Let's break down the problem: ### **Given Data** - Force, \( P = 75 \) N - Mass of rod, \( m = 4 \) kg - Length of rod, \( L = 450 \) mm \( = .45 \) m - The force is applied perpendicular at point B (the end of the rod). - Hinge at C (other end). ### **Step 1: Calculate Moment of Inertia (I)** For a rod rotating about one end: \[ I = \frac{1}{3}mL^2 \] \[ I = \frac{1}{3} \times 4 \times (.45)^2 \] \[ I = \frac{1}{3} \times 4 \times .2025 \] \[ I = \frac{1}{3} \times .81 = .27 \ \text{kg} \cdot \text{m}^2 \] ### **Step 2: Calculate Torque (\(\tau\))** Torque, \( \tau = F \times r \), where \( r = L \) (since force is applied at the end): \[ \tau = 75 \times .45 = 33.75 \ \text{Nm} \] ### **Step 3: Calculate Angular Acceleration (\(\alpha\))** \[ \tau = I\alpha \implies \alpha = \frac{\tau}{I} \] \[ \alpha = \frac{33.75}{.27} \] \[ \alpha = 125 \ \text{rad/s}^2 \] ### **Step 4: Match With Options** Looking at your options: - \( \alpha = 44 \) rad/s² - \( \alpha = 107 \) rad/s² - \( \alpha = 90 \) rad/s² - \( \alpha = 120 \) rad/s² **Closest correct answer is:** \[ \boxed{\alpha = 120 \ \text{rad/s}^2} \] --- **Final Answer:** > **The correct option is** \(\boxed{\alpha = 120 \ \text{rad/s}^2}\)