Let's break down the problem step by step: --- **Given:** - Mass of bullet (\( m_b \)): 1 oz = 1/16 lb (since 16 oz = 1 lb) - Mass of beam (\( m \)): 18 lb - Length of beam (\( L \)): 5.8 ft - The beam is initially at rest and pivots about point \( A \). - The bullet is fired horizontally with velocity \( v_0 \) and embeds itself at the tip \( B \). - The maximum angle of rotation of the beam is \( 90^\circ \). We are to **find the velocity \( v_0 \) of the bullet** such that the beam (with the bullet embedded at \( B \)) swings up to a vertical position (\( 90^\circ \) from its initial position). --- ### **Step 1: Conservation of Linear Momentum (Immediately After Impact)** Since the bullet embeds itself in the beam, this is a perfectly inelastic collision. We consider conservation of angular momentum about point \( A \) at the moment of impact (since external torques about \( A \) are zero during the collision): - **Initial angular momentum (about \( A \))**: Only the bullet has angular momentum before impact. \[ L_{initial} = m_b v_0 L \] - **Final angular momentum (about \( A \))**: The beam and bullet rotate together with angular speed \( \omega \) just after impact. \[ L_{final} = I_{total} \omega \] where \( I_{total} = I_{beam} + I_{bullet} \). #### **Moment of inertia calculations:** - Beam (about end \( A \)): \[ I_{beam} = \frac{1}{3} m L^2 \] - Bullet (as a point mass at \( B \)): \[ I_{bullet} = m_b L^2 \] So, \[ I_{total} = \frac{1}{3} m L^2 + m_b L^2 = \left( \frac{1}{3} m + m_b \right) L^2 \] #### **Equating angular momenta:** \[ m_b v_0 L = \left( \frac{1}{3} m + m_b \right) L^2 \omega \] \[ \omega = \frac{m_b v_0}{\left( \frac{1}{3} m + m_b \right) L} \] --- ### **Step 2: Conservation of Energy (Swing to 90°)** After the collision, the system (beam + bullet) swings up to \( 90^\circ \) (vertical). - **Initial kinetic energy (just after collision):** \[ KE = \frac{1}{2} I_{total} \omega^2 \] - **Potential energy gain (from horizontal to vertical):** - Center of mass of beam rises by \( \Delta y_{beam} = \frac{L}{2} \) - Bullet rises by \( \Delta y_{bullet} = L \) - Gravitational potential energy increase: \[ \Delta PE = m g \frac{L}{2} + m_b g L \] #### **Equate kinetic and potential energy:** \[ \frac{1}{2} I_{total} \omega^2 = m g \frac{L}{2} + m_b g L \] --- ### **Step 3: Substitute Values and Solve for \( v_0 \)** Plug in \( \omega \) from Step 1: \[ \frac{1}{2} I_{total} \left( \frac{m_b v_0}{I_{total} L} \right)^2 = m g \frac{L}{2} + m_b g L \] \[ \frac{1}{2} I_{total} \cdot \frac{m_b^2 v_0^2}{I_{total}^2 L^2} = m g \frac{L}{2} + m_b g L \] \[ \frac{1}{2} \frac{m_b^2 v_0^2}{I_{total} L^2} = m g \frac{L}{2} + m_b g L \] \[ \frac{m_b^2 v_0^2}{2 I_{total} L^2} = m g \frac{L}{2} + m_b g L \] \[ v_0^2 = \frac{2 I_{total} L^2}{m_b^2} \left( m g \frac{L}{2} + m_b g L \right) \] Now plug in the values: - \( m_b = \frac{1}{16} \) lb \( = \frac{1}{16} / 32.2 = 0.00194 \) slugs - \( m = 18 \) lb \( = 18 / 32.2 = 0.559 \) slugs - \( L = 5.8 \) ft - \( g = 32.2 \) ft/s² Calculate \( I_{total} \): \[ I_{total} = \frac{1}{3} m L^2 + m_b L^2 \] \[ = \frac{1}{3} \times 0.559 \times (5.8)^2 + 0.00194 \times (5.8)^2 \] \[ = 0.186 \times 33.64 + 0.00194 \times 33.64 \] \[ = 6.26 + 0.0652 = 6.33 \text{ slug ft}^2 \] Calculate the potential energy term: \[ m g \frac{L}{2} + m_b g L = 0.559 \times 32.2 \times \frac{5.8}{2} + 0.00194 \times 32.2 \times 5.8 \] \[ = 0.559 \times 32.2 \times 2.9 + 0.00194 \times 32.2 \times 5.8 \] \[ = 0.559 \times 93.38 + 0.00194 \times 187 \] \[ = 52.23 + 0.363 = 52.593 \text{ ft-lb} \] Plug all into the equation: \[ v_0^2 = \frac{2 \times 6.33 \times (5.8)^2}{(0.00194)^2} \times 52.593 \] Calculate \( 2 \times 6.33 \times 33.64 = 426.11 \) \[ v_0^2 = \frac{426.11}{(0.00194)^2} \times 52.593 \] \[ (0.00194)^2 = 0.00000376 \] \[ \frac{426.11}{0.00000376} = 113,354,787 \] \[ v_0^2 = 113,354,787 \times 52.593 = 5,958,618,183 \] \[ v_0 = \sqrt{5,958,618,183} \approx 77,240 \text{ ft/s} \] --- ### **Final Answer** \[ \boxed{v_0 \approx 77,200 \text{ ft/s}} \] --- ### **Summary of Steps** 1. **Conservation of Angular Momentum** to relate \( v_0 \) and \( \omega \). 2. **Conservation of Energy** to relate \( \omega \) to the energy required to swing the beam and bullet to the vertical. 3. **Algebraic Manipulation** to solve for \( v_0 \). 4. **Plug in Numbers** and calculate the final answer. If you need any step explained in more detail or want the answer in other units, let me know!