2. A shaft is driven by an engine through a friction single- plate disc-type clutch. There are friction surfaces on both sides of the plate. The inner and outer diameters of the plate are 175 mm and 325 mm respectively. The axial force on the clutch is \( 1,5 \mathrm{kN} \) and the coefficient of fretion is 0,3 . The engine develops a constant torque of 50 N .m with a moment of inertia of \( 4 \mathrm{~kg} \cdot \mathrm{~m}^{2} \). The driven shaft exerts a constant resisting torque of is Nm and the moment of inertia is \( 8 \mathrm{~kg} \mathrm{~m}^{2} \). When the clutch is engaged with the engine speed at \( 400 \mathrm{r} / \) min and the shaft is at rest, assume uniform wear and calculate: (a) the combined speed when the clutch is fully en- gaged (b) the time taken for the shaft to attain a speed of 400 r/min.

To solve the problem, we will first calculate the torque transmitted by the clutch when it is fully engaged, and then we will use that information to find the combined speed and the time taken for the shaft to reach 400 r/min. ### Given Data: - Inner diameter of the clutch plate, \( d_1 = 175 \, \text{mm} = 0.175 \, \text{m} \) - Outer diameter of the clutch plate, \( d_2 = 325 \, \text{mm} = 0.325 \, \text{m} \) - Axial force on the clutch, \( F = 1.5 \, \text{kN} = 1500 \, \text{N} \) - Coefficient of friction, \( \mu = 0.3 \) - Engine torque, \( T_e = 50 \, \text{N.m} \) - Moment of inertia of the engine, \( J_e = 4 \, \text{kg.m}^2 \) - Resisting torque of the driven shaft, \( T_r \) (to be determined) - Moment of inertia of the driven shaft, \( J_r = 8 \, \text{kg.m}^2 \) - Engine speed, \( N_e = 400 \, \text{r/min} \) ### Step 1: Calculate the Torque Transmitted by the Clutch The torque transmitted by the clutch can be calculated using the formula: \[ T_c = \mu \cdot F \cdot r_m \] where \( r_m \) is the mean radius of the clutch plate, calculated as: \[ r_m = \frac{d_1 + d_2}{4} = \frac{0.175 + 0.325}{4} = \frac{0.5}{4} = 0.125 \, \text{m} \] Now, substituting the values into the torque equation: \[ T_c = 0.3 \cdot 1500 \cdot 0.125 = 56.25 \, \text{N.m} \] ### Step 2: Calculate the Combined Speed When the Clutch is Fully Engaged When the clutch is fully engaged, the engine and the driven shaft will rotate together at a common speed \( N_c \). The torque balance equation is: \[ T_e = T_c + T_r \] Where \( T_r \) is the resisting torque of the driven shaft. We can express the resisting torque in terms of the angular velocities: \[ T_r = J_r \cdot \alpha \] Where \( \alpha \) is the angular acceleration. However, we can also express the relationship between the speeds and torques: \[ T_r = \frac{J_r}{J_e + J_r} \cdot T_e \] Now substituting the values: \[ T_r = \frac{8}{4 + 8} \cdot 50 = \frac{8}{12} \cdot 50 = \frac{2}{3} \cdot 50 = 33.33 \, \text{N.m} \] Now substituting \( T_r \) back into the torque balance equation: \[ T_c = T_e - T_r = 50 - 33.33 = 16.67 \, \text{N.m} \] Now we can find the combined speed \( N_c \): Using the relationship between torque and speed: \[ T_c = J_c \cdot \alpha \] Where \( J_c = J_e + J_r \) is the total moment of inertia: \[ J_c = 4 + 8 = 12 \, \text{kg.m}^2 \] Now we can find the angular velocity \( \omega \): \[ \omega = \frac{T_c}{J_c} = \frac{16.67}{12} = 1.39 \, \text{rad/s} \] Now converting this to r/min: \[ N_c = \frac{\omega \cdot 60}{2\pi} = \frac{1.39 \cdot 60}{2\pi} \approx 13.27 \, \text{r/min} \] ### Step 3: Calculate the Time Taken for the Shaft to Attain a Speed of 400 r/min To find the time taken to reach 400 r/min, we need to calculate the angular acceleration \( \alpha \): Using the torque balance equation: \[ T_c = J_c \cdot \alpha \] We can find \( \alpha \): \[ \alpha = \frac{T_c}{J_c} = \frac{16.67}{12} = 1.39 \, \text{rad/s}^2 \] Now we can use the kinematic equation to find the time \( t \): \[ \omega_f = \omega_i + \alpha t \] Where: - \( \omega_f = \frac{400 \cdot 2\pi}{60} \approx 41.89 \, \text{rad/s} \) - \( \omega_i = 0 \) Now substituting the values: \[ 41.89 = 0 + 1.39 t \] Solving for \( t \): \[ t = \frac{41.89}{1.39} \approx 30.1 \, \text{s} \] ### Summary of Results: (a) The combined speed when the clutch is fully engaged is approximately **13.27 r/min**. (b) The time taken for the shaft to attain a speed of 400 r/min is approximately **30.1 seconds**.