Let's break down the problem step by step: ### **Given Data** - **Cube length \( l \) = 0.5 m** - **Cube mass \( m_c \) = 10 kg** - **Rod mass \( m_r \) = 3 kg** - **Rod length \( L \) = 3 m** - **Bullet mass \( m_b \) = 10 g = 0.01 kg** - **Bullet velocity \( v_b \) = 850 m/s** - **Bullet embeds into the center of the cube** ### **Step 1: Find the Moment of Inertia about the Pivot** **Total system = rod + cube + bullet** #### **Moment of inertia of rod about one end:** \[ I_{rod} = \frac{1}{3} m_r L^2 \] \[ I_{rod} = \frac{1}{3} \times 3 \times (3^2) = 9 \text{ kg m}^2 \] #### **Moment of inertia of cube (center at end):** \[ I_{cube} = m_c \times l^2 \] \[ I_{cube} = 10 \times (0.5)^2 = 2.5 \text{ kg m}^2 \] #### **Moment of inertia of bullet (embedded at center of cube):** \[ I_{bullet} = m_b \times l^2 \] \[ I_{bullet} = 0.01 \times (0.5)^2 = 0.0025 \text{ kg m}^2 \] #### **Total moment of inertia:** \[ I_{total} = I_{rod} + I_{cube} + I_{bullet} = 9 + 2.5 + 0.0025 = 11.5025 \text{ kg m}^2 \] --- ### **Step 2: Angular Momentum Conservation** - The bullet hits at a distance of \( 3 + 0.25 = 3.25 \) m from the pivot (since the center of the cube is 0.25 m from the end of the rod). \[ \text{Initial angular momentum} = m_b v_b r \] \[ = 0.01 \times 850 \times 3.25 = 27.625 \text{ kg m}^2/\text{s} \] After collision (all objects rotate together): \[ I_{total} \omega = \text{Initial angular momentum} \] \[ \omega = \frac{27.625}{11.5025} = 2.402 \text{ rad/s} \] --- ### **Step 3: Conservation of Energy** After the impact, the system swings up to a maximum height (all kinetic energy converts to potential energy at the maximum inclination). \[ \frac{1}{2} I_{total} \omega^2 = M_{total} g h \] where: - \( M_{total} = m_r + m_c + m_b = 3 + 10 + 0.01 = 13.01 \) kg - \( h \) is the vertical height gained by the center of mass. #### **Find center of mass (\( x_{cm} \)) from pivot:** \[ x_{cm} = \frac{m_r \times \frac{L}{2} + m_c \times (L + \frac{l}{2}) + m_b \times (L + \frac{l}{2})}{M_{total}} \] \[ = \frac{3 \times 1.5 + 10 \times 3.25 + 0.01 \times 3.25}{13.01} \] \[ = \frac{4.5 + 32.5 + 0.0325}{13.01} \] \[ = \frac{37.0325}{13.01} = 2.846 \text{ m} \] #### **Height gained:** \[ h = x_{cm} (1 - \cos \theta) \] --- ### **Step 4: Set up energy equation** \[ \frac{1}{2} I_{total} \omega^2 = M_{total} g x_{cm}(1 - \cos \theta) \] \[ \frac{1}{2} \times 11.5025 \times (2.402)^2 = 13.01 \times 9.81 \times 2.846 \times (1 - \cos \theta) \] \[ \frac{1}{2} \times 11.5025 \times 5.77 = 13.01 \times 9.81 \times 2.846 \times (1 - \cos \theta) \] \[ 33.24 = 363.39 \times (1 - \cos \theta) \] \[ 1 - \cos \theta = \frac{33.24}{363.39} = 0.0915 \] \[ \cos \theta = 1 - 0.0915 = 0.9085 \] \[ \theta = \cos^{-1}(0.9085) = 24.7^\circ \] --- ### **Step 5: Match with Options** The closest option is **27.16°**. --- ## **Final Answer** **The maximum inclination that the slender rod will swing due to the impact is:** \[ \boxed{27.16^\circ} \]