/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 A wheel rolls 5 revolutions on a... [FREE SOLUTION] | 91Ó°ÊÓ

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A wheel rolls 5 revolutions on a horizontal surface without slipping. If the center of the wheel moves \(3.2 \mathrm{~m}\), what is the radius of the wheel?

Short Answer

Expert verified
The radius of the wheel is approximately 0.102 meters.

Step by step solution

01

Understanding the Problem

We are asked to find the radius of a wheel that rolls 5 revolutions without slipping, given that its center moves 3.2 meters. Understanding rolling without slipping means the distance traveled by the center of the wheel is equal to the circumference of the wheel times the number of revolutions.
02

Identify the Known Values

We know the wheel makes 5 complete revolutions, and the center moves a total distance of 3.2 meters. We need to use the relationship between revolutions and the circumference of the wheel to find the radius.
03

Formula for Total Distance Moved

The total distance moved by the wheel's center in terms of its radius is given by the formula: \[ \text{Distance} = 2\pi r \times \text{revolutions} \]Substitute the known values into this equation to solve for the radius \(r\).
04

Substitute Given Values

Substitute the known values into the formula:\[3.2 = 2 \pi r \times 5\]This will help us solve for the radius \(r\).
05

Solve for the Radius

To isolate \(r\), solve the equation by dividing both sides by \(10\pi\):\[r = \frac{3.2}{10\pi}\]
06

Calculate the Radius

Calculate the radius using \(\pi \approx 3.1416\):\[r = \frac{3.2}{10 \times 3.1416} \approx 0.1019\, \text{m}\]

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Key Concepts

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

Rolling Motion
Rolling motion is a form of motion where an object, typically a wheel or a sphere, moves by turning around an axis while simultaneously moving along a surface. When an object rolls without slipping, it means that every point on the object's edge comes into contact with the surface momentarily at rest but moves in a smooth, non-sliding motion. This distinguishes rolling motion from sliding motion, where there is a slipping action involved.
Rolling motion is essential in many everyday situations, like cars on a road or a ball rolling across a field. In these scenarios, the same point on the wheel or ball contacts the ground after each full rotation, moving the object forward.
The distance traveled by the rolling object can be directly tied to its turning, often calculated in terms of its circumference and the number of rotations or revolutions it completes.
Distance and Circumference Relationship
Understanding the relationship between the distance traveled and the circumference of a wheel is key in solving many physics problems involving rolling motion. This relationship hinges on how many complete revolutions the wheel makes. For a wheel rolling without slipping:
  • Each complete revolution moves the wheel forward by a distance equal to its circumference.
  • The total distance traveled is the product of the number of revolutions and the circumference of the wheel.
For instance, if a wheel with a known radius completes a series of revolutions, you can determine how far its center travels by multiplying the number of revolutions by the wheel's circumference. The formula to relate these quantities is:
\[\text{Distance} = 2\pi r \times \text{revolutions}\]
Here, \(r\) is the radius of the wheel, and \(2\pi r\) is its circumference. This formula holds for any object rolling without slipping, allowing for accurate calculations of distance based on rotation.
Radius Calculation
The radius of a wheel is a crucial element in determining the wheel's overall geometry and how it behaves when rolling. You can calculate the radius if you know the total distance the wheel's center travels and the number of revolutions it completes.
In our problem, the center of the wheel moves a distance of 3.2 meters over 5 revolutions. We use the relationship between distance, circumference, and revolutions:
\[\text{Distance} = 2\pi r \times \text{revolutions}\]
Substituting the given values, we get:
\[3.2 = 2\pi r \times 5\]
Solving for the radius \(r\), we divide both sides by \(10\pi\):
\[r = \frac{3.2}{10\pi}\]
This equation lets us find the radius of the wheel by evaluating it with a known value for \(\pi\), such as 3.1416. In this instance, the radius comes out to approximately 0.1019 meters.
No Slipping Condition
The "no slipping" condition is crucial in rolling motion problems. It ensures that as a wheel or object rolls, every part of it maintains in-contact rotation with the surface, but doesn't slide or slip away from the expected motion path.
This means that the distance traveled by the wheel is exclusively due to its rotation, with no additional movement from sliding. It allows for the direct use of revolutions to compute distance using the circumference formula. When an object slips, some energy that should contribute to forward motion is lost as it battles friction; therefore, slipping can lead to inaccurate distance and speed readings.
In many physics problems, specifying this no slipping condition lets us unambiguously relate the number of wheel revolutions directly to the distance the center of the wheel travels.

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Most popular questions from this chapter

A fixed 0.15-kg solid-disk pulley with a radius of \(0.075 \mathrm{~m}\) is acted on by a net torque of \(6.4 \mathrm{~m} \cdot \mathrm{N}\). What is the angular acceleration of the pulley?

A hoop starts from rest at a height \(1.2 \mathrm{~m}\) above the base of an inclined plane and rolls down under the influence of gravity. What is the linear speed of the hoop's center of mass just as the hoop leaves the incline and rolls onto a horizontal surface? (Neglect friction.)

To start her lawn mower, Julie pulls on a cord that is wrapped around a pulley. The pulley has a moment of inertia about its central axis of \(I=0.550 \mathrm{~kg} \cdot \mathrm{m}^{2}\) and a radius of \(5.00 \mathrm{~cm}\). There is an equivalent frictional torque impeding her pull of \(\tau_{\mathrm{f}}=0.430 \mathrm{~m} \cdot \mathrm{N}\). To accelerate the pulley at \(\alpha=4.55 \mathrm{rad} / \mathrm{s}^{2},\) (a) how much torque does Julie need to apply to the pulley? (b) How much tension must the rope exert?

An ice skater spinning with outstretched arms has an angular speed of \(4.0 \mathrm{rad} / \mathrm{s}\). She tucks in her arms, decreasing her moment of inertia by \(7.5 \% .\) (a) What is the resulting angular speed? (b) By what factor does the skater's kinetic energy change? (Neglect any frictional effects.) (c) Where does the extra kinetic energy come from?

(a) When a disk rolls without slipping, should the product \(r \omega\) be (1) greater than, (2) equal to, or (3) less than \(v_{\mathrm{CM}}\) ? (b) A disk with a radius of \(0.15 \mathrm{~m}\) rotates through \(270^{\circ}\) as it travels \(0.71 \mathrm{~m}\). Does the disk roll without slipping? Prove your answer.

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