Let's break down the problem and collect the given data: ### **Given Data:** - Mass of ball (\( m_B \)): \( 1.2\, \text{lb} \) - Mass of rod (\( m_{rod} \)): \( 8.7\, \text{lb} \) - Distance from fixed point A to impact point (\( d_1 \)): \( 1.0\, \text{ft} \) - Length from A to end of rod (\( d_2 \)): \( 2.1\, \text{ft} \) - Initial velocity of ball (\( v_{B1} \)): \( 35\, \text{ft/s} \) - Coefficient of restitution (\( e \)): \( 0.56 \) - The rod is thin and pivoted at A. ### **Required:** - **Angular velocity of the rod just after impact (\( \omega_2 \)).** --- ### **Step 1: Find the Moment of Inertia of the Rod about A** For a thin rod of mass \( m \) and length \( L \) pivoted at one end: - \( I_A = \frac{1}{3} m L^2 \) First, convert the mass from lb to slugs (since \( 1\, \text{lb} = 1/32.2\, \text{slugs} \)): - \( m_{rod} = \frac{8.7}{32.2} = 0.270\, \text{slugs} \) - \( L = d_2 = 2.1\, \text{ft} \) So: \[ I_A = \frac{1}{3} (0.270) (2.1)^2 = \frac{1}{3} \times 0.270 \times 4.41 = 0.397\, \text{slug-ft}^2 \] --- ### **Step 2: Conservation of Angular Momentum about Point A (Just Before and After Impact)** - Initial angular momentum about A (just before impact): \[ H_{A1} = m_B v_{B1} d_1 \] Convert \( m_B \) to slugs: \[ m_B = \frac{1.2}{32.2} = 0.0373\, \text{slugs} \] \[ H_{A1} = 0.0373 \times 35 \times 1.0 = 1.3055\, \text{slug-ft}^2/\text{s} \] - Just after impact, the angular momentum about A: \[ H_{A2} = m_B v_{B2} d_1 + I_A \omega_2 \] --- ### **Step 3: Coefficient of Restitution (e)** \[ e = \frac{\text{Relative velocity after impact}}{\text{Relative velocity before impact}} \] The relative velocity (at the impact point) after impact: \[ e = \frac{v_{B2} - v_{rod,2}}{v_{rod,1} - v_{B1}} \] But initial velocity of rod at impact point is \( 0 \): \[ e = \frac{v_{B2} - v_{rod,2}}{-v_{B1}} \] \[ v_{rod,2} = \omega_2 d_1 \] So: \[ e = \frac{v_{B2} - (\omega_2 d_1)}{-v_{B1}} \] \[ v_{B2} - \omega_2 d_1 = -e v_{B1} \] \[ v_{B2} = \omega_2 d_1 - e v_{B1} \] --- ### **Step 4: Substitute \( v_{B2} \) in the Angular Momentum Equation** \[ H_{A1} = m_B v_{B2} d_1 + I_A \omega_2 \] \[ 1.3055 = 0.0373 [(\omega_2 d_1 - e v_{B1})] d_1 + 0.397 \omega_2 \] Plug in values: \[ 1.3055 = 0.0373 [\omega_2 (1.0) - 0.56 \times 35] \times 1.0 + 0.397 \omega_2 \] \[ = 0.0373 \omega_2 - 0.0373 \times 0.56 \times 35 + 0.397 \omega_2 \] \[ = 0.0373 \omega_2 - 0.0373 \times 19.6 + 0.397 \omega_2 \] \[ = 0.0373 \omega_2 - 0.731 + 0.397 \omega_2 \] \[ = (0.0373 + 0.397) \omega_2 - 0.731 \] \[ = 0.4343 \omega_2 - 0.731 \] \[ 1.3055 = 0.4343 \omega_2 - 0.731 \] \[ 0.4343 \omega_2 = 1.3055 + 0.731 = 2.0365 \] \[ \omega_2 = \frac{2.0365}{0.4343} = 4.69 \, \text{rad/s} \] --- ## **Final Answer:** \[ \boxed{\omega_2 = 4.69 \, \text{rad/s}} \] This is the **angular velocity of the rod just after impact**.

