Let's break down the problem and solve for the rotational kinetic energy after the collision. --- ## **Given Data** - **Rod mass, \( M = 127\,g = .127\,kg \)** - **Rod length, \( L = 37\,cm = .37\,m \)** - **Putty mass, \( m = 16\,g = .016\,kg \)** - **Putty speed, \( V = 8\,m/s \)** - **Angle, \( \theta = 22^\circ \)** - **Impact position, \( D = \frac{L}{4} = \frac{.37}{4} = .0925\,m \)** - **Rod pivots about its left end** --- ## **Step 1: Initial Angular Momentum** Only the perpendicular component of the putty's velocity contributes to angular momentum: \[ V_\perp = V \sin{\theta} \] Angular momentum before collision (about the hinge): \[ L_{\text{initial}} = m V_\perp D = m V \sin{\theta} D \] --- ## **Step 2: Moment of Inertia After Collision** - **Rod about end:** \( I_{\text{rod}} = \frac{1}{3} M L^2 \) - **Putty (point mass at \( D \) from hinge):** \( I_{\text{putty}} = m D^2 \) Total moment of inertia: \[ I_{\text{total}} = I_{\text{rod}} + I_{\text{putty}} \] --- ## **Step 3: Angular Velocity After Collision** Apply conservation of angular momentum (since the putty sticks): \[ L_{\text{initial}} = I_{\text{total}} \omega \] \[ \omega = \frac{L_{\text{initial}}}{I_{\text{total}}} \] --- ## **Step 4: Rotational Kinetic Energy After Collision** \[ K = \frac{1}{2} I_{\text{total}} \omega^2 \] --- ## **Step 5: Plug in Numbers** ### **Calculation Steps** #### 1. \( V_\perp \): \[ V_\perp = 8 \sin{22^\circ} \approx 8 \times .3746 = 2.9968\, m/s \] #### 2. \( L_{\text{initial}} \): \[ L_{\text{initial}} = .016 \times 2.9968 \times .0925 \approx .004438\, \text{kg}\cdot \text{m}^2/\text{s} \] #### 3. \( I_{\text{rod}} \): \[ I_{\text{rod}} = \frac{1}{3} \times .127 \times (.37)^2 \approx \frac{1}{3} \times .127 \times .1369 \approx .0058\, \text{kg}\cdot \text{m}^2 \] #### 4. \( I_{\text{putty}} \): \[ I_{\text{putty}} = .016 \times (.0925)^2 \approx .016 \times .008556 = .000137\, \text{kg}\cdot \text{m}^2 \] #### 5. \( I_{\text{total}} \): \[ I_{\text{total}} = .0058 + .000137 = .00594\, \text{kg}\cdot \text{m}^2 \] #### 6. \( \omega \): \[ \omega = \frac{.004438}{.00594} \approx .747\, \text{rad/s} \] #### 7. **Rotational Kinetic Energy:** \[ K = \frac{1}{2} \times .00594 \times (.747)^2 \approx .00265\, \text{J} \] --- ## **Final Answer** ### **Rotational kinetic energy after the collision:** \[ \boxed{.0027\, \text{J}} \] *(rounded to two significant digits)*