Let’s break this problem into clear steps: --- ## **Part (a): Required Equations** ### **Step 1: Conservation of Angular Momentum (about B, just before and after impact)** Since the system is pinned at B and the impact is instantaneous, **external impulses about B are zero**. Use conservation of angular momentum about point B: \[ \sum H_B^{\text{before}} = \sum H_B^{\text{after}} \] #### **Angular Momentum Before Impact** - Bar assembly is at rest, so its angular momentum = 0. - Only the putty (D) has angular momentum about B: - Distance BC = L - Perpendicular distance from B to line of v₀ at C = L (since C is vertically below B). - \( H_B^{\text{before}} = m_D \cdot v_0 \cdot L \) #### **Angular Momentum After Impact** - Now the putty and the assembly rotate together at angular velocity ω: - Total moment of inertia about B: - For each bar (slender rod, mass m, length L, **about one end**): \( I_{\text{bar, B}} = \frac{1}{3}mL^2 \) - 2 bars: \( 2 \times \frac{1}{3}mL^2 = \frac{2}{3}mL^2 \) - Putty (point mass at distance L): \( I_{\text{putty}} = m_D L^2 \) - **Total:** \( I_{\text{total}} = \frac{2}{3}mL^2 + m_D L^2 \) - Total angular momentum after: \( H_B^{\text{after}} = I_{\text{total}} \cdot \omega \) #### **Set up the equation:** \[ m_D v_0 L = \left( \frac{2}{3} m L^2 + m_D L^2 \right) \omega \] \[ \boxed{ \omega = \frac{m_D v_0}{\frac{2}{3}mL + m_D L} } \] --- ### **Step 2: Reactions at B Immediately After Impact** Immediately after impact, the **impulsive reaction at B** can be found by considering the change in linear momentum of the system: - The system's center of mass will have some velocity right after the impact. - The total impulsive force at B (\( \vec{J}_B \)) must account for the change in linear momentum of the bars+putty. #### **Linear Impulse-Momentum (in x and y)** \[ \sum \vec{J}_B + \sum \vec{J}_{\text{ext}} = \Delta \vec{P}_{\text{system}} \] But only B applies an impulse, and only the putty has initial momentum. - Initial linear momentum: Only putty, downward: \( m_D v_0 \) (y-direction) - Final linear momentum: System (bars+putty) moves with velocity of center of mass (\( \vec{V}_{\text{cm}} \)) \[ \vec{J}_B = m_{\text{total}} \vec{V}_{\text{cm}} - m_D v_0 \, \hat{j} \] Where \( m_{\text{total}} = 2m + m_D \). #### **Velocity of Center of Mass (\( \vec{V}_{\text{cm}} \))** - After impact, the system rotates with angular velocity ω about B. - Find the center of mass location and its velocity (\( \vec{V}_{\text{cm}} = \vec{\omega} \times \vec{r}_{\text{cm}} \)). ##### **Location of Center of Mass (relative to B):** Let’s take the coordinate system with B at (0,0), y downward, x right. - Bar BA: at 60° left of vertical (i.e., angle θ = +60° from vertical, or 120° from x-axis). - Bar BC: at 60° right of vertical (θ = -60° from vertical). - Center of each bar is at distance L/2 from B. Bar BA center: \[ x_1 = -\frac{L}{2} \sin(60^\circ) = -\frac{L}{2} \cdot \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{4}L \] \[ y_1 = \frac{L}{2} \cos(60^\circ) = \frac{L}{2} \cdot \frac{1}{2} = \frac{L}{4} \] Bar BC center: \[ x_2 = +\frac{L}{2} \sin(60^\circ) = +\frac{\sqrt{3}}{4}L \] \[ y_2 = \frac{L}{4} \] Putty at C: \[ x_3 = L \sin(60^\circ) = \frac{\sqrt{3}}{2}L \] \[ y_3 = L \cos(60^\circ) = \frac{1}{2}L \] Total mass: \( M = 2m + m_D \) So, \[ x_{\text{cm}} = \frac{m