Let's break down the problem and solve it step by step! ### **Step 1: Understanding the System** - **Bar:** Mass = 12 kg, length = 0.51 m (from the figure) - **Disc:** Mass = 4 kg, radius = 0.085 m, attached at the end of the bar - **Spring:** Stiffness \( k = 1520 \) N/m, attached at distance 0.34 m from pivot - **Assume:** Oscillation is about the pivot (left end). Ignore the 3 kg mass because it's not connected to the bar (according to the figure). ### **Step 2: Calculating the Equivalent Mass Moment of Inertia (\( I \))** #### **For the Bar (about the pivot):** A uniform bar pivoted at one end: \[ I_{\text{bar}} = \frac{1}{3} m L^2 \] - \( m = 12 \) kg - \( L = 0.51 \) m \[ I_{\text{bar}} = \frac{1}{3} \times 12 \times (0.51)^2 = 1.02 \text{ kg} \cdot \text{m}^2 \] #### **For the Disc (using parallel axis theorem):** \[ I_{\text{disc}} = I_{\text{center}} + m d^2 \] - \( I_{\text{center}} = \frac{1}{2} m r^2 \) - \( m = 4 \) kg, \( r = 0.085 \) m - \( d = 0.51 \) m (distance from pivot to disc center) \[ I_{\text{disc}} = \frac{1}{2} \times 4 \times (0.085)^2 + 4 \times (0.51)^2 \] \[ = 2 \times 0.007225 + 4 \times 0.2601 \] \[ = 0.01445 + 1.0404 \] \[ = 1.05485 \text{ kg} \cdot \text{m}^2 \] #### **Total Moment of Inertia:** \[ I_{\text{total}} = I_{\text{bar}} + I_{\text{disc}} = 1.02 + 1.05485 = 2.07485 \text{ kg} \cdot \text{m}^2 \] --- ### **Step 3: Calculating the Restoring Torque by the Spring** The spring is attached at 0.34 m from the pivot. For a small angular displacement \( \theta \): - Vertical displacement at spring = \( x = 0.34 \theta \) - Restoring force by spring = \( F = k x = k \cdot 0.34 \theta \) - Restoring torque = \( F \cdot 0.34 = k \cdot (0.34)^2 \theta \) So, \[ \text{Restoring torque} = k a^2 \theta \] where \( a = 0.34 \) m --- ### **Step 4: Setting Up the Equation of Motion** \[ I \cdot \ddot{\theta} + k a^2 \theta = 0 \] This is a standard form for angular simple harmonic motion, so: \[ \omega_n = \sqrt{\frac{k a^2}{I}} \] where: - \( k = 1520 \) N/m - \( a = 0.34 \) m - \( I = 2.07485 \) kg·m² --- ### **Step 5: Calculating the Natural Frequency** \[ \omega_n = \sqrt{ \frac{1520 \times (0.34)^2}{2.07485} } \] \[ = \sqrt{ \frac{1520 \times 0.1156}{2.07485} } \] \[ = \sqrt{ \frac{175.712}{2.07485} } \] \[ = \sqrt{84.75} \] \[ = 9.21 \text{ rad/s} \] **Natural frequency in Hz:** \[ f_n = \frac{\omega_n}{2\pi} = \frac{9.21}{2\pi} \approx 1.47 \text{ Hz} \] --- ## **Summary/Final Answer** **The natural frequency of free vibration of the system is approximately:** \[ \boxed{1.47 \text{ Hz}} \] If you have any step you don't understand, please ask for clarification!