hw 27 question 1 part a ### Part A**Problem Statement:**What is the tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed is 2.00 revolutions per second?**Answer Input Box:**- \( a_t = \) [Enter your answer here] \(\text{ m/s}^2\)**Submission Options:**- **Submit** button to enter your answer.- **Request Answer** link for additional help or hints.### ExplanationThis question involves calculating the tangential acceleration, which is linked to the change in angular velocity over time. Remember to convert the angular speed from revolutions per second to radians per second before performing any calculations, as standard units for angular velocity involve radians. The tangential acceleration can then be found using the formula:\[ a_t = r \cdot \alpha \]where \( r \) is the radius of the disk and \( \alpha \) is the angular acceleration. Note that additional information may be needed to solve the problem completely, such as the radius of the disk or the rate of change of the angular velocity. A compact disk, which has a diameter of 12.0 cm, speeds up uniformly from zero to 4.50 rev/s in 3.20 s.

Answered: Image /qna-images/answer/475eed89-4d6b-4d23-aaf4-5119c3665c8d.jpgThe tangential acceleration of the disk is 0.52 m/s2.

Answered: What is the tangential acceleration of…

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What is the tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed is 2.00 rev/s? ΑΣφ ? m/s?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Angular speed, acceleration and displacement

Angular acceleration is defined as the rate of change in angular velocity with respect to time. It has both magnitude and direction. So, it is a vector quantity.

Angular Position

Before diving into angular position, one should understand the basics of position and its importance along with usage in day-to-day life. When one talks of position, it’s always relative with respect to some other object. For example, position of earth with respect to sun, position of school with respect to house, etc. Angular position is the rotational analogue of linear position.

Question

hw 27 question 1 part a

### Part A

**Problem Statement:**

What is the tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed is 2.00 revolutions per second?

**Answer Input Box:**

- \( a_t = \) [Enter your answer here] \(\text{ m/s}^2\)

**Submission Options:**

- **Submit** button to enter your answer.
- **Request Answer** link for additional help or hints.

### Explanation

This question involves calculating the tangential acceleration, which is linked to the change in angular velocity over time. Remember to convert the angular speed from revolutions per second to radians per second before performing any calculations, as standard units for angular velocity involve radians. The tangential acceleration can then be found using the formula:

\[ a_t = r \cdot \alpha \]

where \( r \) is the radius of the disk and \( \alpha \) is the angular acceleration. Note that additional information may be needed to solve the problem completely, such as the radius of the disk or the rate of change of the angular velocity.

A compact disk, which has a diameter of 12.0 cm, speeds up uniformly from zero to 4.50 rev/s in 3.20 s.

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