Sure! Let's follow a **two-step process** to derive the given result for the **acceleration of the center of mass (\(\ddot{x}_{cm}\))** for a body rolling down an incline **without slipping**: --- ## **Step 1: Force and Motion Equations** ### **Translational Motion (Newton's Second Law)** For a body of mass \( m \) rolling down an incline of angle \( \theta \): - The component of gravity down the plane: \( mg \sin\theta \) - Friction force up the plane: \( F \) So, Newton's second law along the incline: \[ m\ddot{x}_{cm} = mg \sin\theta - F \] **Explanation:** The net force down the plane equals mass times acceleration of the center of mass. --- ## **Step 2: Rotational Motion and No-Slip Constraint** ### **Rotational Motion (Moment About Center of Mass)** - Torque due to friction: \( F \cdot a \), where \( a \) is the radius. - Moment of inertia about the center of mass: \( I_{cm} \). So, \[ I_{cm} \ddot{\varphi} = F a \] But from the no-slip condition, \( \ddot{x}_{cm} = a\ddot{\varphi} \), so \( \ddot{\varphi} = \frac{\ddot{x}_{cm}}{a} \): \[ I_{cm} \frac{\ddot{x}_{cm}}{a} = F a \implies F = \frac{I_{cm}}{a^2} \ddot{x}_{cm} \] **Explanation:** The friction force provides the torque responsible for rotation, and the no-slip condition relates angular and linear accelerations. --- ### **Combine Both Equations** Substitute \( F \) into the translational equation: \[ m\ddot{x}_{cm} = mg \sin\theta - \frac{I_{cm}}{a^2} \ddot{x}_{cm} \] \[ \left( m + \frac{I_{cm}}{a^2} \right)\ddot{x}_{cm} = mg \sin\theta \] \[ \ddot{x}_{cm} = \frac{mg \sin\theta}{m + \frac{I_{cm}}{a^2}} \] Given \( I_{cm} = m k_{cm}^2 \) (where \( k_{cm} \) is the radius of gyration): \[ \ddot{x}_{cm} = \frac{mg \sin\theta}{m + m \left(\frac{k_{cm}^2}{a^2}\right)} = \frac{g \sin\theta}{1 + \left(\frac{k_{cm}^2}{a^2}\right)} \] --- ## **Final Answer** \[ \boxed{ \ddot{x}_{cm} = \frac{mg \sin\theta}{m + \left(\frac{I_{cm}}{a^2}\right)} = \frac{mg \sin\theta}{m + m \left(\frac{k_{cm}^2}{a^2}\right)} = \frac{g \sin\theta}{1 + \left(\frac{k_{cm}^2}{a^2}\right)} } \] **This matches the required result.**