Math Problem Statement

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 You are correct. Your receipt no. is 149-346 Help: Receipt Previous Tries Angle of the displacement of P during this interval?

Solution

To find the angle of the displacement of point PP 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=36R = 36 cm
  • Displacement during one-half revolution: 1.341.34 m

Displacement Calculation:

When the wheel rolls through one-half of a revolution, the point PP initially at the contact point with the ground moves to the top of the wheel. The path traced by PP 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 PP is equal to half the circumference of the wheel: Horizontal Displacement=12×2πR=πR=π×36cm=113.04cm=1.1304m\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): Vertical Displacement=2R=72cm=0.72m\text{Vertical Displacement} = 2R = 72 \, \text{cm} = 0.72 \, \text{m}

  3. Angle of the Displacement: The displacement vector's angle θ\theta relative to the horizontal axis can be calculated using trigonometry: tan(θ)=Vertical DisplacementHorizontal Displacement\tan(\theta) = \frac{\text{Vertical Displacement}}{\text{Horizontal Displacement}} tan(θ)=0.721.1304\tan(\theta) = \frac{0.72}{1.1304}

    θ=tan1(0.721.1304)\theta = \tan^{-1}\left(\frac{0.72}{1.1304}\right)

Let's calculate the angle θ\theta.

Calculation:

θ=tan1(0.721.1304)tan1(0.6369)32.45\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 PP during this interval is approximately 32.45 degrees.

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


Here are 5 related questions you might consider:

  1. What is the total distance traveled by point PP 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 PP?
  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.

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Math Problem Analysis

Mathematical Concepts

Circular motion
Trigonometry

Formulas

Horizontal Displacement = \( \pi R \)
Vertical Displacement = \( 2R \)
Angle of Displacement = \( \tan^{-1}\left(\frac{\text{Vertical Displacement}}{\text{Horizontal Displacement}}\right) \)

Theorems

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Suitable Grade Level

High School