91Ó°ÊÓ

An incompressible fluid with density \(\rho\) is in a horizontal test tube of inner cross-sectional area \(A .\) The test tube spins in a horizontal circle in an ultracentrifuge at an angular speed \omega. Gravitational forces are negligible. Consider a volume element of the fluid of area \(A\) and thickness \(d r^{\prime}\) a distance \(r^{\prime}\) from the rotation axis. The pressure on its inner surface is \(p\) and on its outer surface is \(p+d p .\) (a) Apply Newton's second law to the volume element to show that \(d p=\rho \omega^{2} r^{\prime} d r^{\prime}\) . (b) If the surface of the fluid is at a radius \(r_{0}\) where the pressure is \(p_{0},\) show that the pressure \(p\) at a distance \(r \geq r_{0}\) is \(p=p_{0}+\rho \omega^{2}\left(r^{2}-r_{0}^{2}\right) / 2 .\) (c) An object of volume \(V\) and density \(\rho_{\mathrm{ob}}\) has its center of mass at a distance \(R_{\mathrm{cmob}}\) from the axis. Show that the net horizontal force on the object is \(\rho V \omega^{2} R_{\mathrm{cm}},\) where \(R_{\mathrm{cm}}\) is the distance from the axis to the center of mass of the displaced fluid. (d) Explain why the object will move inward if \(\rho R_{\mathrm{cm}}>\rho_{\mathrm{ob}} R_{\mathrm{cmob}}\) and outward if \(\rho R_{\mathrm{cm}}<\rho_{\mathrm{ob}} R_{\mathrm{cmob}} .\) (e) For small objects of uniform density, \(R_{\mathrm{cm}}=R_{\mathrm{cmob}}\) . What happens to a mixture of small objects of this kind with different densities in an ultracentrifuge?

Step by step solution

01

Define Forces on the Volume Element

We start by considering a small volume element of fluid within the test tube. This element is at a distance \( r' \) from the axis of rotation with a thickness \( dr' \) and area \( A \). According to Newton's second law, the net force acting towards the center (centripetal force) is due to the pressure difference over the element, given by \( A(dp) \). Since the element rotates in a circle, we apply \( F = ma \) with centripetal acceleration \( a = \omega^2 r' \). The force is \( ho A dr' \omega^2 r' \).

02

Equate Forces for Pressure Change

Balancing the forces, we have the equation due to centripetal force: \[ A(dp) = ho A dr' \omega^2 r'. \]Simplifying gives the differential form for the change in pressure,\[ dp = \rho \omega^2 r' dr', \]which shows the pressure increment along \( r' \). This solves part (a).

03

Solve for Total Pressure Change

We integrate the differential pressure equation over a distance from \( r_0 \) to \( r \) to find the pressure \( p \) at a radius \( r \) from the rotation axis. The integration gives:\[ p - p_0 = \int_{r_0}^{r} \rho \omega^2 r' dr' = \rho \omega^2 \left[ \frac{r'^2}{2} \right]_{r_0}^{r} = \frac{\rho \omega^2}{2} (r^2 - r_0^2). \]Thus, \( p = p_0 + \frac{\rho \omega^2}{2} (r^2 - r_0^2) \), which solves part (b).

04

Calculate Net Force on the Object

The object displaces a volume \( V \) of fluid and creates a net force due to pressure differences acting at its center of mass, at distance \( R_{\mathrm{cmob}} \) from the axis. The net force is equal to the difference in pressure times the area, or the same form as before: \[ F_{\text{net}} = \rho V \omega^2 R_{\mathrm{cm}}. \]This expression assumes \( R_{\mathrm{cm}} \) as the center of mass of displaced fluid, solving part (c).

05

Discuss Movement Direction of the Object

According to equilibrium conditions, the object will move radially based on density and radial position. If \( \rho R_{\mathrm{cm}} > \rho_{\mathrm{ob}} R_{\mathrm{cmob}} \), the fluid's centripetal term is larger, pulling the object inward. Conversely, if \( \rho R_{\mathrm{cm}} < \rho_{\mathrm{ob}} R_{\mathrm{cmob}} \), the object's terms dominate, and it is pushed outward, solving part (d).

06

Behavior of Small Uniform Objects

For small objects with uniform density, \( R_{\mathrm{cm}} = R_{\mathrm{cmob}} \). In a mixture of such objects, the distinction purely becomes based on density. Objects with higher densities than the fluid will be pushed outward while lighter objects move inward due to the centrifugal effects during rotation, solving part (e).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Velocity

Angular velocity is a measure expressing how quickly an object rotates or revolves around an axis. It is a vector quantity, meaning it has both magnitude and direction. In this exercise, we deal with a test tube spinning in a circle at a constant angular velocity denoted by \( \omega \). This angular velocity is crucial for calculating the forces acting upon the fluid in the test tube.
- Angular velocity \( \omega \) is measured in radians per second (rad/s). - It determines the rotational speed and direction. - In a centrifuge context, a higher \( \omega \) results in a greater centrifugal force, affecting the fluid's behavior and pressure profile.
Understanding angular velocity helps in analyzing how it contributes to centrifugal forces that act on elements within the rotating system. Hence, it plays a fundamental role in describing the physics of any rotational system in this problem.

Incompressible Fluid

In physics, an incompressible fluid is one whose density remains constant regardless of changes in pressure. This idea simplifies many calculations because, while pressure may vary throughout the fluid due to external forces, its density does not.
- Incompressibility implies \( \frac{d\rho}{dp} = 0 \), meaning density \( \rho \) does not change with pressure. - This is an excellent approximation for liquids, especially under conditions such as those in this rotational problem.
In our exercise, you must assume that the fluid does not change its volume as it spins within the test tube. Thus, while the pressure might vary due to the centrifugal force from the rotation, the fluid's density \( \rho \) remains unchanged. This characteristic is essential when deriving pressure differences or calculating forces acting on the fluid.

Pressure Gradient

A pressure gradient refers to the rate of pressure change occurring over a specific distance. It’s a fundamental concept when analyzing fluid motion, particularly in a centrifuge. The pressure gradient drives fluid flow and is directly related to external forces such as gravity or, in this case, centrifugal forces.
- The pressure difference over a small element of the incompressible fluid is given by \( dp = \rho \omega^2 r' dr' \). - It indicates that within the spinning fluid, pressure increases as the radial distance \( r' \) increases from the axis of rotation. - The gradient exists because of the centripetal action derived from rotation at angular velocity \( \omega \).
Understanding this concept is essential in solving for how pressure varies with radius in the centrifugal field. By using this gradient, one can derive the pressure at different points, necessary for further calculations regarding the forces acting on objects immersed in the fluid.

Buoyancy in Centrifugal Field

Buoyancy typically refers to the upward force that a fluid exerts on an object submerged in it, influenced by gravity. In a centrifugal field, buoyancy is altered due to the centrifugal forces acting outwardly from the axis of rotation.
- Objects experience a net force from the pressure gradient, moving inward or outward depending on their density relative to the fluid. - For an object at distance \( R_{\text{cmob}} \), the radial force is \( \rho V \omega^2 R_{\text{cm}} \), which dictates movement direction based on density comparisons. - If \( \rho R_{\text{cm}} > \rho_{\text{ob}} R_{\text{cmob}} \), the object is pulled inward. If the situation is reversed, it moves outward.
This concept highlights the behavior of differing densities within the same rotating fluid. Objects denser than the fluid move outward, demonstrating a reverse effect of traditional buoyancy. It's a crucial dynamic for understanding how particles separate under centrifugation, as seen in ultracentrifuge applications.