A dentist using a dental drill brings it from rest to maximum operating speed of 352,000 rpm in 4.0 s. Assume that the drill accelerates at a constant rate during this time. (a) What is the angular acceleration of the drill in rev/s2? | rev/s² (b) Find the number of revolutions the drill bit makes during the 4.0 s time interval. rev
A dentist using a dental drill brings it from rest to maximum operating speed of 352,000 rpm in 4.0 s. Assume that the drill accelerates at a constant rate during this time. (a) What is the angular acceleration of the drill in rev/s2? | rev/s² (b) Find the number of revolutions the drill bit makes during the 4.0 s time interval. rev
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
![**Instructions for Problem Solving: Dental Drill Acceleration**
A dentist using a dental drill brings it from rest to a maximum operating speed of 352,000 rpm in 4.0 seconds. Assume that the drill accelerates at a constant rate during this time.
**(a) Question:**
What is the angular acceleration of the drill in \(\text{rev/s}^2\)?
- **Answer field:** [Input box for answer] rev/s\(^2\)
**(b) Question:**
Find the number of revolutions the drill bit makes during the 4.0-second time interval.
- **Answer field:** [Input box for answer] rev
**Explanation:**
To solve these problems, use the following concepts:
1. **Angular acceleration:**
- Angular acceleration (\( \alpha \)) is the rate of change of angular velocity.
- Formula: \( \alpha = \frac{\Delta \omega}{\Delta t} \), where \( \Delta \omega \) is the change in angular velocity and \( \Delta t \) is the time interval.
2. **Conversion from rpm to rev/s:**
- Angular velocity in rev/s can be found by converting rpm to rev/s.
- Use: 1 rpm = \(\frac{1}{60}\) rev/s.
3. **Finding revolutions:**
- The number of revolutions (θ) made by the drill can be determined using the equation for motion under constant acceleration:
- \( \theta = \omega_i t + \frac{1}{2} \alpha t^2 \), where \( \omega_i \) is the initial angular velocity (which is 0 in this case as it starts from rest).
These calculations will lead to the solution for each part of the problem.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6ac9d050-1fff-4e75-b553-243d63aa5f64%2Fae9277e6-a0db-44b1-ac08-97011e4ab2d6%2Fjvy1prs_processed.png&w=3840&q=75)
Transcribed Image Text:**Instructions for Problem Solving: Dental Drill Acceleration**
A dentist using a dental drill brings it from rest to a maximum operating speed of 352,000 rpm in 4.0 seconds. Assume that the drill accelerates at a constant rate during this time.
**(a) Question:**
What is the angular acceleration of the drill in \(\text{rev/s}^2\)?
- **Answer field:** [Input box for answer] rev/s\(^2\)
**(b) Question:**
Find the number of revolutions the drill bit makes during the 4.0-second time interval.
- **Answer field:** [Input box for answer] rev
**Explanation:**
To solve these problems, use the following concepts:
1. **Angular acceleration:**
- Angular acceleration (\( \alpha \)) is the rate of change of angular velocity.
- Formula: \( \alpha = \frac{\Delta \omega}{\Delta t} \), where \( \Delta \omega \) is the change in angular velocity and \( \Delta t \) is the time interval.
2. **Conversion from rpm to rev/s:**
- Angular velocity in rev/s can be found by converting rpm to rev/s.
- Use: 1 rpm = \(\frac{1}{60}\) rev/s.
3. **Finding revolutions:**
- The number of revolutions (θ) made by the drill can be determined using the equation for motion under constant acceleration:
- \( \theta = \omega_i t + \frac{1}{2} \alpha t^2 \), where \( \omega_i \) is the initial angular velocity (which is 0 in this case as it starts from rest).
These calculations will lead to the solution for each part of the problem.
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