The initial angular velocity of a heavy flywheel rotating on its axis is `w_0`. Its angular velocity decreases due to friction. At the end of the first minute its angular velocity is `0.8 omega_(0)`. What is its angular velocity at the end of third minute, if the frictional force is constant ?

Explanation

To solve the problem step by step, we will analyze the situation using the equations of motion for rotational motion.

### Step 1: Understanding the Problem
We know that the initial angular velocity of the flywheel is \( \omega_0 \). After the first minute (60 seconds), the angular velocity decreases to \( 0.8 \omega_0 \). We need to find the angular velocity at the end of the third minute (180 seconds).

### Step 2: Identify the Angular Deceleration
Since the frictional force is constant, the angular deceleration (denoted as \( \alpha \)) will also be constant. We can use the following equation of motion for rotational systems:

\[ \omega_f = \omega_i + \alpha t \]

Where:
- \( \omega_f \) is the final angular velocity,
- \( \omega_i \) is the initial angular velocity,
- \( \alpha \) is the angular acceleration (or deceleration in this case),
- \( t \) is the time.

### Step 3: Apply the Equation for the First Minute
At the end of the first minute (60 seconds), we have:

\[ \omega_f = 0.8 \omega_0 \]
\[ \omega_i = \omega_0 \]
\[ t = 60 \text{ seconds} \]

Substituting these values into the equation:

\[ 0.8 \omega_0 = \omega_0 + \alpha \cdot 60 \]

### Step 4: Solve for Angular Deceleration
Rearranging the equation gives:

\[ 0.8 \omega_0 - \omega_0 = \alpha \cdot 60 \]
\[ -0.2 \omega_0 = \alpha \cdot 60 \]
\[ \alpha = -\frac{0.2 \omega_0}{60} \]

### Step 5: Apply the Equation for the Third Minute
Now we want to find the angular velocity at the end of the third minute (180 seconds). We will use the same equation:

\[ \omega_f = \omega_i + \alpha t \]

Where:
- \( \omega_i = \omega_0 \)
- \( t = 180 \text{ seconds} \)
- \( \alpha = -\frac{0.2 \omega_0}{60} \)

Substituting these values into the equation:

\[ \omega_f = \omega_0 + \left(-\frac{0.2 \omega_0}{60}\right) \cdot 180 \]

### Step 6: Simplify the Expression
Calculating the term:

\[ \omega_f = \omega_0 - \frac{0.2 \omega_0 \cdot 180}{60} \]
\[ = \omega_0 - \frac{0.2 \cdot 3 \omega_0}{1} \]
\[ = \omega_0 - 0.6 \omega_0 \]
\[ = 0.4 \omega_0 \]

### Final Answer
Thus, the angular velocity at the end of the third minute is:

\[ \omega_f = 0.4 \omega_0 \]