**Step 1: Conceptual Introduction** When a moving ball strikes a pivoted thin rod, the impact causes the rod to start rotating. To analyze this, we use principles of conservation of angular momentum and the coefficient of restitution. The initial momentum of the ball transfers to the rod during impact, resulting in a change in the rod's angular velocity. The coefficient of restitution accounts for the elasticity of the collision, influencing the relative velocities after impact. By applying these principles, we can determine the angular velocity of the rod immediately after collision. --- **Step 2: Explanation of the Formulae Used** - **Angular Momentum about the Pivot Point (A):** Before impact, only the ball has linear momentum, which translates to angular momentum about point A: \[ H_{A1} = m_B v_{B1} d_1 \] After impact, both the ball and the rod contribute to the angular momentum: \[ H_{A2} = m_B v_{B2} d_1 + I_A \omega_2 \] where \( v_{B2} \) is the ball's velocity after impact, and \( \omega_2 \) is the rod's angular velocity after impact. - **Coefficient of Restitution:** It relates the relative velocities of the colliding bodies before and after impact: \[ e = \frac{\text{relative velocity after impact}}{\text{relative velocity before impact}} \] Since the rod is initially at rest at the point of impact: \[ e = \frac{v_{B2} - v_{rod,2}}{-v_{B1}} \] Rearranged to find the post-impact ball velocity: \[ v_{B2} = \omega_2 d_1 - e v_{B1} \] - **Moment of Inertia of the Rod:** For a thin rod pivoted at one end: \[ I_A = \frac{1}{3} m_{rod} L^2 \] --- **Step 3: Step-by-Step Calculation** 1. **Convert masses from pounds to slugs:** \[ m_{B} = \frac{1.2}{32.2} \approx 0.0373\, \text{slugs} \] \[ m_{rod} = \frac{8.7}{32.2} \approx 0.270\, \text{slugs} \] 2. **Calculate the initial angular momentum of the ball about point A:** \[ H_{A1} = m_B v_{B1} d_1 = 0.0373 \times 35 \times 1.0 = 1.3055\, \text{slug-ft}^2/\text{s} \] 3. **Compute the moment of inertia of the rod:** \[ I_A = \frac{1}{3} \times 0.270 \times (2.1)^2 = \frac{1}{3} \times 0.270 \times 4.41 \approx 0.397\, \text{slug-ft}^2 \] 4. **Express the post-impact velocities using the coefficient of restitution:** \[ v_{B2} = \omega_2 \times 1.0 - 0.56 \times 35 = \omega_2 - 19.6 \] 5. **Apply conservation of angular momentum about A:** \[ 1.3055 = m_B v_{B2} d_1 + I_A \omega_2 \] Substitute \( v_{B2} \): \[ 1.3055 = 0.0373 (\omega_2 - 19.6) \times 1.0 + 0.397 \omega_2 \] 6. **Simplify and solve for \( \omega_2 \):** \[ 1.3055 = 0.0373 \omega_2 - 0.0373 \times 19.6 + 0.397 \omega_2 \] \[ 1.3055 = 0.0373 \omega_2 - 0.731 + 0.397 \omega_2 \] \[ 1.3055 + 0.731 = (0.0373 + 0.397) \omega_2 \] \[ 2.0365 = 0.4343 \omega_2 \] \[ \omega_2 = \frac{2.0365}{0.4343} \approx 4.69\, \text{rad/s} \] --- **Final Answer:** \[ \boxed{\omega_2 \approx 4.69\, \text{rad/sec}} \] This is the angular velocity of the rod immediately after impact.

