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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 is approximately 147.25 m/s.

Step by step solution

01

Understanding Terminal Speed

Terminal speed is the constant speed that a freely falling object eventually reaches when the resistance from the medium through which it is falling prevents further acceleration. At terminal speed, the drag force is equal to the gravitational force on the object.
02

Formula for Terminal Speed

The formula for the terminal speed \( v_t \) of a spherical object falling through a fluid can be given by: \[ v_t = \sqrt{\frac{2mg}{\rho C_d A}} \]where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity, \( \rho \) is the density of the fluid, \( C_d \) is the drag coefficient, and \( A \) is the cross-sectional area of the object.
03

Cross-Sectional Area Calculation

The cross-sectional area \( A \) of the sphere can be calculated using the formula:\[ A = \pi r^2 \]where \( r \) is the radius of the sphere. Given the radius \( r = 3.00 \) cm, we first convert it to meters: \( r = 0.03 \) m.Therefore,\[ A = \pi (0.03)^2 = 2.827 \, \mathrm{cm}^2 \approx 2.827 \times 10^{-3} \, \mathrm{m}^2 \].
04

Inserting Known Values

Now, insert all known values into the terminal speed formula:- \( m = 6.00 \, \text{kg} \)- \( g = 9.81 \, \text{m/s}^2 \)- \( \rho = 1.20 \, \text{kg/m}^3 \)- \( C_d = 1.60 \)- \( A = 2.827 \times 10^{-3} \text{m}^2 \).This gives us:\[ v_t = \sqrt{\frac{2 imes 6.00 imes 9.81}{1.20 \times 1.60 \times 2.827 \times 10^{-3}}} \].
05

Calculating the Terminal Speed

Calculate the terminal speed using the inserted values:\[ v_t = \sqrt{\frac{117.72}{1.60 \times 3.3924 \times 10^{-3}}} \approx \sqrt{\frac{117.72}{0.005428}} \approx \sqrt{21686} \approx 147.25 \, \text{m/s}. \]
06

Conclusion

The terminal speed of the spherical ball is approximately \( 147.25 \) meters per second.

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Key Concepts

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

Drag Coefficient
When talking about a freely falling object in a fluid, we often mention its drag coefficient. The drag coefficient, denoted by \( C_d \), is a dimensionless number that characterizes the object's resistance to air or fluid flow. Think of it as a measure of how "slippery" or "sticky" the object is in the air.

A higher drag coefficient typically means that the object experiences more resistance as it moves through the fluid. This increase in resistance results in a slower movement. Conversely, an object with a lower drag coefficient will likely cut through the air more efficiently, leading to a faster terminal speed.

The shape and surface texture of an object significantly influence its drag coefficient. For example:
  • A smooth, streamlined object tends to have a lower \( C_d \).
  • An irregular or rough surface might increase \( C_d \).
Understanding \( C_d \) is essential when calculating terminal speeds because it directly impacts how quickly an object settles into its maximum speed during free fall.
Cross-Sectional Area
The cross-sectional area \( A \), in the context of a falling spherical object, is quite influential in determining its terminal speed. This area represents the size of the "face" of the object as it falls through the air.

Calculating the cross-sectional area of a sphere involves using the radius \( r \) of the sphere in the formula: \( A = \pi r^2 \). In our example, with a radius of 3.00 cm (or 0.03 m), the cross-sectional area comes out to approximately \( 2.827 \times 10^{-3} \) square meters.

The larger the cross-sectional area, the greater the surface for air resistance to act upon. This means that a larger area tends to slow down the object, lowering its terminal speed.
  • For spheres, increasing the radius increases the cross-sectional area significantly since it involves squaring the radius in the formula.
  • Acceleration and terminal speed are inversely related to this area.
Understanding cross-sectional area helps in comprehending how different shapes and sizes of objects will behave similarly or differently when falling through a fluid like air.
Density of Air
The density of air, denoted as \( \rho \), plays a crucial role in determining an object's terminal speed. Air density is the mass of air per unit volume and is influenced by factors like temperature and altitude. For most calculations near the Earth's surface, air density is approximately \( 1.20 \, \text{kg/m}^3 \).

Higher air density means there are more air molecules for a falling object to interact with, thereby increasing the drag force.
  • As density increases, the drag force acting against the object grows, and the terminal speed decreases.
  • In regions with thinner air, like at high altitudes, objects fall faster due to reduced air resistance.
Understanding air density is critical when predicting how fast an object will fall to the ground, especially in varied atmospheric conditions. It intricately ties into both the drag coefficient and cross-sectional area to define terminal speed.
Gravitational Force
Gravitational force is the force exerted by the Earth (or another large body) on an object due to gravity. This force is pivotal in the concept of terminal speed as it drives the object downward.

For an object with mass \( m \), the gravitational force \( F_g \) can be calculated using the formula: \( F_g = mg \), where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \) on Earth's surface). In this exercise, we have a mass of 6.00 kg, leading to a gravitational force of \( 6.00 \times 9.81 = 58.86 \, \text{N} \).

At terminal speed, the gravitational force is balanced by the drag force, meaning:
  • There is no net acceleration.
  • The object continues to fall at constant speed.
This equilibrium helps determine the terminal speed and is why understanding gravitational force is fundamental to comprehending free-falling motion in fluids.

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Most popular questions from this chapter

In about 1915, Henry Sincosky of Philadelphia suspended himself from a rafter by gripping the rafter with the thumb of each hand on one side and the fingers on the opposite side (Fig. 6-21). Sincosky's mass was \(79 \mathrm{~kg}\). If the coefficient of static friction between hand and rafter was \(0.70\), what was the least magnitude of the normal force on the rafter from each thumb or opposite fingers? (After suspending himself, Sincosky chinned himself on the rafter and then moved hand-over-hand along the rafter. If you do not think Sincosky's grip was remarkable, try to repeat his stunt.)

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