Problem 75 - College Physics

Question

The viscous force on an oil drop is measured to be equal to \(3.0 \times 10^{-13} \mathrm{~N}\) when the drop is falling through air with a speed of \(4.5 \times 10^{-4} \mathrm{~m} / \mathrm{s}\). If the radius of the drop is \(2.5 \times 10^{-6} \mathrm{~m}\), what is the viscosity of air?

Step-by-Step Solution

Answer

The viscosity of air is \(1.69 \times 10^{-5} \, \mathrm{Pa} \, \mathrm{s}\).

Step 1: Understand the problem

The task is to determine the viscosity of air based on the viscous force acting on an oil droplet, the speed at which the droplet is falling, and the radius of the droplet. Stokes' law is used, which states that the force of viscosity \(F\) is equal to \(6 \pi \eta r v\), where \(\eta\) is the viscosity, \(r\) is the radius of the droplet and \(v\) is the velocity.

Step 2: Calculate viscosity

To solve for viscosity \(\eta\), need to rearrange the formula of Stokes' law to: \(\eta = F / (6 \pi r v)\). Here, \(F = 3 \times 10^{-13}\) N, \(r = 2.5 \times 10^{-6}\) m, and \(v = 4.5 \times 10^{-4}\) m/s. By substituting these values into the formula, we can calculate \(\eta\).

Step 3: Substitute the values into the formula

The values can now be inserted into the formula like so: \(\eta = (3.0 \times 10^{-13}) / (6 \pi \times 2.5 \times 10^{-6} \times 4.5 \times 10^{-4})\).

Step 4: Compute the final value

By performing the calculations, the viscosity of air \(\eta\) is found to be \(1.69 \times 10^{-5} \, \mathrm{Pa} \, \mathrm{s}\).

Key Concepts

Viscosity of Air

Viscosity, often described as the "thickness" of a fluid, refers to its resistance to flow. When we talk about the viscosity of air, we are referring to how resistant air is to motion or how "sticky" it behaves when particles move through it. Unlike liquids, air is a gas, and its viscosity is relatively much lower.

In our everyday experience, air feels easy to move through and doesn't offer much resistance. However, for very small particles like oil droplets, air's viscosity is significant enough to influence their movement. This property becomes crucial in fields like aerodynamics and meteorology.

Factors influencing air viscosity include temperature and pressure. Generally, as the temperature increases, the viscosity of air also increases because molecules move more vigorously, affecting the flow resistance.

Viscous Force

Viscous force is a type of friction that occurs in fluids. When an object moves through a fluid, such as air or oil, the fluid molecules exert a force opposing this motion, known as the viscous force. This force depends on how quickly the object moves, its size, and the fluid's viscosity.

Stokes' Law helps to determine this force quantitatively for small spherical objects (like our oil drop) moving through a viscous fluid. The law gives the formula for viscous force as:

• \( F = 6 \pi \eta r v \), where:
• \( F \) is the viscous force,
• \( \eta \) is the fluid's viscosity,
• \( r \) is the radius of the spherical object, and
• \( v \) is the object's velocity through the fluid.

Understanding viscous force is key for designing efficient systems in engineering and technology, such as automotive and aerospace components, where fluid interactions constantly occur.

Calculation of Viscosity

The calculation of viscosity is crucial for understanding how fluids behave, especially in engineering and physical sciences. Using Stokes' Law, we can determine the viscosity of a fluid by measuring the viscous force acting on a moving object, like our given oil drop.

In the example problem, we rearranged Stokes' formula to solve for viscosity \( \eta \):

• \( \eta = \frac{F}{6 \pi r v} \)

Given the values:

• Force \( F = 3.0 \times 10^{-13} \) N,
• Radius \( r = 2.5 \times 10^{-6} \) m, and
• Velocity \( v = 4.5 \times 10^{-4} \) m/s,

we substitute these into the equation:

• \( \eta = \frac{3.0 \times 10^{-13}}{6 \pi \times 2.5 \times 10^{-6} \times 4.5 \times 10^{-4}} \)

Finally, performing the calculations helps us ascertain the viscosity of air is approximately \( 1.69 \times 10^{-5} \, \mathrm{Pa} \, \mathrm{s} \). This result is essential in predicting how objects will behave when moving through the air.