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Act like a helpful tutor and exlain me :Chapter 1 — System Modeling Problem 1.4 — The following figure shows a field-controlled motor. The motor inertia is I; and the load inertia is I,. The motor and load are connected through a spring with rotational stiffness (kr). We assume there is no translation motion between the motor and the load but just rotational motion. Task 1: Obtain a set of differential equations to model the entire system. Hint #1: Similar to spring's linear dynamics (Force = Linear Stiffness x Compression, F = kAx) , for spring's rotational dynamics, Torque = Rotational Stiffness x Rotational Compression. Hint #2: Find the equation of motion for motor and load using equation (Iw? = T) separately. Hint #3: There should be 3 equations in total. Task 2 (Optional): Describe the system using only a single differential equation model with 8; as output and vy as input. + 0, Ry hr I 000 Motor Load 1 | oom so +] kr c T 0 0,

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Act like a helpful tutor and exlain me :Uploaded ImageChapter 1 — System Modeling Problem 1.4 — The following figure shows a field-controlled motor. The motor inertia is I; and the load inertia is I,. The motor and load are connected through a spring with rotational stiffness (kr). We assume there is no translation motion between the motor and the load but just rotational motion. Task 1: Obtain a set of differential equations to model the entire system. Hint #1: Similar to spring's linear dynamics (Force = Linear Stiffness x Compression, F = kAx) , for spring's rotational dynamics, Torque = Rotational Stiffness x Rotational Compression. Hint #2: Find the equation of motion for motor and load using equation (Iw? = T) separately. Hint #3: There should be 3 equations in total. Task 2 (Optional): Describe the system using only a single differential equation model with 8; as output and vy as input. + 0, Ry hr I 000 Motor Load 1 | oom so +] kr c T 0 0,

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Asked by: Alok Kumar
Created at: 2026-01-26 04:17:00

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# Explanation of 1.4 – System Modeling Let's break down the **system modeling** and guide you through setting up the differential equations step by step. --- ## **Understanding the System** You have: - **Field-controlled motor** (with inertia \( I_1 \)) - **Load** (with inertia \( I_2 \)) - **Rotational spring** (with stiffness \( k_T \)) connecting motor and load - Input: Field voltage (\( v_f \)) - Output: Load angular position (\( \theta_2 \)) - **No translational motion**—only rotational ### **Key Variables** - \( \theta_1 \): Angular position of the motor - \( \theta_2 \): Angular position of the load - \( v_f \): Field voltage input to the motor --- ## **Task 1: Obtain Set of Differential Equations** ### **Step 1: Electrical Dynamics (Field Circuit)** From the circuit: \[ v_f = R_f i_f + L_f \frac{di_f}{dt} \] Where: - \( R_f \): Field resistance - \( L_f \): Field inductance - \( i_f \): Field current ### **Step 2: Mechanical Dynamics (Motor and Load)** #### **A. Motor (Inertia \( I_1 \))** - **Equation of motion**: \( I_1 \ddot{\theta}_1 = T - T_s \) - \( T \): Torque produced by the motor (proportional to field current, \( i_f \)) - \( T_s \): Torque due to the spring - **Spring torque** (like spring force): \[ T_s = k_T (\theta_1 - \theta_2) \] - **Motor torque**: \( T = K_f i_f \), with \( K_f \) as the torque constant So, \[ I_1 \ddot{\theta}_1 = K_f i_f - k_T (\theta_1 - \theta_2) \] #### **B. Load (Inertia \( I_2 \))** - **Equation of motion**: \( I_2 \ddot{\theta}_2 = T_s \) (assuming no damping or external load torque) So, \[ I_2 \ddot{\theta}_2 = k_T (\theta_1 - \theta_2) \] --- ### **Summary of Differential Equations** 1. **Field Circuit:** \[ v_f = R_f i_f + L_f \frac{di_f}{dt} \] 2. **Motor:** \[ I_1 \ddot{\theta}_1 = K_f i_f - k_T (\theta_1 - \theta_2) \] 3. **Load:** \[ I_2 \ddot{\theta}_2 = k_T (\theta_1 - \theta_2) \] --- ## **Task 2 (Optional): Single Differential Equation (with \( \theta_2 \) as Output, \( v_f \) as Input)** ### **Goal**: Express everything in terms of \( \theta_2 \) and \( v_f \). **Outline of Steps:** 1. Express \( i_f \) from the field circuit equation. 2. Relate \( \theta_1 \) to \( \theta_2 \) using the load and motor equations. 3. Substitute back to eliminate intermediate variables. #### **Step-by-step Outline** 1. **From load equation:** \( I_2 \ddot{\theta}_2 = k_T (\theta_1 - \theta_2) \) \( \Rightarrow \theta_1 = \theta_2 + \frac{I_2}{k_T} \ddot{\theta}_2 \) 2. **Substitute \( \theta_1 \) into motor equation:** \( I_1 \ddot{\theta}_1 = K_f i_f - k_T (\theta_1 - \theta_2) \) But \( (\theta_1 - \theta_2) = \frac{I_2}{k_T} \ddot{\theta}_2 \) 3. **Combine and substitute \( i_f \) from field circuit equation.** This leads to a **third-order** single differential equation relating \( \theta_2 \) and \( v_f \). --- ## **Summary Table** | Equation Type | Equation | |-------------------|--------------------------------------------------------------------------| | Field Circuit | \( v_f = R_f i_f + L_f \frac{di_f}{dt} \) | | Motor | \( I_1 \ddot{\theta}_1 = K_f i_f - k_T (\theta_1 - \theta_2) \) | | Load | \( I_2 \ddot{\theta}_2 = k_T (\theta_1 - \theta_2) \) | --- ### **Diagram Reference** - \( v_f \) is the input voltage, \( \theta_2 \) is the output (load position). --- ## **Summary** - **Task 1**: Three equations—one for the field circuit, one for the motor, and one for the load. - **Task 2**: Combine the equations to relate \( v_f \) (input) to \( \theta_2 \) (output) by eliminating intermediate variables. Let me know if you want to see the single combined equation for Task 2 worked out explicitly!

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