Let's analyze the question: > "A slab is rotating about a fixed axis passing through point G which is also its center of mass. Calculate the linear momentum about point P on the slab." ### Key Concept - **Linear momentum about a point** is defined as \( \vec{L}_P = \sum_i \vec{r}_i \times m_i \vec{v}_i \). - If the axis passes through the center of mass (G), and the slab is only rotating (not translating), the **linear momentum of the entire slab is zero** because the velocity of the center of mass is zero. ### Options 1. \( 1 \ \text{kg} \cdot \text{m}^2/\text{s} \) 2. \( 2 \ \text{kg} \cdot \text{m}^2/\text{s} \) 3. \( 1.5 \ \text{kg} \cdot \text{m}^2/\text{s} \) 4. \( \ \text{kg} \cdot \text{m}^2/\text{s} \) ### **Correct Option** **The answer is:** #### \( \boxed{ \ \text{kg} \cdot \text{m}^2/\text{s}} \) #### **Reason:** Since the center of mass (G) is stationary (no translational velocity), the linear momentum of the slab about any point (including P) is **zero**. --- **Final Answer:** ** kg·m²/s**