**Step 1: Conceptual Introduction** When a moving ball strikes a pivoted thin rod, the collision causes the rod to rotate. To analyze this, we rely on the principles of conservation of angular momentum about the pivot point and the coefficient of restitution, which measures the elasticity of the collision. The initial linear momentum of the ball is converted into angular momentum about the pivot during impact. The coefficient of restitution relates the relative velocities of the ball and rod at the point of impact before and after collision. By applying these principles, we can determine the rod's angular velocity immediately after impact. --- **Step 2: Explanation of the Formulae Used** - **Angular Momentum Conservation:** The total angular momentum about the pivot point before impact equals that after impact, considering the contributions from the ball's linear momentum and the rod's rotational inertia. \[ H_{A1} = H_{A2} \] where \[ H_{A1} = m_B v_{B1} d_1 \] (initial angular momentum of the ball) and \[ H_{A2} = m_B v_{B2} d_1 + I_A \omega_2 \] (total angular momentum after impact, with \(v_{B2}\) the ball's velocity after impact and \(\omega_2\) the rod's angular velocity). - **Coefficient of Restitution:** It relates the relative velocities of the colliding bodies before and after impact: \[ e = \frac{\text{relative velocity after}}{\text{relative velocity before}} \] Since the rod is initially at rest at the impact point, the relationship simplifies to: \[ v_{B2} = \omega_2 d_1 - e v_{B1} \] - **Moment of Inertia of the Rod:** For a thin rod pivoted at one end: \[ I_A = \frac{1}{3} m_{rod} L^2 \] --- **Step 3: Step-by-Step Calculation** **1. Convert masses from pounds to slugs:** \[ m_B = \frac{1.2\, \text{lb}}{32.2} \approx 0.0373\, \text{slugs} \] \[ m_{rod} = \frac{8.7\, \text{lb}}{32.2} \approx 0.270\, \text{slugs} \] **2. Calculate initial angular momentum of the ball about point A:** \[ H_{A1} = m_B v_{B1} d_1 = 0.0373 \times 35 \times 1.0 = 1.3055\, \text{slug-ft}^2/\text{s} \] **3. Compute the moment of inertia of the rod:** \[ I_A = \frac{1}{3} \times 0.270 \times (2.1)^2 = \frac{1}{3} \times 0.270 \times 4.41 \approx 0.397\, \text{slug-ft}^2 \] **4. Express the post-impact velocities using the coefficient of restitution:** \[ v_{B2} = \omega_2 \times 1.0 - 0.56 \times 35 = \omega_2 - 19.6 \] **5. Apply conservation of angular momentum about A:** \[ 1.3055 = m_B v_{B2} d_1 + I_A \omega_2 \] Substitute \( v_{B2} \): \[ 1.3055 = 0.0373 (\omega_2 - 19.6) \times 1.0 + 0.397 \omega_2 \] **6. Simplify and solve for \(\omega_2\):** \[ 1.3055 = 0.0373 \omega_2 - 0.0373 \times 19.6 + 0.397 \omega_2 \] \[ 1.3055 = 0.0373 \omega_2 - 0.731 + 0.397 \omega_2 \] Combine \(\omega_2\) terms: \[ 1.3055 + 0.731 = (0.0373 + 0.397) \omega_2 \] \[ 2.0365 = 0.4343 \omega_2 \] Solve for \(\omega_2\): \[ \boxed{\omega_2 \approx \frac{2.0365}{0.4343} \approx 4.69\, \text{rad/sec}} \] --- **Final Answer:** \[ \boxed{\omega_2 \approx 4.69\, \text{rad/sec}} \] This is the angular velocity of the rod immediately after impact.

