Let's break the problem into three clear steps with explanations for each step. --- ## **Step 1: Analyze the Forces and Torques** **Given:** - Mass \( m = 8.50 \) kg - Combined radius of gyration \( k = 120 \) mm \( = 0.12 \) m - Force \( P = 20 \) N - Coefficients of friction: \( \mu_s = 0.25 \), \( \mu_k = 0.2 \) - Drum radius \( r = 80 \) mm \( = 0.08 \) m - Disk radius \( R = 160 \) mm \( = 0.16 \) m The force \( P \) is applied tangentially at the drum's radius. We need to find whether the disk slips or rolls without slipping. ### **Frictional Force for Rolling Without Slipping:** The friction force must prevent slipping and is given by: \[ f_{\text{max}} = \mu_s N \] where \( N = mg \). If the friction required for rolling without slipping exceeds \( f_{\text{max}} \), then slipping occurs and kinetic friction applies. **Explanation:** This step helps us decide whether to use static or kinetic friction for our calculations. --- ## **Step 2: Write the Equations of Motion** ### **Linear Motion:** \[ \sum F = ma \] \[ P - f = ma \] ### **Rotational Motion about Center of Mass:** \[ \sum M_G = I_G \alpha \] Torque due to \( P \) and \( f \) (friction at the ground): \[ P \cdot r - f \cdot R = I_G \alpha \] Where \( I_G = m k^2 \). ### **Rolling Without Slipping Condition:** \[ a = R \alpha \] **Explanation:** These equations relate the applied force, friction, and rotational inertia to the linear and angular acceleration of the disk. --- ## **Step 3: Solve for Angular Acceleration** Let's solve for \( \alpha \): From Step 2: 1. \( P - f = ma \) 2. \( P r - f R = I_G \alpha \) 3. \( a = R \alpha \) Substitute \( a = R\alpha \) and \( I_G = m k^2 \): 1. \( P - f = m R \alpha \) 2. \( P r - f R = m k^2 \alpha \) Let’s solve for \( f \) from equation 1: \[ f = P - mR\alpha \] Substitute \( f \) into equation 2: \[ P r - (P - mR\alpha) R = m k^2 \alpha \] \[ P r - P R + m R^2 \alpha = m k^2 \alpha \] \[ P(r - R) = m (k^2 - R^2) \alpha \] \[ \alpha = \frac{P(r - R)}{m (k^2 - R^2)} \] Plug in the values: - \( P = 20 \) N - \( r = 0.08 \) m - \( R = 0.16 \) m - \( m = 8.5 \) kg - \( k = 0.12 \) m \[ \alpha = \frac{20 (0.08 - 0.16)}{8.5 \left[ (0.12)^2 - (0.16)^2 \right]} \] \[ \alpha = \frac{20 \times (-0.08)}{8.5 \left( 0.0144 - 0.0256 \right)} \] \[ \alpha = \frac{-1.6}{8.5 \times (-0.0112)} \] \[ \alpha = \frac{-1.6}{-0.0952} \] \[ \alpha = 16.807 \text{ rad/s}^2 \] Now, check the friction force: \[ f = P - mR\alpha = 20 - 8.5 \times 0.16 \times 16.807 \] \[ = 20 - 22.861 = -2.861 \text{ N} \] Since the friction force is negative (i.e., acts in the opposite direction), check its magnitude: \[ |f| = 2.861 \text{ N} \] Max static friction: \[ f_{\text{max}} = \mu_s mg = 0.25 \times 8.5 \times 9.81 = 20.866 \text{ N} \] Since \( |f| < f_{\text{max}} \), rolling without slipping is possible. --- ## **Final Answer:** **The angular acceleration of the disk is:** \[ \boxed{16.807 \ \text{rad/s}^2} \] --- **Summary of Steps:** 1. **Analyzed forces and determined friction needed for rolling.** 2. **Wrote equations of motion for translation and rotation.** 3. **Solved for angular acceleration and checked friction requirement.**