s; ω 2 = 23.97 rad/s; Δt=.42s 4. Angular Momentum before: L= ω 1 *I 1 ; L= 0.0049 6 kg m 2 s Angular Momentum after: L= ω 2 *(I 1 + I 2 )= 0.00 503 kg m 2 s Percent Difference: 0.00503 kg m 2 s 0.00496 kg m 2 s ( 0.00503 kg m 2 s + 0.00496 kg m 2 s )/ 2 =1.4% 5. α =-0.4842 rad/s 2 Δω= Δt*α= -0.2034 rad/s ΔL= Δω* I 1 = -2.127 E-5 kg m 2 s The change in momentum due to friction, and not to the dropping of the 2 nd aluminum disk should be -2.127E-5 kg m 2 s . However, because our angular momentum after was higher than the angular momentum before dropping the 2 nd aluminum disk (This was most likely to an accidental torque placed on the 2 nd aluminum disk when dropping the 2 nd aluminum disk.), our results for the portion of difference in angular momentum that
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