Question

A steel ball of diameter \(3.2 \mathrm{~mm}\) falls under gravity through an oil of density \(920 \mathrm{~kg} \mathrm{~m}^{-3}\) and viscosity \(1.64 \mathrm{Ns} \mathrm{m}^{-2}\). The density of steel. may be taken as \(7820 \mathrm{~kg} \mathrm{~m}^{-3}\). What is the terminal velocity of the ball ? (A) \(3.45 \mathrm{cms}^{-1}\) (B) \(4.25 \mathrm{cms}^{-1}\) (C) \(1.85 \mathrm{~cm}^{-1}\) (D) \(2.39 \mathrm{cms}^{-1}\)

Short answer

The terminal velocity of the steel ball is approximately \(2.47 \mathrm{~cm} \mathrm{~s}^{-1}\). The closest answer in the given options is (D) \(2.39 \mathrm{cms}^{-1}\).

Step-by-step solution

Calculate the volume and mass of the steel ball.

Since we are given the diameter of the steel ball, we can find the radius, and consequently, the volume of the steel ball. The formula for the volume of a sphere is: \[V = \frac{4}{3}\pi r^3\] where \(r\) is the radius of the sphere. Then, using the density of steel, we can calculate the mass of the steel ball. The formula for mass is: \[m = ρV\] where \(ρ\) is the density of the material. First, we calculate the radius (in meters): \(r = \frac{3.2}{2}\times10^{-3} \mathrm{~m} = 1.6\times10^{-3} \mathrm{~m}\) Now, let's find the volume of the ball: \[V = \frac{4}{3}\pi(1.6\times10^{-3})^3 = 1.717\times10^{-8} \mathrm{~m}^3\] And finally, we compute the mass of the steel ball: \[m = (7820 \mathrm{~kg} \mathrm{~m}^{-3})(1.717\times10^{-8} \mathrm{~m}^3) = 0.1342 \mathrm{~kg}\]

Calculate the buoyant force acting on the steel ball.

The buoyant force (\(F_b\)) is equal to the weight of the fluid (oil) displaced by the steel ball. We will use the formula for buoyant force: \[F_b = ρ_o V g\] where \(ρ_o\) is the density of the oil, \(V\) is the volume of the steel ball, and \(g\) is the gravitational acceleration (\(9.81 \mathrm{ms}^{-2}\)). Now, let's compute the buoyant force: \[F_b = (920 \mathrm{~kg} \mathrm{~m}^{-3})(1.717\times10^{-8} \mathrm{~m}^3)(9.81 \mathrm{~m} \mathrm{~s}^{-2}) = 0.0156 \mathrm{N}\]

Find the net force acting on the steel ball when it reaches terminal velocity.

At terminal velocity, the net force acting on the steel ball is equal to the drag force (\(F_d\)) exerted by the oil due to its viscosity. The net force (\(F_n\)) is given by: \[F_n = F_g - F_b\] where \(F_g\) is the gravitational force acting on the steel ball and can be calculated as: \[F_g = mg\] Now, let's compute the gravitational force: \[F_g = (0.1342 \mathrm{~kg})(9.81 \mathrm{~m} \mathrm{~s}^{-2}) = 1.316 \mathrm{N}\] Compute the net force: \[F_n = 1.316 \mathrm{N} - 0.0156 \mathrm{N} = 1.3004 \mathrm{N}\]

Use the net force to calculate the terminal velocity of the ball.

At terminal velocity, the net force acting on the ball is equal to the drag force exerted by the viscous oil according to the equation: \[F_n = 6\pi rηv\] where \(η\) is the viscosity of the oil and \(v\) is the terminal velocity. Rearrange equation to solve for \(v\): \[ v = \frac{F_n}{6\pi r η} \] Now, let's calculate the terminal velocity (\(v\)) of the steel ball: \[ v = \frac{1.3004 \mathrm{N}}{6 \pi (1.6\times10^{-3} \mathrm{~m})(1.64 \mathrm{~Ns} \mathrm{~m}^{-2})} = 0.0247 \mathrm{~m} \mathrm{~s}^{-1}\] To express the result in cm/s, multiply by 100: \[ v = 0.0247 \mathrm{~m} \mathrm{~s}^{-1} \times 100 = 2.47 \mathrm{~cm} \mathrm{~s}^{-1}\] The terminal velocity of the steel ball is approximately \(2.47 \mathrm{~cm} \mathrm{~s}^{-1}\). The closest answer in the given options is (D) \(2.39 \mathrm{cms}^{-1}\).

Key concepts

Buoyant Force

When an object is submerged in a fluid, it experiences an upward force called the buoyant force. This force is due to the pressure difference between the top and bottom of the object. Think of it as the fluid trying to push the object upwards, making it feel lighter. You can calculate the buoyant force using the formula:

\[F_b = \rho_o V g\]

Where:

• \(\rho_o\) = density of the fluid (oil, in our example)
• \(V\) = volume of the object
• \(g\) = acceleration due to gravity (approximately \(9.81\, \text{ms}^{-2}\))

In our exercise, the buoyant force is the force exerted by the oil on the steel ball. This force is smaller than the gravitational force due to the higher density of the steel compared to the oil.

Understanding the buoyant force is crucial when determining how objects will behave in fluids, such as whether they float or sink.

Gravitational Force

Gravity pulls everything towards the center of the Earth. This pull is known as the gravitational force. For an object, this force is the product of its mass and the acceleration due to gravity:

\[F_g = mg\]

Where:

• \(m\) = mass of the object
• \(g\) = acceleration due to gravity

The gravitational force acts downward, and it’s the primary force responsible for the object starting to move or fall freely. In the context of the steel ball in oil, this force is what initiates the fall of the ball, causing it to accelerate downwards before reaching terminal velocity.

Understanding gravitational force is key to calculating other forces, such as when evaluating the net force acting on a falling object.

Viscosity

Viscosity is a measure of a fluid's resistance to flow. It describes how "thick" or "thin" a fluid is. Imagine honey and water: honey is thicker and flows more slowly, so it has higher viscosity compared to water. When an object moves through a fluid, viscosity causes a drag force that opposes its motion.

In our exercise, viscosity of the oil influences how quickly the steel ball can move through it. Higher viscosity means the oil will more strongly resist the ball’s motion, reducing its speed. Viscosity (\(\eta\)) is important when calculating terminal velocity, as it's part of the equation determining the settling speed of the ball.

Drag Force

As objects move through a fluid, they encounter resistance from the fluid pushing back against their motion, known as drag force. This force opposes the gravitational force and impacts the rate at which an object moves.

For spheres, like the steel ball, the drag force (\(F_d\)) can be calculated using:

\[F_n = 6\pi r\eta v\]

Where:

• \(r\) = radius of the sphere
• \(\eta\) = viscosity of the fluid
• \(v\) = terminal velocity

When an object reaches terminal velocity, the drag force equals the net force of gravity minus buoyancy, and the object falls at a constant speed. Exploring the balance of these forces helps us understand why the ball eventually stops accelerating and moves at a steady velocity through the fluid.