Let's break down the solution step by step: --- ## **Step 1: Equivalent Inertia Calculation (A)** We need to express all rotating and translating inertias to the motor shaft (input side). ### **1. Motor Inertia (Direct):** \[ J_{motor} = I_m \] ### **2. Drive Wheels Inertia (Reflected):** There are 4 wheels, each with inertia \(I_{dw}\), reflected to the motor shaft through the gear ratio \(N\): \[ J_{dw,eq} = 4 \times I_{dw} \times N^2 \] ### **3. Mass of Drive Chains (Reflected):** There are 2 chains, each with mass \(m_c\), moving at the same speed as the drive wheels. The equivalent rotational inertia at the wheel is \(m_c r_{d}^2\), reflected to the motor shaft: \[ J_{chain,eq} = 2 \times m_c \times r_d^2 \times N^2 \] ### **4. Load Mass (Reflected):** The load mass \(m_L\) is moving linearly with the belt, but we convert to equivalent rotational inertia at the drive wheel, then to the motor shaft: \[ J_{load,eq} = m_L r_d^2 \times N^2 \] ### **Total Equivalent Inertia (A):** \[ A = I_m + 4I_{dw}N^2 + 2m_c r_d^2 N^2 + m_L r_d^2 N^2 \] --- #### **Plug in the numbers:** - \(I_m = 0.04\) kg·m² - \(I_{dw} = 0.012\) kg·m² - \(N = 1/6\) - \(m_c = 4\) kg - \(m_L = 9\) kg - \(r_d = 0.076\) m \[ N^2 = \left(\frac{1}{6}\right)^2 = \frac{1}{36} \] \[ A = 0.04 + 4(0.012)\left(\frac{1}{36}\right) + 2(4)(0.076^2)\left(\frac{1}{36}\right) + 9(0.076^2)\left(\frac{1}{36}\right) \] Calculate each term: - \(4 \times 0.012 = 0.048\) - \(0.048 \times \frac{1}{36} = 0.00133\) - \(0.076^2 = 0.005776\) - \(2 \times 4 = 8\) - \(8 \times 0.005776 = 0.0462\) - \(0.0462 \times \frac{1}{36} = 0.001283\) - \(9 \times 0.005776 = 0.051984\) - \(0.051984 \times \frac{1}{36} = 0.001444\) Sum up: \[ A = 0.04 + 0.00133 + 0.001283 + 0.001444 = \boxed{0.0441} \text{ kg·m}^2 \] --- ## **Step 2: Equivalent Load Torque (B)** The load torque at the drive shaft is \(T_L\). Reflected to the motor shaft (input side) through the gear ratio \(N\): \[ B = N T_L \] Plug in the values: - \(N = 1/6\) - \(T_L = 11\) N·m \[ B = \frac{1}{6} \times 11 = \boxed{1.83} \text{ N·m} \] --- ## **Step 3: Final Answers** \[ \boxed{A = 0.0441 \ \text{kg·m}^2} \] \[ \boxed{B = 1.83 \ \text{N·m}} \] --- ### **Summary Table** | Parameter | Value | |-----------|---------------| | A | 0.0441 kg·m² | | B | 1.83 N·m | --- ## **Equation of Motion** \[ A \dot{\omega}_m = T_i - B \operatorname{sgn}(\omega_m) \] --- **Fill in the blanks:** - The value of A is **0.0441 kg·m²** - The value of B is **1.83 N·m**

# Step-by-Step Solution Explanation This section breaks down the calculations for equivalent inertia (A) and equivalent load torque (B) in the conveyor system. ## Step 1: Equivalent Inertia Calculation (A) ### 1. Motor Inertia (Direct) The motor inertia is given directly as: \[ J_{motor} = I_m \] ### 2. Drive Wheels Inertia (Reflected) To reflect the inertia of the drive wheels back to the motor shaft, we use the gear ratio \(N\): \[ J_{dw,eq} = 4 \times I_{dw} \times N^2 \] ### 3. Mass of Drive Chains (Reflected) The chains also contribute to inertia, reflected to the motor shaft: \[ J_{chain,eq} = 2 \times m_c \times r_d^2 \times N^2 \] ### 4. Load Mass (Reflected) The load mass is reflected similarly: \[ J_{load,eq} = m_L \times r_d^2 \times N^2 \] ### Total Equivalent Inertia (A) Combining all contributions gives: \[ A = I_m + 4I_{dw}N^2 + 2m_c r_d^2 N^2 + m_L r_d^2 N^2 \] ### Numerical Calculation Substituting the values: - \(I_m = 0.04\) kg·m² - \(I_{dw} = 0.012\) kg·m² - \(N = \frac{1}{6}\) (since the reducer reduces speed by a factor of 6) - \(m_c = 4\) kg - \(m_L = 9\) kg - \(r_d = 0.076\) m Calculate \(N^2\): \[ N^2 = \left(\frac{1}{6}\right)^2 = \frac{1}{36} \] Now compute \(A\): \[ A = 0.04 + 4(0.012)\left(\frac{1}{36}\right) + 2(4)(0.076^2)\left(\frac{1}{36}\right) + 9(0.076^2)\left(\frac{1}{36}\right) \] Calculating each term: - \(4 \times 0.012 = 0.048\) - \(0.048 \times \frac{1}{36} \approx 0.00133\) - \(0.076^2 = 0.005776\) - \(2 \times 4 \times 0.005776 = 0.046208\) - \(0.046208 \times \frac{1}{36} \approx 0.001283\) - \(9 \times 0.005776 = 0.051984\) - \(0.051984 \times \frac{1}{36} \approx 0.001444\) Summing up all contributions: \[ A = 0.04 + 0.00133 + 0.001283 + 0.001444 \approx 0.0441 \text{ kg·m}^2 \] **Final Inertia Value:** \[ \boxed{A = 0.0441 \text{ kg·m}^2} \] --- ## Step 2: Equivalent Load Torque (B) The load torque \(T_L\) at the drive shaft needs to be reflected to the motor shaft: \[ B = N \times T_L \] ### Numerical Calculation Given: - \(N = \frac{1}{6}\) - \(T_L = 11\) N·m Calculating \(B\): \[ B = \frac{1}{6} \times 11 \approx 1.83 \text{ N·m} \] **Final Torque Value:** \[ \boxed{B = 1.83 \text{ N·m}} \] --- ## Final Answers ### Summary of Results - **Inertia (A):** \[ \boxed{0.0441 \text{ kg·m}^2} \] - **Torque (B):** \[ \boxed{1.83 \text{ N·m}} \] --- ## Equation of Motion The equation of motion for the conveyor system in terms of motor velocity \( \dot{\omega}_m \) and motor torque \( T_i \) is given by: \[ A \dot{\omega}_m = T_i - B \operatorname{sgn}(\omega_m) \] --- ## Fill in the Blanks - The value of \( A \) is **0.0441 kg·m²** - The value of \( B \) is **1.83 N·m**