To find the air pressure as a function of the distance \( r \) from the rotation axis in a long horizontal cylinder that is closed at one end and rotated with a constant angular velocity, we can apply some principles from fluid dynamics and thermodynamics. The scenario involves a rotating fluid, which leads to a pressure gradient due to the centrifugal force acting on the air inside the cylinder.

Understanding the Forces at Play

When the cylinder rotates, the air inside experiences a centrifugal force that pushes it outward. This force increases with distance from the rotation axis, leading to a variation in pressure. The pressure at any point in the cylinder can be described using the hydrostatic equilibrium condition, which states that the change in pressure with respect to distance is related to the density of the fluid and the forces acting on it.

Applying the Hydrostatic Equation

We start with the hydrostatic pressure equation, which in cylindrical coordinates can be expressed as:

• \( \frac{dP}{dr} = -\rho \cdot a \)

Here, \( P \) is the pressure, \( \rho \) is the density of the air, and \( a \) is the acceleration due to the centrifugal force. The centrifugal acceleration at a distance \( r \) from the axis of rotation is given by:

• \( a = \omega^2 r \)

where \( \omega \) is the angular velocity of the cylinder. Substituting this into our hydrostatic equation gives:

• \( \frac{dP}{dr} = -\rho \cdot \omega^2 r \)

Relating Density and Pressure

To proceed, we need to relate the density \( \rho \) to the pressure \( P \). Using the ideal gas law, we have:

• \( P = \rho R T \)

where \( R \) is the specific gas constant for air. Rearranging this gives:

• \( \rho = \frac{P}{RT} \)

Substituting and Integrating

Now we can substitute this expression for \( \rho \) back into our hydrostatic equation:

• \( \frac{dP}{dr} = -\frac{P}{RT} \cdot \omega^2 r \)

This is a separable differential equation. We can rearrange it as:

• \( \frac{dP}{P} = -\frac{\omega^2 r}{RT} dr \)

Integrating both sides gives:

• \( \ln P = -\frac{\omega^2}{2RT} r^2 + C \)

Exponentiating both sides results in:

• \( P = P_0 e^{-\frac{\omega^2}{2RT} r^2} \)

Determining the Constant of Integration

To find the constant \( C \), we can use the boundary condition at the open end of the cylinder, where \( r = R \) (the radius of the cylinder) and the pressure is equal to the outside pressure \( P_0 \). Thus, we can express the pressure as:

• \( P(R) = P_0 \)

Substituting this into our equation allows us to solve for \( C \) and ultimately gives us the pressure as a function of \( r \):

• \( P(r) = P_0 e^{-\frac{\omega^2}{2RT} (r^2 - R^2)} \)

Final Expression

In summary, the air pressure inside the rotating cylinder as a function of the distance \( r \) from the rotation axis is given by:

• \( P(r) = P_0 e^{-\frac{\omega^2}{2RT} (r^2 - R^2)} \)

This equation illustrates how the pressure decreases exponentially with increasing distance from the rotation axis, influenced by the angular velocity and temperature of the air. Understanding this relationship is crucial in applications involving rotating systems, such as in engineering and atmospheric sciences.