Let's solve the problem step by step: --- ### **Step 1: Draw the Free-Body Diagram and Write Equations** - **Given:** - Radius of drum \( r_1 = 80 \) mm \( = 0.08 \) m - Radius of disk \( r_2 = 160 \) mm \( = 0.16 \) m - Mass \( m = 8.5 \) kg - Radius of gyration \( k = 120 \) mm \( = 0.12 \) m - Force \( P = 20 \) N - Coefficients: \( \mu_s = 0.25 \), \( \mu_k = 0.2 \) - **Forces acting:** - Tension \( P \) upward on drum at radius \( r_1 \) - Friction force \( f \) at the ground, at radius \( r_2 \), direction opposes motion - **Moment of inertia:** \[ I = m k^2 = 8.5 \times (0.12)^2 = 0.1224 \text{ kg} \cdot \text{m}^2 \] **Explanation:** We identify all forces and moments acting on the disk and calculate the moment of inertia using the radius of gyration. --- ### **Step 2: Check for Rolling Without Slipping** For rolling without slipping, friction must be enough to prevent slipping: - Maximum static friction: \( f_{s,\text{max}} = \mu_s N = \mu_s m g \) \[ f_{s,\text{max}} = 0.25 \times 8.5 \times 9.81 = 20.86 \text{ N} \] **Now, write equations of motion:** - Translational: \( P - f = m a \) - Rotational (about center): \( P r_1 + f r_2 = I \alpha \) - For rolling without slipping: \( a = r_2 \alpha \) **Explanation:** We calculate the maximum static friction and set up the equations for translation and rotation, considering non-slipping motion. --- ### **Step 3: Solve Simultaneous Equations** From the above: - \( P - f = m a \) - \( P r_1 + f r_2 = I \alpha \) - \( a = r_2 \alpha \implies \alpha = \frac{a}{r_2} \) **Substitute \(\alpha\) in the torque equation:** \[ P r_1 + f r_2 = I \frac{a}{r_2} \] Now, solve for \( f \): \[ P - f = m a \implies f = P - m a \] \[ P r_1 + (P - m a) r_2 = I \frac{a}{r_2} \] \[ P r_1 + P r_2 - m a r_2 = I \frac{a}{r_2} \] \[ P (r_1 + r_2) - m a r_2 = I \frac{a}{r_2} \] \[ P (r_1 + r_2) = m a r_2 + I \frac{a}{r_2} \] \[ P (r_1 + r_2) = a \left( m r_2 + \frac{I}{r_2} \right) \] \[ a = \frac{P (r_1 + r_2)}{m r_2 + \frac{I}{r_2}} \] Plug in values: - \( P = 20 \) N - \( r_1 = 0.08 \) m - \( r_2 = 0.16 \) m - \( m = 8.5 \) kg - \( I = 0.1224 \) kg·m² Calculate denominator: \[ m r_2 = 8.5 \times 0.16 = 1.36 \] \[ \frac{I}{r_2} = \frac{0.1224}{0.16} = 0.765 \] \[ \text{Denominator} = 1.36 + 0.765 = 2.125 \] \[ r_1 + r_2 = 0.08 + 0.16 = 0.24 \] \[ P(r_1 + r_2) = 20 \times 0.24 = 4.8 \] \[ a = \frac{4.8}{2.125} = 2.259 \text{ m/s}^2 \] Now, angular acceleration: \[ \alpha = \frac{a}{r_2} = \frac{2.259}{0.16} = 14.119 \text{ rad/s}^2 \] **Check friction force:** \[ f = P - m a = 20 - (8.5 \times 2.259) = 20 - 19.202 = 0.798 \text{ N} \] Since \( 0.798 \text{ N} < 20.86 \text{ N} \), rolling without slipping is valid. --- ### **Final Answer:** \[ \boxed{14.119 \ \text{rad}/\text{s}^2} \] --- **Summary of Steps:** 1. Identified forces, moments, and calculated moment of inertia. 2. Checked for rolling without slipping and wrote equations. 3. Solved equations and verified friction condition; calculated final angular acceleration.