Let's solve for the **accessory moment of inertia \( I_a \)** and the **percent difference** for the cylindrical ring and disk, as asked. --- ## Step 1: Calculate Theoretical \( I_a \) ### **For the Cylindrical Ring** Theoretical moment of inertia for a cylindrical ring: \[ I_{\text{ring}} = \frac{1}{2} M (R_1^2 + R_2^2) \] - \( M = 4.234 \) kg - \( R_1 = .11 \) m - \( R_2 = .124 \) m \[ I_{\text{ring}} = \frac{1}{2} \times 4.234 \times \left[ (.11)^2 + (.124)^2 \right] \] \[ I_{\text{ring}} = 2.117 \times (.0121 + .015376) \] \[ I_{\text{ring}} = 2.117 \times .027476 = .0581 \text{ kg·m}^2 \] **Explanation:** We used the formula for a thick ring, plugging in the mass and the two radii. --- ### **For the Disk** Theoretical moment of inertia for a solid disk: \[ I_{\text{disk}} = \frac{1}{2} M R^2 \] - \( M = 4.730 \) kg - \( R = .125 \) m \[ I_{\text{disk}} = \frac{1}{2} \times 4.730 \times (.125)^2 \] \[ I_{\text{disk}} = 2.365 \times .015625 = .0369 \text{ kg·m}^2 \] **Explanation:** We used the disk formula, inputting the mass and radius. --- ## Step 2: Calculate Accessory \( I_a \) and Percent Difference ### **Accessory \( I_a \):** \[ I_a = I_{\text{system}} - I_{\text{support}} \] From the table: - \( I_{\text{support (only)}} = 1.49 \times 10^{-4} \) kg·m² - \( I_{\text{system, cyl. ring}} = 4.32 \times 10^{-4} \) kg·m² - \( I_{\text{system, disk}} = 8.2 \times 10^{-4} \) kg·m² #### **Cylindrical Ring:** \[ I_a = 4.32 \times 10^{-4} - 1.49 \times 10^{-4} = 2.83 \times 10^{-4} \text{ kg·m}^2 \] #### **Disk:** \[ I_a = 8.2 \times 10^{-4} - 1.49 \times 10^{-4} = 6.71 \times 10^{-4} \text{ kg·m}^2 \] **Explanation:** We subtracted the experimental inertia of the empty support from the system with accessory. --- ### **Percent Difference** \[ \text{Percent Difference} = \frac{|\text{Experimental} - \text{Theoretical}|}{\text{Theoretical}} \times 100\% \] #### **Cylindrical Ring:** \[ \text{Percent Difference} = \frac{|2.83 \times 10^{-4} - .0581|}{.0581} \times 100\% \approx 99.5\% \] #### **Disk:** \[ \text{Percent Difference} = \frac{|6.71 \times 10^{-4} - .0369|}{.0369} \times 100\% \approx 98.2\% \] **Explanation:** The large percent differences indicate the experimental values are much lower than theoretical, possibly due to significant friction, measurement, or calculation errors. --- ## **Final Answers** - **Cylindrical Ring:** - Theoretical \( I_a = .0581 \) kg·m² - Experimental \( I_a = 2.83 \times 10^{-4} \) kg·m² - Percent Difference \( \approx 99.5\% \) - **Disk:** - Theoretical \( I_a = .0369 \) kg·m² - Experimental \( I_a = 6.71 \times 10^{-4} \) kg·m² - Percent Difference \( \approx 98.2\% \)