/*! 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 14 (II) A 61-cm-diameter wheel acce... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(II) A 61-cm-diameter wheel accelerates uniformly about its center from 120 rpm to 280 rpm in 4.0 s. Determine \((a)\) its angular acceleration, and \((b)\) the radial and tangential components of the linear acceleration of a point on the edge of the wheel 2.0 s after it has started accelerating.

Short Answer

Expert verified
(a) Angular acceleration is approximately \(2.67\) rad/s². (b) Tangential acceleration is approximately \(2.55\) m/s², and radial acceleration is approximately \(43.18\) m/s².

Step by step solution

01

Convert rpm to rad/s

The angular velocities are given in revolutions per minute (rpm). First convert them to radians per second (rad/s) using the formula: \ \[\omega = \frac{\text{rpm} \times 2\pi}{60}\]. \ For initial angular velocity \(\omega_i = 120\) rpm: \ \[\omega_i = \frac{120 \times 2\pi}{60} = 4\pi \] rad/s. \ For final angular velocity \(\omega_f = 280\) rpm: \ \[\omega_f = \frac{280 \times 2\pi}{60} = \frac{28\pi}{3} \] rad/s.
02

Calculate Angular Acceleration

Use the formula for angular acceleration \(\alpha\) when angular velocity changes uniformly: \ \[\alpha = \frac{\omega_f - \omega_i}{t} \]. \ Substitute in the values: \ \[\alpha = \frac{\frac{28\pi}{3} - 4\pi}{4} = \frac{8\pi}{3} \] rad/s².
03

Calculate Tangential Acceleration After 2 Seconds

Tangential acceleration \(a_t\) is given by \(a_t = \alpha \cdot r\), where \(r\) is the radius of the wheel. \ The radius \(r\) is half of the diameter, so \(r = \frac{61}{2} = 30.5\) cm = \(0.305\) m. \ Substitute the values: \ \[a_t = \frac{8\pi}{3} \times 0.305 = \frac{2.44\pi}{3} \approx 2.55 \] m/s².
04

Calculate Radial Acceleration After 2 Seconds

Radial acceleration \(a_r\) is given by \(a_r = \omega^2 \cdot r\). First, find the angular velocity \(\omega\) at \(t = 2\) s using \(\omega = \omega_i + \alpha t\): \ \[\omega = 4\pi + \frac{8\pi}{3} \times 2 = \frac{20\pi}{3} \] rad/s. \ Now calculate \(a_r\): \ \[a_r = \left(\frac{20\pi}{3}\right)^2 \times 0.305 = \frac{400\pi^2}{9} \times 0.305 \approx 43.18 \] m/s².

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Radial Acceleration
Radial acceleration is the component of acceleration that points towards the center of the circular path an object follows. It is also called centripetal acceleration and plays a key role in circular motion.
This acceleration is important because it ensures that an object continues moving along a circular path, rather than flying off in a straight line. It can be calculated using the formula:
  • \(a_r = \omega^2 \cdot r\)
where \(\omega\) is the angular velocity and \(r\) is the radius of the circular path.
In our example, we calculated \(a_r\) for the wheel after it had been accelerating for 2 seconds. We first found the angular velocity at this time using \(\omega = \omega_i + \alpha t\), which allowed us to plug into our radial acceleration formula. The result we obtained shows how quickly the wheel gains speed in maintaining its circular trajectory as it rotates.
Tangential Acceleration
Tangential acceleration refers to the change in speed of an object moving along a circular path. This type of acceleration changes the object's speed but does not affect its direction. It's directly related to the angular acceleration and can be expressed as:
  • \(a_t = \alpha \cdot r\)
where \(\alpha\) is the angular acceleration and \(r\) is the radius of the path.
In the problem we considered, after the initial 2 seconds, the tangential acceleration was calculated using the previously found angular acceleration and the given radius of the wheel.
Thus, tangential acceleration describes how the linear speed of the point on the wheel's edge is increasing due to its rotation. It shows the growth of velocity across time as the wheel accelerates uniformly.
Linear Acceleration
Linear acceleration is what you might think of as the everyday acceleration we experience when we speed up or slow down in a straight line. It's essentially a combination of radial and tangential accelerations when we're discussing rotational motion.
While linear acceleration in circular motion can typically involve both radial and tangential components, each with distinct originating points, in this context, it's useful to describe how tangential acceleration specifically affects linear speed.
The overall effect of these two types of acceleration results in changes to both the velocity magnitude and direction as the wheel spins. Essentially, it expands our understanding of how circular motion behaves under uniform acceleration conditions, and how different components contribute to the motion we observe. Thus, while radial acceleration keeps the object moving in a curve, the tangential acceleration changes its speed along that path. Together, they inform the complete picture of how the wheel's motion evolves.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.