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Chapter 1: Problem 24

Rolling tire A tire of radius \(R\) rolls in a straight line without slipping. Its center moves with constant speed \(V .\) A small pebble lodged in the tread of the tire touches the road at \(t=0 .\) Find the pebble's position, velocity, and acceleration as functions of time.

Short Answer

Expert verified
The pebble follows a cycloidal path with position, velocity, and acceleration as derived.

Step by step solution

01

Understanding the Motion

The tire rolls without slipping, so the point of contact on the road is instantaneously at rest. The wheel's center moves with velocity \(V\), and as it rolls, the pebble follows a cycloidal path. We will find its position, velocity, and acceleration as functions of time, \(t\).
02

Calculating the Pebble's Position

The pebble's position as a function of time, \(t\), can be represented parametrically as: \[\begin{align*}x(t) &= R \cdot (\omega t - \sin(\omega t)), \y(t) &= R \cdot (1 - \cos(\omega t))\end{align*}\]where \(\omega = \frac{V}{R}\) is the angular speed of the wheel.
03

Deriving the Pebble's Velocity

To derive the pebble's velocity, we differentiate its position with respect to time:\[\begin{align*}v_x(t) &= \frac{d}{dt}[R(\omega t - \sin(\omega t))] = R\omega (1 - \cos(\omega t)), \v_y(t) &= \frac{d}{dt}[R(1 - \cos(\omega t))] = R\omega \sin(\omega t)\end{align*}\]
04

Deriving the Pebble's Acceleration

The pebble's acceleration is found by differentiating the velocity function with respect to time:\[\begin{align*}a_x(t) &= \frac{d}{dt}[R\omega(1 - \cos(\omega t))] = R\omega^2 \sin(\omega t), \a_y(t) &= \frac{d}{dt}[R\omega \sin(\omega t)] = R\omega^2 \cos(\omega t)\end{align*}\]
05

Summarizing Results

The pebble's position, velocity, and acceleration as functions of time are:- Position: \(x(t) = R(\omega t - \sin(\omega t)), \; y(t) = R(1 - \cos(\omega t))\)- Velocity: \(v_x(t) = R\omega (1 - \cos(\omega t)), \; v_y(t) = R\omega \sin(\omega t)\)- Acceleration: \(a_x(t) = R\omega^2 \sin(\omega t), \; a_y(t) = R\omega^2 \cos(\omega t)\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rolling Motion
Rolling motion occurs when an object moves while rotating along a surface. In this context, we have a tire that rolls without slipping. This means the linear speed of the tire's center is related to its rotation such that there is no relative motion between the tire and the ground at the point of contact. Hence, the contact point is momentarily stationary.
This synchronization defines rolling without slipping:
  • The center of the tire moves at a constant speed, denoted as \( V \).
  • The angular speed is given by \( \omega = \frac{V}{R} \), where \( R \) is the tire radius.
Due to this relationship, the movements of points on the tire, like a pebble in the tread, describe intricate paths, known as cycloidal paths.
Parametric Equations
Parametric equations offer an elegant way to express motion paths in terms of a parameter, usually time. In this scenario, the pebble's motion is best modeled using parametric equations, taking advantage of the periodic and cyclical nature of cycloidal paths.
Here's how it works:
  • The horizontal position \( x(t) \) is given as \( R(\omega t - \sin(\omega t)) \).
  • The vertical position \( y(t) \) is defined by \( R(1 - \cos(\omega t)) \).
These equations incorporate the cosine and sine functions to account for the cyclical rolling pattern of the pebble around the tire's center. It captures both displacement in the direction of travel and height variation as the pebble rotates with the tire.
Differentiation in Physics
Differentiation is a powerful tool used to analyze motion, providing insights into how position, velocity, and acceleration change over time. To analyze the motion of the pebble on the rolling tire, we differentiate the parametric equations.
Here's the process:
  • Velocity:
    By differentiating the position functions with respect to time:
    • The horizontal velocity component \( v_x(t) \) becomes \( R\omega(1 - \cos(\omega t)) \).
    • The vertical velocity component \( v_y(t) \) becomes \( R\omega \sin(\omega t) \).
  • Acceleration:
    Further differentiation gives the acceleration components:
    • The horizontal acceleration component \( a_x(t) \) is \( R\omega^2 \sin(\omega t) \).
    • The vertical acceleration component \( a_y(t) \) is \( R\omega^2 \cos(\omega t) \).
This technique shows how changes in velocity lead to specific pathways in acceleration, illustrating the dynamic nature of cycloidal motion seen in objects like rolling tires with embedded pebbles.

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