Give step-by-step solution with explanation and final answer:Problem 1.3 — For the geared system shown below, assume that shaft inertias and the
gear inertias I,1,,I; are negligible. The motor and lo

Question

Give step-by-step solution with explanation and final answer:Problem 1.3 — For the geared system shown below, assume that shaft inertias and the
gear inertias I,1,,I; are negligible. The motor and load inertias in kg-m?are I, =
0.03 and Is = 0.15. The speed ratios are 2 = 2 = 1.6. Derive the system model in terms
2 ws
of the speed wj3, with the applied torque T as the input.
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VIPSolutions · Accepted Answer

Let's solve the problem step by step:

---

## **Step 1: System Description**

- **Given:**
  - Shaft and gear inertias ($I_1, I_2, I_3$) are negligible.
  - Motor inertia ($I_4$) = .03 kg·m²
  - Load inertia ($I_5$) = .15 kg·m²
  - Speed ratios: $\frac{\omega_1}{\omega_2} = \frac{\omega_2}{\omega_3} = 1.6$
  - Input: Applied torque $T$
  - Output: Speed $\omega_3$

---

## **Step 2: Equivalent Inertia Referenced to Shaft 3 ($\omega_3$)**

### **A. Gear Ratios**

Let:
- $N_1 = \frac{\omega_1}{\omega_2} = 1.6$
- $N_2 = \frac{\omega_2}{\omega_3} = 1.6$

So,
$$
\frac{\omega_1}{\omega_3} = N_1 \times N_2 = 1.6 \times 1.6 = 2.56
$$

---

### **B. Reflecting Inertia to Shaft 3**

- **Motor Inertia ($I_4$), reflected to shaft 3:**
  $$
  I_{4,\,eq} = I_4 \left( \frac{\omega_4}{\omega_3} \right)^2
  $$
  But $\omega_4 = \omega_1$, so:
  $$
  I_{4,\,eq} = I_4 \left( \frac{\omega_1}{\omega_3} \right)^2 = .03 \times (2.56)^2 = .03 \times 6.5536 = .1966
  $$

- **Load Inertia ($I_5$), already at shaft 3:**  
  $I_5 = .15$

- **Total Equivalent Inertia at shaft 3:**
  $$
  J_{eq} = I_{4,\,eq} + I_5 = .1966 + .15 = .3466~\text{kg}\cdot\text{m}^2
  $$

---

## **Step 3: Equation of Motion**

The equation of motion about shaft 3 (ignoring friction, as not specified):

$$
J_{eq} \frac{d\omega_3}{dt} = T_{eq}
$$

where $T_{eq}$ is the equivalent torque at shaft 3. The applied torque $T$ is applied at shaft 1.

### **Torque Transformation**

Torque transforms **inversely** to speed ratio:
$$
T_3 = T \left( \frac{\omega_3}{\omega_1} \right)
$$
But $\frac{\omega_3}{\omega_1} = \frac{1}{2.56}$:

$$
T_3 = \frac{T}{2.56}
$$

---

## **Step 4: Final System Model**

$$
J_{eq} \frac{d\omega_3}{dt} = \frac{T}{2.56}
$$

$$
.3466 \frac{d\omega_3}{dt} = \frac{T}{2.56}
$$

$$
\boxed{
\frac{d\omega_3}{dt} = \frac{T}{.3466 \times 2.56} = \frac{T}{.8873}
}
$$

---

### **Final Answer (Boxed):**

$$
\boxed{
\frac{d\omega_3}{dt} = \frac{T}{.8873}
}
$$

where:
- $\omega_3$ = speed of the output shaft (shaft 3)
- $T$ = applied torque at the motor (shaft 1)

---

**Summary of Steps:**
1. Calculated speed ratios across the gears.
2. Reflected all inertias to shaft 3.
3. Transformed input torque to shaft 3.
4. Wrote the equation of motion in terms of $\omega_3$ and $T$.

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