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What is the terminal speed of a \(6.00 \mathrm{~kg}\) spherical ball that has a radius of \(3.00 \mathrm{~cm}\) and a drag coefficient of \(1.60 ?\) The density of the air through which the ball falls is \(1.20 \mathrm{~kg} / \mathrm{m}^{3}\).

Short Answer

Expert verified
The terminal speed of the ball is approximately 147.12 m/s.

Step by step solution

01

Gather and Understand Key Formula

To find the terminal speed of an object, we use the formula for terminal velocity: \[ v_t = \sqrt{\frac{2mg}{\rho C_d A}} \]. Here, \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (approximately \( 9.81 \ \mathrm{m/s^2} \)), \( \rho \) is the density of the fluid (air in this case), \( C_d \) is the drag coefficient, and \( A \) is the cross-sectional area of the object.
02

Calculate Cross-Sectional Area

The object is a spherical ball, so its cross-sectional area \( A \) is calculated using the formula for the area of a circle, \( A = \pi r^2 \), where \( r \) is the radius. The radius is given as 3.00 cm, which is equivalent to 0.03 m. Thus, \( A = \pi (0.03)^2 \approx 2.827 \times 10^{-3} \ \mathrm{m^2} \).
03

Substitute Values into Terminal Speed Formula

Now that we have all the necessary values, substitute them into the terminal speed formula: mass \( m = 6.00 \ \mathrm{kg} \), gravity \( g = 9.81 \ \mathrm{m/s^2} \), air density \( \rho = 1.20 \ \mathrm{kg/m^3} \), drag coefficient \( C_d = 1.60 \), and cross-sectional area \( A \approx 2.827 \times 10^{-3} \ \mathrm{m^2} \).
04

Compute Terminal Speed

Substitute the values into the formula: \[ v_t = \sqrt{\frac{2 \times 6.00 \times 9.81}{1.20 \times 1.60 \times 2.827 \times 10^{-3}}} \]. Simplifying inside the square root gives \( v_t = \sqrt{\frac{117.72}{0.0054384}} \approx \sqrt{21648.6} \approx 147.12 \ \mathrm{m/s} \).
05

Interpret and Conclude

After calculating, the terminal speed of the ball is approximately 147.12 meters per second. This velocity is the constant speed the ball reaches when the force of gravity is balanced by the drag force exerted by the air.

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Key Concepts

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

Drag Coefficient
The drag coefficient, denoted as \(C_d\), is a dimensionless number that describes how easily an object can move through a fluid, like air or water. It gauges the drag force, which opposes the object's motion. The value of \(C_d\) is determined by the shape and roughness of the object, as well as the flow conditions. A lower drag coefficient indicates less resistance, while a higher \(C_d\) suggests more resistance to motion. Understanding drag coefficient is crucial for calculating terminal velocity, as it directly influences the speed at which the gravitational force equals the drag force. In practical applications, designers modify shapes to achieve optimal \(C_d\) values, enhancing performance.
  • For streamlined objects, such as cars and airplanes, the drag coefficient is minimized.
  • Blunt objects, like a flat plate, have a higher drag coefficient due to greater air resistance.
Efficiently assessing \(C_d\) helps in predicting the terminal speed of objects like balls, cars, and even falling raindrops.
Cross-Sectional Area
The cross-sectional area, represented as \(A\), is the surface area of the cut made perpendicular to the object's motion. For a spherical ball, it is calculated using the formula \(A = \pi r^2\), where \(r\) is the radius of the sphere. This area is essential in determining the drag force experienced by an object. A larger cross-sectional area leads to a higher drag force as there is more surface for the air to exert its resistance upon.
  • For a circle, which represents the cross-section of a sphere, the equation involves the radius squared, multiplying by \(\pi\) (approximately 3.14159).
  • In practical situations, knowledge of cross-sectional area aids in navigation and reducing energy consumption by minimizing areas exposed to fluid flow.
When analyzing problems or optimizing designs, it’s important to accurately calculate \(A\) to achieve precise predictions and efficient outcomes.
Density of Air
The density of air, denoted as \(\rho\), refers to the mass of air per unit volume. Typically measured in kilograms per cubic meter (kg/m³), it varies based on atmospheric conditions such as temperature, humidity, and pressure. In the context of terminal velocity, air density plays a crucial role, as it is part of the drag force calculation. Higher air density means increased drag force acting against the object's motion, slowing it down.
  • At sea level, average air density is about \(1.225 \ \mathrm{kg/m^3}\) but can slightly vary.
  • In thinner air at high altitudes, objects experience less drag, potentially increasing terminal speed.
Understanding air density is key, not only for terminal velocity computations but also in fields like aerodynamics, meteorology, and aviation, where it affects lift, drag, and performance.
Acceleration Due to Gravity
Acceleration due to gravity, symbolized as \(g\), is the rate at which an object speeds up as it falls freely near the earth's surface. Its approximate value is \(9.81 \ \mathrm{m/s^2}\). This constant value plays a vital role in numerous physics calculations, including the derivation of terminal velocity. When an object falls, gravity accelerates it downward, counteracted by resistive forces such as drag.
  • The value of \(g\) can vary slightly, depending on geographic location and altitude.
  • In physics problems, \(g\) is often rounded to \(10 \ \mathrm{m/s^2}\) for simplicity in calculations.
Knowing \(g\) is essential for predicting the motion of falling objects and understanding gravitational influences in various scientific and engineering fields."

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Most popular questions from this chapter

A student wants to determine the coefficients of static friction and kinetic friction between a box and a plank. She places the box on the plank and gradually raises one end of the plank. When the angle of inclination with the horizontal reaches \(30^{\circ}\), the box starts to slip, and it then slides \(2.5 \mathrm{~m}\) down the plank in \(4.0 \mathrm{~s}\) at constant acceleration. What are (a) the coefficient of static friction and (b) the coefficient of kinetic friction between the box and the plank?

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of \(0.25\) with the floor. If the train is initially moving at a speed of \(48 \mathrm{~km} / \mathrm{h}\), in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?

A person pushes horizontally with a force of \(220 \mathrm{~N}\) on a \(55 \mathrm{~kg}\) crate to move it across a level floor. The coefficient of kinetic friction between the crate and the floor is \(0.35 .\) What is the magnitude of (a) the frictional force and (b) the acceleration of the crate?

A circular curve of highway is designed for traffic moving at \(60 \mathrm{~km} / \mathrm{h}\). Assume the traffic consists of cars without negative lift. (a) If the radius of the curve is \(150 \mathrm{~m}\), what is the correct angle of banking of the road? (b) If the curve were not banked, what would be the minimum coefficient of friction between tires and road that would keep traffic from skidding out of the turn when traveling at \(60 \mathrm{~km} / \mathrm{h} ?\)

A bedroom bureau with a mass of \(45 \mathrm{~kg}\), including drawers and clothing, rests on the floor. (a) If the coefficient of static friction between the bureau and the floor is \(0.45\), what is the magnitude of the minimum horizontal force that a person must apply to start the bureau moving? (b) If the drawers and clothing, with \(17 \mathrm{~kg}\) mass, are removed before the bureau is pushed, what is the new minimum magnitude?

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