Let's solve the problem step by step. --- ## **Given Data** - Length of rod, \( L = 1 \) m - Mass of rod, \( m = 1 \) kg - Force applied at \( B \), \( P = 1 \) kN \( = 100 \) N - Rod is supported by pin at \( C \), located \( \frac{L}{3} \) from \( A \) - Rod is at rest at the instant shown. --- ## **Step 1: Free Body Diagram & Moments** Let's define: - \( A \): Top of rod - \( B \): Bottom of rod (where force is applied) - \( C \): Pin support (\( \frac{L}{3} \) from \( A \), \( \frac{2L}{3} \) from \( B \)) **Forces acting:** - Weight \( mg \) acting at the center (midpoint, \( \frac{L}{2} \) from \( A \)) - Applied force \( P \) at \( B \) (horizontal) - Reactions at \( C \): \( C_x \), \( C_y \) --- ## **(a) Angular Acceleration of the Rod (\( \alpha_{AB} \))** ### **Moment of Inertia about C** For a rod of length \( L \) and mass \( m \), the moment of inertia about a point \( d \) from the center is: \[ I = I_{\text{center}} + m d^2 = \frac{1}{12} mL^2 + m d^2 \] Here, \( d = | \text{distance from center to C} | = \left| \frac{L}{2} - \frac{L}{3} \right| = \frac{L}{6} \) So, \[ I_C = \frac{1}{12} mL^2 + m \left( \frac{L}{6} \right)^2 = \frac{1}{12} mL^2 + \frac{1}{36} mL^2 = \frac{1}{9} mL^2 \] ### **Moment Equation about C** Sum of moments about \( C \): \[ \sum M_C = I_C \alpha \] Moments due to: - \( P \) at \( B \): Perpendicular distance from \( C \) to \( B \) is \( \frac{2L}{3} \) - Weight passes through the center, which is at a distance \( \frac{L}{6} \) from \( C \) \[ P \cdot \frac{2L}{3} - mg \cdot \frac{L}{6} = I_C \alpha \] Plug in values: - \( P = 100 \) N - \( m = 1 \) kg - \( g = 9.81 \) m/s² - \( L = 1 \) m - \( I_C = \frac{1}{9} \) kg·m² \[ 100 \cdot \frac{2 \times 1}{3} - 1 \cdot 9.81 \cdot \frac{1}{6} = \frac{1}{9} \alpha \] \[ = 100 \cdot \frac{2}{3} - 9.81 \cdot \frac{1}{6} = \frac{1}{9} \alpha \] \[ = 666.67 - 1.635 = \frac{1}{9} \alpha \] \[ = 665.035 = \frac{1}{9} \alpha \] \[ \alpha = 9 \times 665.035 = 5985.32 \ \text{rad/s}^2 \] **Final Answer:** \[ \boxed{\alpha_{AB} = 5985 \ \text{rad/s}^2} \] --- ## **(b) Components of the Reaction at \( C \) (\( C_x \), \( C_y \))** ### **Equations of Motion** 1. **Sum of Forces in \( x \):** \[ C_x - P = m a_{Gx} \] 2. **Sum of Forces in \( y \):** \[ C_y - mg = m a_{Gy} \] 3. **Acceleration of Center of Mass \( G \):** - \( a_{Gx} = \alpha \cdot r_{G/C} \) (horizontal because rotation is about \( C \)) - \( r_{G/C} = \frac{L}{2} - \frac{L}{3} = \frac{L}{6} \) - Acceleration is perpendicular to rod, so horizontal at this instant. \[ a_{Gx} = \alpha \cdot \frac{L}{6} \] \[ a_{Gx} = 5985 \cdot \frac{1}{6} = 997.5 \ \text{m/s}^2 \] So, \[ C_x - 100 = 1 \times 997.5 \] \[ C_x = 1997.5 \ \text{N} \] - \( a_{Gy} = \) (since angular acceleration is horizontal at instant shown) \[ C_y - 9.81 = 1 \times \] \[ C_y = 9.81 \ \text{N} \] **Final Answers:** \[ \boxed{C_x = 1997.5 \ \text{N}} \\ \boxed{C_y = 9.81 \ \text{N}} \] --- ## **(c) Acceleration of Centre of Gravity (\( a_G \)) and Point \( A \) (\( a_A \))** ### **Centre of Gravity (\( a_G \))** - \( a_{Gx} = \alpha \cdot r_{G/C} = 5985 \cdot \frac{1}{6} = 997.5 \) m/s² (right) - \( a_{Gy} = \) \[ \boxed{a_G = 997.5 \ \text{m/s}^2 \text{ (right)}} \] --- ### **Point \( A \) (\( a_A \))** - Distance from \( C \) to \( A \) is \( -\frac{L}{3} \) (upwards from \( C \)) - Acceleration of \( A \) relative to \( C \) is \( a_{A/C} = \alpha \cdot r_{A/C} \) (perpendicular to rod, i.e., horizontal at this instant) So, \[ a_{A, x} = \alpha \cdot \left( -\frac{L}{3} \right) = 5985 \cdot \left( -\frac{1}{3} \right) = -1995 \ \text{m/s}^2 \] - (Negative means to the left) \[ \boxed{a_A = 1995 \ \text{m/s}^2 \text{ (left)}} \] --- ## **Summary Table** | Quantity | Value | Direction | |------------------------|-------------|--------------| | Angular acceleration (\( \alpha_{AB} \)) | 5985 rad/s² | - | | Reaction at C, \( C_x \) | 1997.5 N | Right | | Reaction at C, \( C_y \) | 9.81 N | Up | | Acceleration of G, \( a_G \) | 997.5 m/s² | Right | | Acceleration of A, \( a_A \) | 1995 m/s² | Left | --- ### **Let me know if you need further clarification or a diagram!**