**Step 1: Conceptual Introduction** When a moving ball strikes a pivoted thin rod, the collision causes the rod to rotate about its pivot point. To analyze this impact, we use the principles of conservation of angular momentum and the coefficient of restitution, which quantifies the elasticity of the collision. The initial linear momentum of the ball is transferred into angular momentum of the rod, resulting in its rotation. The coefficient of restitution relates the relative velocities of the ball and rod at the point of impact before and after collision, helping us determine the velocities post-impact. By combining these principles, we can calculate the angular velocity of the rod immediately after impact. --- **Step 2: Explanation of the Formulae Used** The core idea involves conserving angular momentum about the pivot point (A). Before impact, only the ball has linear momentum, which translates into angular momentum about A: \( H_{A1} = m_B v_{B1} d_1 \). After impact, both the ball and the rod contribute to the total angular momentum: \( H_{A2} = m_B v_{B2} d_1 + I_A \omega_2 \), where \( v_{B2} \) is the ball's velocity after impact, and \( \omega_2 \) is the rod's angular velocity. The coefficient of restitution relates the velocities pre- and post-impact at the impact point, accounting for energy loss or elasticity: \( e = \frac{v_{B2} - v_{rod,2}}{- v_{B1}} \). The moment of inertia \( I_A \) for a thin rod pivoted at one end is \( \frac{1}{3} m_{rod} L^2 \), which determines how the rod resists angular acceleration. --- **Step 3: Step-by-Step Calculation with Explanation** **1. Convert masses from pounds to slugs:** Masses in pounds are converted to slugs because the standard units in dynamics calculations involve slugs for mass. \[ m_B = \frac{1.2\, \text{lb}}{32.2} \approx 0.0373\, \text{slugs} \] \[ m_{rod} = \frac{8.7\, \text{lb}}{32.2} \approx 0.270\, \text{slugs} \] This step ensures correct unit consistency for subsequent calculations. **2. Calculate initial angular momentum of the ball about point A:** The initial angular momentum is obtained by multiplying the mass of the ball, its velocity, and the distance from the pivot at impact: \[ H_{A1} = m_B v_{B1} d_1 = 0.0373 \times 35 \times 1.0 = 1.3055\, \text{slug-ft}^2/\text{s} \] This value represents the rotational effect of the ball's linear momentum about the pivot before impact. **3. Compute the moment of inertia of the rod:** For a thin rod pivoted at one end, the moment of inertia is \[ I_A = \frac{1}{3} m_{rod} L^2 = \frac{1}{3} \times 0.270 \times (2.1)^2 = \frac{1}{3} \times 0.270 \times 4.41 \approx 0.397\, \text{slug-ft}^2 \] This quantity indicates how much resistance the rod has to change its rotational state. **4. Express the post-impact velocities using the coefficient of restitution:** The velocity of the ball after impact relates to the rod's angular velocity via: \[ v_{B2} = \omega_2 \times 1.0 - e v_{B1} = \omega_2 - 0.56 \times 35 = \omega_2 - 19.6 \] This captures how the relative velocities change due to the collision's elasticity. **5. Apply conservation of angular momentum and solve for \(\omega_2\):** Set the initial angular momentum equal to the sum after impact: \[ 1.3055 = m_B v_{B2} d_1 + I_A \omega_2 \] Substitute for \( v_{B2} \): \[ 1.3055 = 0.0373 (\omega_2 - 19.6) + 0.397 \omega_2 \] Expand: \[ 1.3055 = 0.0373 \omega_2 - 0.0373 \times 19.6 + 0.397 \omega_2 \] Calculate: \[ 1.3055 = 0.0373 \omega_2 - 0.731 + 0.397 \omega_2 \] Combine \(\omega_2\) terms: \[ 1.3055 + 0.731 = (0.0373 + 0.397) \omega_2 \] \[ 2.0365 = 0.4343 \omega_2 \] Solve for \(\omega_2\): \[ \omega_2 = \frac{2.0365}{0.4343} \approx 4.69\, \text{rad/sec} \] This value represents the angular velocity of the rod immediately after impact. --- **Final Answer:** \[ \boxed{\omega_2 \approx 4.69\, \text{rad/sec}} \] This is the angular velocity of the rod just after the collision.