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VIPSolutions · Accepted Answer

## **Objective**

To determine the angular acceleration of the bar $ \alpha_{AB} $ (in rad/s²) for the given system of a uniform disk and a rod AB, with all parameters provided.

---

## **Given Data**

- Length of rod, $ L = 2 $ ft  
- Radius of disk, $ R = .1 $ ft  
- Weight of disk, $ P = 6 $ lb  
- Weight of rod, $ P_1 = 5 $ lb  
- Angle, $ \theta = .7 $ rad  
- Angular velocity, $ \dot{\theta} = -.6 $ rad/s

---

## **Step 1: Identify Forces and Free-Body Diagram**

- Disk rolls without slipping at point $ A $.
- Rod AB is pin-connected at $ A $.
- The wall is smooth (no friction at wall except at the contact point for rolling).

---

## **Step 2: Convert Weights to Masses**

Weight, $ W = mg \Rightarrow m = \frac{W}{g} $, where $ g = 32.2 $ ft/s².

$$
m_{\text{disk}} = \frac{6}{32.2} = .186 \ \text{slugs}
$$
$$
m_{\text{rod}} = \frac{5}{32.2} = .155 \ \text{slugs}
$$

---

## **Step 3: Moments of Inertia**

- **Disk about center:** $ I_{\text{disk}} = \frac{1}{2} m_{\text{disk}} R^2 $
- **Rod about A:** $ I_{\text{rod}} = \frac{1}{3} m_{\text{rod}} L^2 $

$$
I_{\text{disk}} = \frac{1}{2} \times .186 \times (.1)^2 = .00093 \ \text{slug-ft}^2
$$
$$
I_{\text{rod}} = \frac{1}{3} \times .155 \times (2)^2 = .207 \ \text{slug-ft}^2
$$

---

## **Step 4: Kinematics — Rolling Constraint**

If disk rolls without slipping at $ A $:

$$
v_A = R \cdot \dot{\theta}
$$
But since the disk and rod are attached at $ A $, angular acceleration of the disk $ \alpha_{\text{disk}} $ and rod $ \alpha_{AB} $ must be related.

---

## **Step 5: Equations of Motion (Dynamic Equilibrium)**

Take moments about point $ A $ for the system (rod + disk):

$$
\sum M_A = I_{\text{total}} \cdot \alpha_{AB}
$$

Where:
$$
I_{\text{total}} = I_{\text{disk}} + I_{\text{rod}}
$$

Calculate $ I_{\text{total}} $:

$$
I_{\text{total}} = .00093 + .207 = .20793 \ \text{slug-ft}^2
$$

---

## **Step 6: Calculate Moments Due to Weights**

- **Rod:** Weight acts at center ($ L/2 $ from A, angle $ \theta $)
- **Disk:** Weight acts at center ($ R $ from A, angle $ \theta $)

$$
M_{W,\text{rod}} = P_1 \cdot (L/2) \cdot \sin(\theta)
$$
$$
M_{W,\text{disk}} = P \cdot R \cdot \sin(\theta)
$$

Plug in values:

$$
M_{W,\text{rod}} = 5 \times (2/2) \times \sin(.7)
= 5 \times 1 \times .6442 = 3.221 \ \text{lb-ft}
$$

$$
M_{W,\text{disk}} = 6 \times .1 \times \sin(.7)
= 6 \times .1 \times .6442 = .386 \ \text{lb-ft}
$$

$$
M_{\text{total}} = 3.221 + .386 = 3.607 \ \text{lb-ft}
$$

Convert lb-ft to slug-ft²/s² (since 1 lb = 1 slug-ft/s²):

$$
M_{\text{total}} = 3.607 \ \text{slug-ft}^2/\text{s}^2
$$

---

## **Step 7: Apply Newton's 2nd Law for Rotation**

$$
\sum M_A = I_{\text{total}} \cdot \alpha_{AB}
$$

$$
\alpha_{AB} = \frac{M_{\text{total}}}{I_{\text{total}}}
$$

$$
\alpha_{AB} = \frac{3.607}{.20793}
$$

$$
\alpha_{AB} = 17.34 \ \text{rad/s}^2
$$

---

## **Step 8: Direction**

Since the angular velocity is negative, if you want to assign the direction, check the sense of moments and initial condition; if not, just provide the magnitude.

---

## **Final Answer**

### **The angular acceleration of the bar $ \alpha_{AB} $ is $ \boxed{17.3 \ \text{rad/s}^2} $ (rounded to 1 decimal place).**

---

**Summary:**  
The solution involved calculating the moments of inertia for the rod and disk, summing moments about the pin, and solving for angular acceleration using Newton's 2nd law for rotation. The answer is $ 17.3 \ \text{rad/s}^2 $ (to 1 decimal place).

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