Problem 23 - Arihant AIEEE Physics

Question

A horizontal cylinder open from one end and closed from the other, is rotated with a constant angular velocity '\omega' about a vertical axis passing through its open end. If outside air pressure is \(P_{0}\), the temperature is \(T\), and the molar mass of air is \(m\), then find the air pressure as a function of the distance \(r^{\prime}\) from the rotational axis, : (Assume molar mass is independent of \(r\) ): (a) \(P=P_{0} e^{\left(w \omega^{2} r^{2} / 2 R T\right)}\) (b) \(P=P_{\alpha} e^{\left(m \omega^{2} r^{2} / R T\right)}\) (c) \(P=P_{0} e\) m \(\omega \mathrm{r}^{2} / \mathrm{RT} \quad\) (d) \(P=P_{0} e^{-m \omega^{2} r^{2} / 2 R T}\)

Step-by-Step Solution

Answer

The pressure as a function of distance is given by option (d): \(P = P_0 e^{-\frac{m \omega^2 r^2}{2RT}}\).

Step 1: Understand the Context

We're dealing with a horizontal cylinder open at one end, rotating with angular velocity \(\omega\). We need to find how air pressure \(P\) varies with the distance \(r\) from the rotational axis, considering temperature \(T\) and the molar mass \(m\) of the air.

Step 2: Apply the Physics Principles

Because of rotation, air inside experiences a centrifugal effect, altering air pressure along the radius. We'll apply a form of the hydrostatic pressure differential, where: \(dP = -\rho \omega^2 r \, dr\), with \(\rho\) as the air density.

Step 3: Substitute Known Relationships

The molar form of the ideal gas law gives \(\rho = \frac{Pm}{RT}\). Substituting this into our differential form, we have \(dP = -\frac{Pm}{RT} \omega^2 r \, dr\).

Step 4: Integrate the Differential Equation

Rearrange terms for integration: \(\frac{1}{P} \, dP = -\frac{m \omega^2 r}{RT} \, dr\). Integrate both sides: \(\int \frac{1}{P} \, dP = -\int \frac{m \omega^2 r}{RT} \, dr\).

Step 5: Solve the Integral

The left integral results in \(\ln P\), and the right integral gives \(\frac{-m \omega^2 r^2}{2RT}\). Thus, \(\ln P = \ln P_0 - \frac{m \omega^2 r^2}{2RT}\).

Step 6: Apply Exponential to Solve for P

Exponentiate both sides: \(P = P_0 \cdot e^{-\frac{m \omega^2 r^2}{2RT}}\). This shows pressure \(P\) decreases exponentially with \(r\).

Step 7: Verify and Review Options

The derived expression matches option (d): \(P = P_0 e^{-\frac{m \omega^2 r^2}{2RT}}\). Confirm by reviewing context and ensuring assumptions hold correctly.

Key Concepts

Centrifugal Force

Centrifugal force is a concept often explored in rotational dynamics. When an object is in a rotating reference frame, like the rotating cylinder in our exercise, it experiences a force that seems to push it outward from the center of rotation. This is the centrifugal force, which is not a real force but appears due to the inertia of the objects within the rotating frame.

The centrifugal force formula is often given as:

• \( F_c = m \omega^2 r \)

where \( m \) is the mass, \( \omega \) is the angular velocity, and \( r \) is the distance from the axis of rotation.
This force affects the pressure distribution in our cylinder since the air molecules are pushed outward as the cylinder rotates, causing a gradient in air pressure. The farther from the axis, the higher the outward force experienced by the air, contributing to a pressure decrease moving radially outward.

Ideal Gas Law

The ideal gas law is an essential principle in physics that relates the pressure, volume, and temperature of a gas to its amount in moles. The equation for the ideal gas law is:

• \( PV = nRT \)

In this equation, \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

For our problem, the ideal gas law is expressed in terms of density \( \rho \) and can be written as:

• \( \rho = \frac{Pm}{RT} \)

This adaptation helps us understand how the density of air changes with pressure and temperature inside the cylinder. By incorporating the centrifugal force concept, the variation in air density can be connected to changing pressure across the radius of the cylinder.

Pressure Distribution

Pressure distribution inside a rotating system refers to how the pressure changes from one point to another within that system. In our rotating cylinder, pressure distribution is directly influenced by the centrifugal force.

As the air inside the cylinder rotates, it experiences an outward force, causing the pressure to change with the radius \( r \). This alteration in pressure can be described by the differential equation:

• \( dP = -\rho \omega^2 r \, dr \)

We incorporate the ideal gas law to express \( \rho \) as a function of pressure, leading to a differential form suitable for integration. This pressure change results in an exponential decrease, as derived in the solution steps, and specifically observed in the form \( P = P_0 e^{-\frac{m \omega^2 r^2}{2RT}} \).
This relationship shows that pressure decreases with distance from the rotational axis, aligning with physical expectations regarding centrifugal effects.

Angular Velocity

Angular velocity is a fundamental concept that describes the rate of rotation for an object. It is usually denoted by \( \omega \) and given in radians per second.
Angular velocity tells us how quickly an object spins around its axis, with implications for centrifugal forces and pressure distribution in systems like our cylinder.

• The formula for centrifugal force incorporates angular velocity:\( F_c = m \omega^2 r \)
• Thus, as \( \omega \) increases, the centrifugal force acting on air particles in our problem also increases.

In the context of our exercise, angular velocity is crucial because it helps determine how rapidly the pressure decreases with the radius. The pressure equation includes a term \( \omega^2 \), which shows how sensitive the pressure distribution is to rotational speed.

Understanding \( \omega \) enables us to grasp how variations in the rotation speed of the cylinder influence the air pressure profile dramatically.