A cylinder rotating at an angular speed of 50 rev/s is brought in contact with an identical stationary cylinder. Because of the kinetic friction, torques act on the two cylinders, accelerating the stationary one and decelerating the moving one. If the common magnitude of the acceleration and deceleration be one revolution per second square, how long will it take before the two cylinders have equal angular speed?

Explanation

To solve the problem, we need to analyze the motion of both cylinders and determine the time it takes for them to reach the same angular speed.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- For Cylinder 1 (the rotating cylinder):
- Initial angular speed, \( \omega_{0_1} = 50 \, \text{rev/s} \)
- Angular acceleration, \( \alpha_1 = -1 \, \text{rev/s}^2 \) (deceleration due to friction)

- For Cylinder 2 (the stationary cylinder):
- Initial angular speed, \( \omega_{0_2} = 0 \, \text{rev/s} \)
- Angular acceleration, \( \alpha_2 = +1 \, \text{rev/s}^2 \) (acceleration due to friction)

2. **Use the Angular Motion Equation**:
The angular motion equation is given by:
\[ \omega = \omega_0 + \alpha t \]
We will apply this equation to both cylinders.

3. **Apply the Equation for Cylinder 1**:
For Cylinder 1:
\[ \omega = 50 + (-1)t \]
Simplifying, we have:
\[ \omega = 50 - t \tag{1} \]

4. **Apply the Equation for Cylinder 2**:
For Cylinder 2:
\[ \omega = 0 + (1)t \]
Simplifying, we have:
\[ \omega = t \tag{2} \]

5. **Set the Angular Velocities Equal**:
Since both cylinders will have the same final angular speed \( \omega \) at time \( t \), we can set equations (1) and (2) equal to each other:
\[ 50 - t = t \]

6. **Solve for Time \( t \)**:
Rearranging the equation:
\[ 50 = 2t \]
Dividing both sides by 2:
\[ t = 25 \, \text{seconds} \]

### Final Answer:
The time it takes for the two cylinders to have equal angular speed is \( t = 25 \, \text{seconds} \).