x_1 + m x_2 + m_D x_3}{2m + m_D} \] \[ y_{\text{cm}} = \frac{m y_1 + m y_2 + m_D y_3}{2m + m_D} \] Plug in values: \[ x_{\text{cm}} = \frac{m(-\frac{\sqrt{3}}{4}L) + m(\frac{\sqrt{3}}{4}L) + m_D(\frac{\sqrt{3}}{2}L)}{2m + m_D} = \frac{0 + m_D \frac{\sqrt{3}}{2}L}{2m + m_D} = \frac{m_D \frac{\sqrt{3}}{2}L}{2m + m_D} \] \[ y_{\text{cm}} = \frac{m(\frac{L}{4}) + m(\frac{L}{4}) + m_D(\frac{L}{2})}{2m + m_D} = \frac{2m \frac{L}{4} + m_D \frac{L}{2}}{2m + m_D} = \frac{\frac{mL}{2} + \frac{m_D L}{2}}{2m + m_D} = \frac{(m + m_D)\frac{L}{2}}{2m + m_D} \] ##### **Velocity of Center of Mass:** Right after impact, only rotation about B: \[ \vec{V}_{\text{cm}} = \omega \hat{k} \times \vec{r}_{\text{cm}} \] So, \[ V_{\text{cm}, x} = -\omega y_{\text{cm}} \] \[ V_{\text{cm}, y} = \omega x_{\text{cm}} \] --- ## **Part (b): Numerical Solutions** Given: \[ v_0 = 1.5\, \text{m/s}, \quad m = 0.7\, \text{kg}, \quad L = 0.4\, \text{m}, \quad m_D = 0.5\, \text{kg} \] ### **Step 1: Angular Velocity** \[ \omega = \frac{m_D v_0}{\frac{2}{3}mL + m_D L} \] First, calculate denominator: \[ \frac{2}{3} m L = \frac{2}{3} \cdot 0.7 \cdot 0.4 = \frac{2 \cdot 0.7 \cdot 0.4}{3} = \frac{0.56}{3} = 0.1867 \] \[ m_D L = 0.5 \cdot 0.4 = 0.2 \] \[ \frac{2}{3} m L + m_D L = 0.1867 + 0.2 = 0.3867 \] So, \[ \omega = \frac{0.5 \cdot 1.5}{0.3867} = \frac{0.75}{0.3867} \approx 1.94\, \text{rad/s} \] --- ### **Step 2: Center of Mass Position** \[ x_{\text{cm}} = \frac{m_D \cdot \frac{\sqrt{3}}{2}L}{2m + m_D} = \frac{0.5 \cdot 0.866 \cdot 0.4}{2 \cdot 0.7 + 0.5} = \frac{0.5 \cdot 0.3464}{1.9} = \frac{0.1732}{1.9} \approx 0.0912\, \text{m} \] \[ y_{\text{cm}} = \frac{(m + m_D)\frac{L}{2}}{2m + m_D} = \frac{(0.7 + 0.5)\cdot 0.2}{1.9} = \frac{1.2 \cdot 0.2}{1.9} = \frac{0.24}{1.9} \approx 0.1263\, \text{m} \] --- ### **Step 3: Center of Mass Velocity** \[ V_{\text{cm}, x} = -\omega y_{\text{cm}} = -1.94 \cdot 0.1263 \approx -0.245\, \text{m/s} \] \[ V_{\text{cm}, y} = \omega x_{\text{cm}} = 1.94 \cdot 0.0912 \approx 0.177\, \text{m/s} \] --- ### **Step 4: Impulsive Reaction at B** \[ \vec{J}_B = M \vec{V}_{\text{cm}} - m_D v_0 \hat{j} \] where \( M = 2m + m_D = 1.9\, \text{kg} \) \[ J_{Bx} = M V_{\text{cm}, x} = 1.9 \cdot (-0.245) = -0.465\, \text{kg m/s} \] \[ J_{By} = M V_{\text{cm}, y} - m_D v_0 = 1.9 \cdot 0.177 - 0.5 \cdot 1.5 = 0.336 - 0.75 = -0.414\, \text{kg m/s} \] --- ## **Final Answers** ### **(a) Equations:** - **Angular velocity:** \[ \boxed{ \omega = \frac{m_D v_0}{\frac{2}{3}mL + m_D L} } \] - **Reactions at B (impulsive):** \[ \boxed{ \begin{aligned} J_{Bx} &= (2m + m_D) V_{\text{cm},x} \\ J_{By} &= (2m + m_D) V_{\text{cm},y} - m_D v_0 \end{aligned} } \] where \[ \begin{aligned} x_{\text{cm}} &= \frac{m_D \frac{\sqrt{3}}{2}L}{2m + m_D} \\ y_{\text{cm}} &= \frac{(m+m_D)\frac{L}{2}}{2m + m_D} \\ V_{\text{cm}, x} &= -\omega y_{\text{cm}} \\ V_{\text{cm}, y} &= \omega x_{\text{cm}} \end{aligned} \] ### **(b) Numerical Answers:** - **Angular velocity right after impact:** \[ \boxed{ \omega \approx 1.94\ \text{rad/s} } \] - **Impulsive reactions at B:** \[ \boxed{ \begin{aligned} J_{Bx} &\approx -0.465\ \text{kg}\cdot\text{m/s}\\ J_{By} &\approx -0.414\ \text{kg}\cdot\text{m/s} \end{aligned} } \] --- If you need further clarification or step-by-step calculations for the center of mass or the moment of inertia, let me know!