Let's break down the solution for both questions as per your requested steps: --- ## **Step 1: Conceptual Introduction** ### **a)** This problem deals with the dynamics of a falling uniform pole that pivots about a point after striking it. We are to find the angular velocity of the pole just after it impacts point A, considering it starts from rest and falls under gravity. **Explanation:** When the pole falls from rest, it gains angular velocity due to gravitational potential energy being converted into rotational kinetic energy. At the moment it hits point A and pivots, its energy state changes, and using energy conservation (neglecting losses), we can relate the change in height of the center of mass to the angular velocity about A. --- ## **Step 2: Formulae in the Solution** - **Conservation of energy:** \(\Delta PE = \Delta KE\) - **Potential Energy Change:** \(mgh\) (where \(h\) is the change in height of the center of mass) - **Rotational Kinetic Energy:** \(KE_{rot} = \frac{1}{2} I_A \omega^2\) - **Moment of Inertia about A (end of rod):** \(I_A = \frac{1}{3}mL^2\) **Explanation:** We use the conservation of energy principle to equate the loss in potential energy (as the pole's center of mass falls) to the gain in rotational kinetic energy about point A. The moment of inertia about the end of a uniform rod is critical here to correctly compute the angular velocity. --- ## **Step 3: Step-by-Step Calculation for Question (a)** **Given:** - Mass of pole, \(m = 10\,kg\) - Length of pole, \(L = 3\,m\) - Gravity, \(g = 9.81\,m/s^2\) - Initial angle, \(\theta_1 = 90^\circ\) - Final angle, \(\theta_2 = 60^\circ\) - Moment of inertia about A, \(I_A = \frac{1}{3}mL^2\) ### **Step 3.1: Calculate change in height of center of mass (\(\Delta h\))** The center of mass is at \(L/2\) from the pivot. - Initial height: \(h_1 = (L/2)\sin 90^\circ = 1.5\,m\) - Final height: \(h_2 = (L/2)\sin 60^\circ = 1.5 \times 0.866 = 1.299\,m\) - Change in height: \(\Delta h = h_1 - h_2 = 1.5 - 1.299 = 0.201\,m\) ### **Step 3.2: Apply Conservation of Energy** \[ m g \Delta h = \frac{1}{2} I_A \omega^2 \] \[ 10 \times 9.81 \times 0.201 = \frac{1}{2} \times \frac{1}{3} \times 10 \times 3^2 \times \omega^2 \] \[ 19.72 = \frac{1}{2} \times \frac{1}{3} \times 10 \times 9 \times \omega^2 \] \[ 19.72 = 15 \omega^2 \] \[ \omega^2 = \frac{19.72}{15} = 1.315 \] \[ \omega = \sqrt{1.315} = 1.147\, \text{rad/s} \] **Explanation:** We first found the vertical drop of the center of mass, then set the loss in potential energy equal to the gain in rotational kinetic energy, solved for \(\omega\), and checked all units and values carefully. --- ## **Step 4: Step-by-Step Calculation for Question (b)** **Given:** - \(AB = 5\,m\) (from Pythagoras: \( \sqrt{3^2 + 4^2} = 5\,m \)) - \(v_A = 2\,m/s\) (downward) - \(v = r\omega\) ### **Step 4.1: Find Angular Velocity \(\omega\) of Rod AB** \[ v_A = AB \cdot \omega \] \[ 2 = 5 \omega \] \[ \omega = \frac{2}{5} = 0.4\, \text{rad/s} \] ### **Step 4.2: Find Velocity of Block B (\(v_B\))** \(B\) moves horizontally. Using the geometry: - \(v_B = \omega \times 4\,m = 0.4 \times 4 = 1.6\,m/s\) (to the left) **Explanation:** We used the relation \(v = r \omega\) for each end. Since A is moving vertically and B horizontally, the respective distances from the instant center (the other end) are used. The calculated angular velocity is substituted to find the velocity of B. --- ## **Final Answers** **a) The angular velocity of the pole just after impact is:** \[ \boxed{1.15\, \text{rad/s}} \] **b) The angular velocity of rod AB is:** \[ \boxed{0.4\, \text{rad/s}} \] **and the velocity of block B is:** \[ \boxed{1.6\, \text{m/s (to the left)}} \]