Problem 57 - College Physics

Question

What speed must a gold sphere of radius 3.00 \(\mathrm{mm}\) have in castor oil for the viscous drag force to be one-fourth of the weight of the sphere? The density of gold is \(19,300 \mathrm{kg} / \mathrm{m}^{3}\) and the viscosity of the oil is 0.986 \(\mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\)

Step-by-Step Solution

Answer

The speed must be approximately 0.095 m/s.

Step 1: Calculate the Volume of the Sphere

To find the volume of the sphere, use the formula for the volume of a sphere: \( V = \frac{4}{3} \pi r^3 \). The radius \( r \) is given as 3 mm, which is equivalent to 0.003 m. So, the volume \( V \) will be:\[ V = \frac{4}{3} \pi (0.003)^3 = 1.13 \times 10^{-7} \ \text{m}^3 \]

Step 2: Calculate the Mass of the Sphere

The mass \( m \) of the sphere can be calculated using its density \( \rho\,=\,19,300 \,\text{kg/m}^3 \) and volume \( V \) from Step 1:\[ m = \rho \times V = 19,300 \times 1.13 \times 10^{-7} = 2.18 \times 10^{-3} \ \text{kg} \]

Step 3: Calculate the Weight of the Sphere

The weight \( W \) of the sphere can be found from its mass using the equation \( W = mg \), where \( g \) is the acceleration due to gravity (approximately \( 9.81 \ \text{m/s}^2 \)).\[ W = 2.18 \times 10^{-3} \times 9.81 = 0.0214 \ \text{N} \]

Step 4: Determine the Desired Drag Force

We're asked to find the speed where the viscous drag force is one-fourth of the sphere's weight. So, \( F_d = \frac{1}{4}W \).\[ F_d = \frac{1}{4} \times 0.0214 = 0.00535 \ \text{N} \]

Step 5: Use Stokes' Law to Find the Speed

Stokes' Law for the drag force on a sphere in a viscous medium is given by \( F_d = 6 \pi \eta r v \), where \( \eta \) is the viscosity and \( v \) is the speed. Rearrange to solve for \( v \):\[ v = \frac{F_d}{6 \pi \eta r} \]Substitute \( \eta = 0.986 \ \text{N} \cdot \text{s} / \text{m}^2 \), \( r = 0.003 \ \text{m} \), and \( F_d = 0.00535 \ \text{N} \) into the formula:\[ v = \frac{0.00535}{6 \pi \times 0.986 \times 0.003} \approx 0.095 \ \text{m/s} \]

Key Concepts

Viscous Drag Force

When an object moves through a fluid, like a sphere moving in castor oil, it experiences resistance from the fluid. This resistance is known as the viscous drag force. It acts opposite to the direction of the object's motion, effectively slowing it down. The amount of drag force depends on several factors, including the object's speed, the viscosity of the fluid, and the shape and size of the object. In our problem, the drag force is required to be one-fourth of the sphere's weight for a gold sphere moving through castor oil. This means the sphere's motion is significantly slowed by the fluid, which exhibits a viscosity value that indicates how 'thick' or 'sticky' the fluid is. By controlling the speed and measuring the drag force, we can understand how the object and fluid interact.

Stokes' Law

Stokes' Law is a critical concept that helps us calculate the viscous drag force on small spherical objects moving through a fluid. It states that the drag force is proportional to the radius of the sphere, the fluid's viscosity, and the sphere's velocity. The mathematical form of Stokes' Law is given by:

• \( F_d = 6 \pi \eta r v \)

Here, \( F_d \) represents the drag force, \( \eta \) is the dynamic viscosity of the fluid, \( r \) is the radius of the sphere, and \( v \) is the velocity of the sphere. In our context, knowing the desired drag force as one-fourth of the sphere's weight allows us to rearrange the formula and find the required speed that achieves this condition. This law is especially useful when dealing with small particles or droplets in a viscous medium.

Sphere Volume Calculation

To find the volume of a sphere, we have a simple and well-known formula:

• \( V = \frac{4}{3} \pi r^3 \)

This formula calculates how much space the sphere occupies. In the given exercise, the gold sphere has a radius of 3 mm, which needs to be converted to meters (0.003 m) for the calculation. After substituting this into the formula, we find that the sphere's volume is approximately \( 1.13 \times 10^{-7} \ \text{m}^3 \). This volume measurement is crucial for determining other related physical properties of the sphere, such as its mass when combined with its density. It's always important to ensure units are consistent, especially when converting measurements from millimeters to meters.

Density and Mass Relationship

Understanding the relationship between density and mass is critical when dealing with physical objects. Density tells us how much mass is contained within a certain volume. The formula to find mass based on density and volume is:

• \( m = \rho \times V \)

In this scenario, knowing the density of gold as \( 19,300 \ \text{kg/m}^3 \) allows us to calculate the mass of a gold sphere once we have its volume. By multiplying the density by the sphere's volume \( 1.13 \times 10^{-7} \ \text{m}^3 \), we get the mass approximately \( 2.18 \times 10^{-3} \ \text{kg} \). This mass is then used to determine the weight of the sphere, further allowing us to calculate and compare forces acting on it. Understanding this relationship is fundamental in physics, aiding in solving various problems involving motion and forces.