# A high-speed drill reaches 2000 rpm in 0.50 s

## Question

A high-speed drill reaches 2000 rpm in 0.50 s.

1. What is the drill’s angular acceleration?

2. Through how many revolutions does it turn during this first

1. The drill’s angular acceleration is 419 rad/s2
2. The number of revolutions it turned during this first is 8.334 rev

## Explanation

### Solution to 1

1. We will start by solving question one (1) from the question

Data: The information from the question

We have angular velocity, v = 2000 rpm [Note: rpm = revolution per minute]

1 revolution = 3600 per rotation = 2 x 1800 per rotation = (2 x π) / 1 rotation = 2π

Additionally, 1 minute = 60 seconds = 60 s

We can now confidently say that the angular velocity which is 2000 rpm can be rewritten as:

angular velocity, ω = 2000 x (2π / 60) = (2000 x 2 x 3.142) / 60 = 209.5 rad/s

The time given to us from the question is t = 0.5 seconds

Note: Don’t forget that the formula for angular velocity is θ(t) = θ(0) + ω(0)t + 0.5 at2 or ω = 2πN/60

where ω is the angular velocity, and N is the number of revolutions per minute.

Unknown: Information we need to find

Angular acceleration, a = ?

Formula: The equation that will help us solve the problem

Remember that a = v / t

Therefore, we will use the formula a = ω / t

Solution

We will substitute our data into the formula

a = ω / t = 209.5 / 0.5 = 419 rad/s2

Therefore, the drill’s angular acceleration is 419 rad/s2.

### Solution to 2

2. Moving on to the second question, we will use the formula below to find the number of revolutions:

θ(t) = θ(0) + ω(0)t + 0.5 at2

Where θ(t) = ?, θ(0) = 0, ω(0)t = 0, a = 419 rad/s2, and t = 0.5 s

Therefore, after substituting the above data into the formula, we will have:

θ(t) = θ(0) + ω(0)t + 0.5 at2 = 0 + 0 + 0.5 x 419 x 0.52 = 52.375 rad/s

By converting the above answer from rad/s to revolution, we will obtain: