1. **Problem statement:** A uniform slender rod of length 3 m and mass 18 kg is pinned at one end and released from rest at 30° below the horizontal. We need to find the initial angular acceleration and the pin reactions. 2. **Known values:** - Length, $L = 3$ m - Mass, $m = 18$ kg - Initial angle, $\theta = 30^\circ$ below horizontal - Gravity, $g = 9.8$ m/s$^2$ 3. **Step 1: Calculate the moment of inertia $I$ of the rod about the pinned end.** For a uniform rod pinned at one end, the moment of inertia is: $$I = \frac{1}{3} m L^2$$ Substitute values: $$I = \frac{1}{3} \times 18 \times 3^2 = \frac{1}{3} \times 18 \times 9 = 54 \text{ kg m}^2$$ 4. **Step 2: Calculate the torque $\tau$ due to gravity about the pivot.** The weight acts at the center of mass, which is at $\frac{L}{2}$ from the pivot. The component of weight causing torque is $mg \sin(\theta)$ because the rod is at $30^\circ$ below horizontal. $$\tau = mg \sin(\theta) \times \frac{L}{2}$$ Substitute values: $$\tau = 18 \times 9.8 \times \sin(30^\circ) \times \frac{3}{2}$$ Since $\sin(30^\circ) = 0.5$: $$\tau = 18 \times 9.8 \times 0.5 \times 1.5 = 132.3 \text{ Nm}$$ 5. **Step 3: Calculate the initial angular acceleration $\alpha$.** Using Newton's second law for rotation: $$\tau = I \alpha$$ Solve for $\alpha$: $$\alpha = \frac{\tau}{I} = \frac{132.3}{54} = 2.45 \text{ rad/s}^2$$ 6. **Step 4: Calculate the pin reactions.** The pin reactions have horizontal ($H$) and vertical ($V$) components. - Vertical forces: $V - mg = 0 \Rightarrow V = mg = 18 \times 9.8 = 176.4$ N - Horizontal forces: The rod is initially at rest horizontally, so $H = 0$ N 7. **Final answers:** - Initial angular acceleration: $\boxed{2.45 \text{ rad/s}^2}$ - Pin reactions: Horizontal $H = 0$ N, Vertical $V = 176.4$ N