Starting from rest a centrifuge accelerates uniformly at the rate of `40 "rad"//s"^2`. What is the number of revolutions made by it when its angular velocity becomes 1800 rev/min ?

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant physics equations.

### Step 1: Identify the given values
- Initial angular velocity, \( \omega_0 = 0 \, \text{rad/s} \) (starting from rest)
- Angular acceleration, \( \alpha = 40 \, \text{rad/s}^2 \)
- Final angular velocity, \( \omega = 1800 \, \text{rev/min} \)

### Step 2: Convert the final angular velocity to radians per second
To convert revolutions per minute (rev/min) to radians per second (rad/s), we use the conversion factor:
\[ \omega = 1800 \, \text{rev/min} \times \frac{2\pi \, \text{rad}}{1 \, \text{rev}} \times \frac{1 \, \text{min}}{60 \, \text{s}} \]
Calculating this gives:
\[ \omega = 1800 \times \frac{2\pi}{60} = 60\pi \, \text{rad/s} \]

### Step 3: Use the kinematic equation for angular motion
We can use the equation:
\[ \omega^2 = \omega_0^2 + 2\alpha\theta \]
Since \( \omega_0 = 0 \), the equation simplifies to:
\[ \omega^2 = 2\alpha\theta \]
Rearranging to solve for \( \theta \):
\[ \theta = \frac{\omega^2}{2\alpha} \]

### Step 4: Substitute the values into the equation
Substituting \( \omega = 60\pi \, \text{rad/s} \) and \( \alpha = 40 \, \text{rad/s}^2 \):
\[ \theta = \frac{(60\pi)^2}{2 \times 40} \]
Calculating \( (60\pi)^2 \):
\[ (60\pi)^2 = 3600\pi^2 \]
Now substituting back:
\[ \theta = \frac{3600\pi^2}{80} = 45\pi^2 \, \text{radians} \]

### Step 5: Calculate the numerical value of \( \theta \)
Using \( \pi \approx 3.14 \):
\[ \theta \approx 45 \times (3.14)^2 \approx 45 \times 9.8596 \approx 443.68 \, \text{radians} \]

### Step 6: Calculate the number of revolutions
The number of revolutions \( n \) can be found by dividing \( \theta \) by \( 2\pi \):
\[ n = \frac{\theta}{2\pi} = \frac{443.68}{2\pi} \]
Calculating this gives:
\[ n \approx \frac{443.68}{6.28} \approx 70.7 \]
Rounding gives approximately \( n \approx 71 \) revolutions.

### Final Answer
The number of revolutions made by the centrifuge is approximately **71 revolutions**.

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