hw 27 question 2 part b 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. ## Physics Problem: Tangential Acceleration of a Rotating Disk### Part A**Problem Statement:**Determine the tangential acceleration of a point on the outer rim of a disk when its angular speed is 2.00 revolutions per second (rev/s).**Input Field:**- \( a_t = \) [Insert your value in the box] m/s\(^2\)**Instructions:**To calculate the tangential acceleration (\( a_t \)), use the relationship between tangential acceleration, angular acceleration, and radius of the disk. You can use the formula:\[ a_t = \alpha \cdot r \]Where:- \( \alpha \) is the angular acceleration (radians per second squared)- \( r \) is the radius of the disk (meters)Note: This exercise assumes you have the necessary context to determine angular acceleration or any other required parameters not specified in the problem statement.**Submission:**- Click the "Submit" button to record your answer.- If you need guidance, select "Request Answer" for assistance.

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Answered: A compact disk, which has a diameter of…

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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.

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 2 part b

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.

## Physics Problem: Tangential Acceleration of a Rotating Disk

### Part A

**Problem Statement:**

Determine the tangential acceleration of a point on the outer rim of a disk when its angular speed is 2.00 revolutions per second (rev/s).

**Input Field:**

- \( a_t = \) [Insert your value in the box] m/s\(^2\)

**Instructions:**

To calculate the tangential acceleration (\( a_t \)), use the relationship between tangential acceleration, angular acceleration, and radius of the disk. You can use the formula:

\[ a_t = \alpha \cdot r \]

Where:
- \( \alpha \) is the angular acceleration (radians per second squared)
- \( r \) is the radius of the disk (meters)

Note: This exercise assumes you have the necessary context to determine angular acceleration or any other required parameters not specified in the problem statement.

**Submission:**

- Click the "Submit" button to record your answer.
- If you need guidance, select "Request Answer" for assistance.

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