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

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**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.
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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