- Part 2 Long Axis Small Radius 0 39 6 14 Rad Slope Of W Part 3 Medium Radius 6 89 0 07 Rad S Slope Of W 1 S 1 (127.34 KiB) Viewed 15 times
- Part 2 Long Axis Small Radius 0 39 6 14 Rad Slope Of W Part 3 Medium Radius 6 89 0 07 Rad S Slope Of W 1 S 2 (149.34 KiB) Viewed 15 times
- Part 2 Long Axis Small Radius 0 39 6 14 Rad Slope Of W Part 3 Medium Radius 6 89 0 07 Rad S Slope Of W 1 S 3 (112.63 KiB) Viewed 15 times
Part #2 long axis, small radius 0 39 * 6.14 rad/ Slope of w() Part #3 medium radius 6.89 + 0,07 / rad/s Slope of w(1) (short axis) OX 37.4 + Slope of w() (long axis) 9. 2 rad/s2 ANALYSIS 1. Use Eq. (3) and Eq. (12) to calculate the static moment of inertia I, and its corresponding uncertainty AT, for both the short and long axis configurations, Is = arta ) = 4431 12 m 글, (a +2,2)=442! (2.55'+3.754) 3 4 15 = room (a. Aa4 BAG)- (1.2x0.5+1.5x0.5) 44%. 6 huzh (6.56 +19.06 759.22 12 - 99.69 2. Use Eq. (6) and Eq. (13) to calculate the dynamic moment of inertia Id and its corresponding uncertainty Ald for all parts (4 values total). Assume Ar=0. es Toamn (96) - 20 X 0.48 9.81 X0.48-5.735 0.029 5.73+ 0.029 = 9.6 4.70 -5.73 I 6.029 5.73 I.G.029. AI a = ng or mor Axtm a= (21 Aw) X + 1) 20x9.8140.48 5.72 x 0.029 a.476 2
DATA Steel block Hanging mass Small pulley Medium pulley M = 113. 8 m = re = 0.48 cm 20 I'm = 1.43 cm Steel block dimensions Smallest → Largest Measurement error Aa= Ab= AC = cm 65 cm C= cm Δα = 0.5 cm Part #1 - short axis, small radius Slope of angular velocity graph wat) Q= 5.73 +0.029 rad/s? Initial and final angular position O;= Og=121.56 Initial and final angular velocity Wi= O wg= 36.94 W Calculate AK, AU, and AE, and use the value of AE to determine the average torque due to friction (see steps #5 & #6 in the manual). 2 - 4 K = 1 lI atent") Ewo? _ wina 4 (759 .275208 트를 0.4. (36.94) 20
Discussion & Conclusion Discuss your results with your group members and list the major points in your discussion below. 3