Let's solve the problem **step by step**: ### **Step 1: List Known Values** - \( L = 1.3 \) ft (length of rod AB) - \( R = 0.1 \) ft (radius of disk) - \( P = 6 \) lb (weight of disk) - \( P_1 = 3 \) lb (weight of rod) - \( \theta = 0.6 \) rad - \( \dot{\theta} = -0.5 \) rad/s (angular velocity of rod) We are to find **angular acceleration** (\( \ddot{\theta} \)) of the bar. --- ### **Step 2: Free Body Diagram and Moments** #### **Moments About A** The sum of moments about A for the rod and disk system: \[ \sum M_A = I_{total} \, \ddot{\theta} \] --- ### **Step 3: Find Masses** - Weight (\( W \)) = mass (\( m \)) × gravity (\( g \)) - \( g = 32.2 \) ft/s² \[ m_{\text{disk}} = \frac{6}{32.2} = 0.186 \text{ slugs} \] \[ m_{\text{rod}} = \frac{3}{32.2} = 0.093 \text{ slugs} \] --- ### **Step 4: Moments of Inertia** - **Disk about A (using parallel axis theorem):** \[ I_{\text{disk, center}} = \frac{1}{2} m R^2 \] \[ d = R \rightarrow I_{\text{disk, A}} = I_{\text{disk, center}} + m R^2 = \frac{1}{2} m R^2 + m R^2 = \frac{3}{2} m R^2 \] - **Rod about A:** \[ I_{\text{rod, A}} = \frac{1}{3} m L^2 \] --- ### **Step 5: Forces and Moments** #### **Gravity Forces** - Disk: \( P = 6 \) lb, acts vertically downward at the center. - Rod: \( P_1 = 3 \) lb, acts vertically downward at its center (\( L/2 \) from A). #### **Moment Arms** - Rod: Distance from A to center of gravity is \( L/2 \). - Disk: Distance from A to center is \( R \). Both angles from vertical are \(\theta\). - Moment due to rod: \( P_1 (L/2) \sin\theta \) - Moment due to disk: \( P R \sin\theta \) Both moments tend to rotate the bar **clockwise** (negative direction if \(\theta\) is measured from horizontal). --- ### **Step 6: Write Moment Equation** \[ \sum M_A = -P_1 \left(\frac{L}{2}\right) \sin\theta - P R \sin\theta \] Set equal to total moment of inertia times angular acceleration: \[ I_{total} \ddot{\theta} = -P_1 \left(\frac{L}{2}\right) \sin\theta - P R \sin\theta \] \[ \ddot{\theta} = \frac{ -P_1 \left(\frac{L}{2}\right) \sin\theta - P R \sin\theta }{ I_{total} } \] --- ### **Step 7: Calculate \( I_{total} \) (in slug·ft²)** - Disk: \[ I_{\text{disk, A}} = \frac{3}{2} m_{\text{disk}} R^2 = \frac{3}{2} \times 0.186 \times (0.1)^2 = \frac{3}{2} \times 0.186 \times 0.01 = 0.00279 \text{ slug·ft}^2 \] - Rod: \[ I_{\text{rod, A}} = \frac{1}{3} m_{\text{rod}} L^2 = \frac{1}{3} \times 0.093 \times (1.3)^2 = 0.031 \times 1.69 = 0.0524 \text{ slug·ft}^2 \] - Total: \[ I_{total} = 0.00279 + 0.0524 = 0.05519 \text{ slug·ft}^2 \] --- ### **Step 8: Plug In Values** \[ \sin(0.6) \approx 0.5646 \] \[ \ddot{\theta} = \frac{ -3 \left(\frac{1.3}{2}\right) \times 0.5646 - 6 \times 0.1 \times 0.5646 }{ 0.05519 } \] Calculate numerators: - \( 3 \times 0.65 = 1.95 \) - \( 1.95 \times 0.5646 = 1.10097 \) - \( 6 \times 0.1 = 0.6 \) - \( 0.6 \times 0.5646 = 0.33876 \) - Total numerator: \(-1.10097 - 0.33876 = -1.43973\) \[ \ddot{\theta} = \frac{-1.43973}{0.05519} = -26.08 \text{ rad/s}^2 \] --- ### **Final Answer** \[ \boxed{-26.1 \text{ rad/s}^2} \] **(rounded to 1 decimal place)** --- **Summary of Steps:** 1. Found moments due to weights. 2. Calculated moments of inertia. 3. Applied moment equation. 4. Plugged in values and solved for angular acceleration. Let me know if you need a more detailed breakdown of each step!