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

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.

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:

θ(t) = 52.375 rad/s x (1rad/2π) = 52.375 rad/s x 0.159134 = 8.334 rev [1 rad = 0.159134 rev]

Therefore, the number of revolutions is 8.334 rev.

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