The uniform 12-foot pole is hinged to the truck bed and perfectly balanced at rest, when the truck accelerates at a constant of 3 fu/s² to the right. Find the angular velocity of the pole when the pole passes the horizontal position. Hints: • Draw the FBD and KD of the pole at an arbitrary angle between vertical and horizontal. Make that angle a variable 0. • Trick that works o Break ac into components ag=ao+aGo+aGo Draw each one acting at G on the KD. o Then Mo = Ica + ma,d can be rewritten as ΣMo = loa+maod,+ mac/otdz + mac/ond3, where each acceleration component has a different d based on the direction of that acceleration component. You should see that dy-0. From the moment equation, find a="something"(g sino + accos 6) Find by integrating do= 0 da to get an answer less than 10 rad/s.

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Each problem requires a proper FBD and separate kinematic diagram. 

 

**Problem Description**

A uniform 12-foot pole is hinged to a truck bed and initially perfectly balanced at rest. When the truck accelerates at a constant rate of 3 ft/s² to the right, determine the angular velocity of the pole (ω) when it passes the horizontal position.

**Hints for Solving**

1. **Free Body Diagram (FBD) and Kinematic Diagram (KD):**
   - Draw the FBD and KD for the pole at an arbitrary angle (θ) between the vertical and horizontal. Let this angle be variable (θ).

2. **Calculation Trick:**
   - Break acceleration into components: \( a_G = a_x \hat{i} + a_{Gx} \hat{i} + a_{Gy} \hat{j} \).
   - Draw each component acting at point G on the KD.
   - The moment equation \( \sum M_0 = I_G \alpha + m a_G d \) can be rewritten as \( \sum M_0 = I_G \alpha + m a_{Gx} d_1 + m a_{x} d_2 + m a_{Gy} d_3 \).
   - Each acceleration component affects a different arm of the moment, \( d_1, d_2, d_3 \). Note that \( d_1 = 0 \).

3. **From the Moment Equation:**
   - Find α by constructing an equation with terms including \( g \sin \theta + a \cos \theta \).

4. **Integration for Angular Velocity (ω):**
   - Solve for ω by integrating \( \omega \frac{d\omega}{d\theta} \) to obtain a result less than 10 rad/s.

**Diagram Explanation**

The image includes a diagram showing a side view of a truck with a 12-foot pole hinged on its bed. The pole forms an arc as it moves from the vertical to the horizontal position, illustrating its path. The truck is moving to the right with an acceleration of 3 ft/s². The motion transitions from a vertical position at 90° to a horizontal position at 0°, marking the trajectory of the end of the pole.
Transcribed Image Text:**Problem Description** A uniform 12-foot pole is hinged to a truck bed and initially perfectly balanced at rest. When the truck accelerates at a constant rate of 3 ft/s² to the right, determine the angular velocity of the pole (ω) when it passes the horizontal position. **Hints for Solving** 1. **Free Body Diagram (FBD) and Kinematic Diagram (KD):** - Draw the FBD and KD for the pole at an arbitrary angle (θ) between the vertical and horizontal. Let this angle be variable (θ). 2. **Calculation Trick:** - Break acceleration into components: \( a_G = a_x \hat{i} + a_{Gx} \hat{i} + a_{Gy} \hat{j} \). - Draw each component acting at point G on the KD. - The moment equation \( \sum M_0 = I_G \alpha + m a_G d \) can be rewritten as \( \sum M_0 = I_G \alpha + m a_{Gx} d_1 + m a_{x} d_2 + m a_{Gy} d_3 \). - Each acceleration component affects a different arm of the moment, \( d_1, d_2, d_3 \). Note that \( d_1 = 0 \). 3. **From the Moment Equation:** - Find α by constructing an equation with terms including \( g \sin \theta + a \cos \theta \). 4. **Integration for Angular Velocity (ω):** - Solve for ω by integrating \( \omega \frac{d\omega}{d\theta} \) to obtain a result less than 10 rad/s. **Diagram Explanation** The image includes a diagram showing a side view of a truck with a 12-foot pole hinged on its bed. The pole forms an arc as it moves from the vertical to the horizontal position, illustrating its path. The truck is moving to the right with an acceleration of 3 ft/s². The motion transitions from a vertical position at 90° to a horizontal position at 0°, marking the trajectory of the end of the pole.
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