A wheel with a radius of 36 cm rolls without slipping along a horizontal floor (see the figure). At time t1 the dot P painted on the rim of the wheel is at the point of contact between the wheel and the floor. At a later time t2, the wheel has rolled through one-half of a revolution.
What is the magnitude of the displacement of P during this interval?
1.34 m
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Angle of the displacement of P during this interval?

To find the angle of the displacement of point \( P \) during the interval when the wheel rolls through one-half of a revolution, let's first visualize and calculate the problem.

### Given:
- **Radius of the wheel**: \( R = 36 \) cm
- **Displacement during one-half revolution**: \( 1.34 \) m

### Displacement Calculation:
When the wheel rolls through one-half of a revolution, the point \( P \) initially at the contact point with the ground moves to the top of the wheel. The path traced by \( P \) during this motion can be visualized as part of a circle and a horizontal displacement.

However, to answer your question, we need to determine the angle of the displacement vector relative to the horizontal axis.

### Analysis:
1. **Horizontal Displacement**: When the wheel rolls without slipping, the horizontal distance covered by the point \( P \) is equal to half the circumference of the wheel:
   \[
   \text{Horizontal Displacement} = \frac{1}{2} \times 2\pi R = \pi R = \pi \times 36 \, \text{cm} = 113.04 \, \text{cm} = 1.1304 \, \text{m}
   \]

2. **Vertical Displacement**: The vertical displacement is equal to the diameter of the wheel (from the bottom to the top):
   \[
   \text{Vertical Displacement} = 2R = 72 \, \text{cm} = 0.72 \, \text{m}
   \]

3. **Angle of the Displacement**: The displacement vector's angle \( \theta \) relative to the horizontal axis can be calculated using trigonometry:
   \[
   \tan(\theta) = \frac{\text{Vertical Displacement}}{\text{Horizontal Displacement}}
   \]
   \[
   \tan(\theta) = \frac{0.72}{1.1304}
   \]

   \[
   \theta = \tan^{-1}\left(\frac{0.72}{1.1304}\right)
   \]

Let's calculate the angle \( \theta \).

### Calculation:
\[
\theta = \tan^{-1}\left(\frac{0.72}{1.1304}\right) \approx \tan^{-1}(0.6369) \approx 32.45^\circ
\]

So, the angle of the displacement of \( P \) during this interval is approximately **32.45 degrees**.

Would you like more details on the calculations or have any questions?

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Here are 5 related questions you might consider:
1. What is the total distance traveled by point \( P \) in one complete revolution?
2. How does the displacement change if the wheel's radius is doubled?
3. What would be the vertical and horizontal displacement if the wheel rolls through a quarter of a revolution?
4. How does the no-slip condition affect the displacement of point \( P \)?
5. What would be the displacement angle if the wheel were rolling up an incline?

**Tip:** Understanding the relationship between circular motion and linear displacement is key in problems involving rolling motion.

Explore how to calculate the angle of displacement for point P on a rolling wheel. Given a radius of 36 cm and a displacement of 1.34 m during one-half of a revolution, learn the principles of circular motion and trigonometry involved in solving this problem.

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