Let's solve the problem step by step: --- ### **a. Position Vector from A to G** The position vector from A to G (\(\vec{r}_{A \to G}\)): - \(A\) is at one end, \(G\) (center of mass) is at the center. - The length from \(A\) to \(G\) is \(L/2\). \[ \boxed{ \vec{r}_{A \to G} = \frac{L}{2} \hat{i} } \] Where \(\hat{i}\) is the unit vector along the beam (horizontal). --- ### **b. Acceleration of the Center of Mass in terms of \(a_A\), \(\omega\), \(\alpha\)** Using rigid body kinematics: \[ \vec{a}_G = \vec{a}_A + \vec{\alpha} \times \vec{r}_{A \to G} + \vec{\omega} \times (\vec{\omega} \times \vec{r}_{A \to G}) \] Initially, \(\omega = 0\): \[ \boxed{ \vec{a}_G = \vec{a}_A + \vec{\alpha} \times \vec{r}_{A \to G} } \] --- ### **c. Horizontal and Vertical Forces** Let \(T\) be the tension at \(A\). #### **Horizontal:** \[ T_x = m a_{Gx} \] #### **Vertical:** \[ T_y - mg = m a_{Gy} \] --- ### **d. Angular Acceleration and Moments Around G** Taking moments about G (center of mass): \[ \sum M_G = I_G \alpha \] The only force creating a moment is the tension at A (\(T\)), at a distance \(\frac{L}{2}\): \[ T_y \cdot \frac{L}{2} = I_G \alpha \] \[ \boxed{ T_y = \frac{2 I_G \alpha}{L} } \] --- ### **e. Expressions for Angular Acceleration and Tension** From part (b): \[ \vec{a}_G = \vec{a}_A + \vec{\alpha} \times \vec{r}_{A \to G} \] Let’s use vertical direction (y): \[ a_{Gy} = a_{Ay} + \alpha \left(\frac{L}{2}\right) \] But at the instant after cable B fails, the bar starts rotating about A (pivot): - At A, \(a_A = 0\) - At G, \(a_{Gy} = a_G = \alpha \left(\frac{L}{2}\right)\) So from Newton's 2nd law (y): \[ T_y - mg = m a_{Gy} \] \[ T_y - mg = m \alpha \frac{L}{2} \] \[ T_y = mg + m \alpha \frac{L}{2} \] But from earlier: \[ T_y = \frac{2 I_G \alpha}{L} \] Set equal: \[ \frac{2 I_G \alpha}{L} = mg + m \alpha \frac{L}{2} \] \[ \frac{2 I_G \alpha}{L} - m \alpha \frac{L}{2} = mg \] \[ \alpha \left( \frac{2 I_G}{L} - \frac{m L}{2} \right) = mg \] \[ \alpha = \frac{mg}{\frac{2 I_G}{L} - \frac{m L}{2}} \] \[ \boxed{ \alpha = \frac{mg}{\frac{2 I_G}{L} - \frac{mL}{2}} } \] Now, tension in chord: \[ T_y = \frac{2 I_G \alpha}{L} } \] --- ### **f. Scalar Values** Given: - \(L = 0.5\,m\) - \(m = 2\,kg\) - \(I_G = 0.042\,kg \cdot m^2\) - \(g = 9.81\,m/s^2\) Plug values into the formula for \(\alpha\): \[ \alpha = \frac{2 \times 9.81}{\frac{2 \times 0.042}{0.5} - \frac{2 \times 0.5}{2}} \] Calculate the denominator: \[ \frac{2 \times 0.042}{0.5} = \frac{0.084}{0.5} = 0.168 \] \[ \frac{2 \times 0.5}{2} = \frac{1.0}{2} = 0.5 \] So: \[ \alpha = \frac{19.62}{0.168 - 0.5} = \frac{19.62}{-0.332} \] \[ \alpha = -59.1\, \text{rad/s}^2 \] The negative sign indicates the direction of rotation (downward rotation). Now, tension: \[ T_y = \frac{2 I_G \alpha}{L} \] \[ T_y = \frac{2 \times 0.042 \times (-59.1)}{0.5} \] \[ = \frac{0.084 \times (-59.1)}{0.5} \] \[ = \frac{-4.9644}{0.5} \] \[ = -9.93\,N \] The negative sign indicates the direction is opposite to the assumed positive direction (upward). --- ## **Final Answers (Summary Table)** | Part | Expression/Value | |------|------------------| | a | \(\vec{r}_{A \to G} = \frac{L}{2} \hat{i}\) | | b | \(\vec{a}_G = \vec{a}_A + \vec{\alpha} \times \vec{r}_{A \to G}\) | | c | \(T_x = m a_{Gx}\), \(T_y - mg = m a_{Gy}\) | | d | \(T_y \frac{L}{2} = I_G \alpha\) | | e | \(\alpha = \frac{mg}{\frac{2 I_G}{L} - \frac{mL}{2}}\), \(T_y = \frac{2 I_G \alpha}{L}\) | | f | \(\alpha = -59.1\,\text{rad/s}^2\), \(T_y = -9.93\,N\) | If you need a more detailed explanation for any step or a diagram, let me know!

Certainly! Here's a concise 7-line theoretical explanation: When the support cable at B fails, the uniform beam becomes a freely pivoting rigid body about point A. The initial angular acceleration can be determined using rotational dynamics, where the net moment about the center of mass relates to the angular acceleration via the moment of inertia. The forces acting include gravity, tension in the remaining support at A, and inertial reactions. By applying Newton's second law for translation and rotation, one can relate the linear accelerations of the center of mass to angular acceleration and the applied forces. The tension in the support cable can be found from the vertical force balance, considering the inertial effects. Calculating these quantities provides insight into the dynamic response of the beam immediately after the support failure.

Certainly! Here are all the final answers summarized clearly: --- ### **Final Answers** **a. Position vector from A to G:** \[ \boxed{ \vec{r}_{A \to G} = \frac{L}{2}\,\hat{i} } \] --- **b. Acceleration of the center of mass G (in terms of \(a_A\), \(\omega\), \(\alpha\)):** \[ \boxed{ \vec{a}_G = \vec{a}_A + \vec{\alpha} \times \vec{r}_{A \to G} } \] --- **c. Horizontal and vertical force relations:** \[ \boxed{ \text{Horizontal:} \quad T_x = m a_{Gx} } \] \[ \boxed{ \text{Vertical:} \quad T_y - mg = m a_{Gy} } \] --- **d. Moment relation around G (center of mass):** \[ \boxed{ T_y \times \frac{L}{2} = I_G \alpha } \] --- **e. Expressions for angular acceleration \(\alpha\) and tension \(T_y\):** \[ \boxed{ \alpha = \frac{mg}{\frac{2 I_G}{L} - \frac{mL}{2}} } \] \[ \boxed{ T_y = \frac{2 I_G \alpha}{L} } \] --- **f. Numerical (scalar) values given data:** - \(L = 0.5\,m\) - \(m = 2\,kg\) - \(I_G = 0.042\,kg\cdot m^2\) - \(g = 9.81\,m/s^2\) **Angular acceleration:** \[ \boxed{ \alpha \approx -59.1\, \text{rad/s}^2 } \] **Tension in the chord:** \[ \boxed{ T_y \approx -9.93\, N } \] (Note: Negative signs indicate the direction of rotation and tension relative to the assumed positive directions.) --- Let me know if you'd like further clarification!