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Chapter 10: Problem 5

Firing tests on a \(10-\mathrm{mm}\) -diameter bullet with a mass of \(20 \mathrm{~g}\) show that in standard air, the bullet speed decreases from \(300 \mathrm{~m} / \mathrm{s}\) to \(180 \mathrm{~m} / \mathrm{s}\) over a distance of \(200 \mathrm{~m}\). Estimate the average drag coefficient of the bullet. Neglect compressibility effects.

Short Answer

Expert verified
The average drag coefficient \(C_d\) is approximately 0.295.

Step by step solution

01

Identify Key Variables

First, define the key variables given in the problem: Initial speed \(v_i = 300 \, \text{m/s}\), final speed \(v_f = 180 \, \text{m/s}\), distance \(d = 200 \, \text{m}\), bullet diameter \(D = 10 \, \text{mm} = 0.01 \, \text{m}\), and bullet mass \(m = 20 \, \text{g} = 0.02 \, \text{kg}\).
02

Calculate Average Deceleration

Use the kinematic equation to find the average deceleration \(a\): \[ v_f^2 = v_i^2 + 2a \cdot d \]Rearranging this for \(a\):\[ a = \frac{v_f^2 - v_i^2}{2d} \]Substitute the known values:\[ a = \frac{(180)^2 - (300)^2}{2 \times 200} = \frac{32400 - 90000}{400} = -144 \cdot 0.5 = -90 \text{ m/s}^2 \]
03

Relate Force to Acceleration

Using Newton's second law, relate the drag force \(F_d\) to the mass and acceleration:\[ F_d = m \cdot a \]Substituting the values:\[ F_d = 0.02 \, \text{kg} \times (-90 \, \text{m/s}^2) = -1.8 \, \text{N} \]
04

Use Drag Force and Velocity to Find Drag Coefficient

The drag force can also be expressed as:\[ F_d = \frac{1}{2} \cdot \rho \cdot v^2 \cdot C_d \cdot A \]where \(\rho = 1.225 \, \text{kg/m}^3\) is the density of air, \(v = \frac{v_i + v_f}{2} = 240 \, \text{m/s}\) is the average velocity, and \(A = \pi \left(\frac{D}{2}\right)^2\) is the cross-sectional area.Calculate \(A\):\[ A = \pi \left(0.005\right)^2 = 7.85 \times 10^{-5} \, \text{m}^2 \]Substitute these into the drag force equation and solve for \(C_d\):\[ 1.8 = \frac{1}{2} \cdot 1.225 \cdot (240)^2 \cdot C_d \cdot 7.85 \times 10^{-5} \]Solve for \(C_d\):\[ C_d = \frac{1.8}{0.5 \times 1.225 \times 57600 \times 7.85 \times 10^{-5}} = 0.295 \]
05

Conclusion

The calculated drag coefficient \(C_d\) for the bullet is approximately 0.295. This is a dimensionless number representing the bullet's aerodynamic efficiency.

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Key Concepts

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

Bullet Trajectory
Understanding bullet trajectory is crucial in ballistics studies. It involves analyzing how a bullet travels through the air after being fired. The path that the bullet follows is known as its trajectory. This trajectory is influenced by several factors:
  • Initial velocity: How fast the bullet is moving when it leaves the gun.
  • Angle of launch: The direction in which the bullet is fired.
  • Environmental factors: Wind, humidity, and temperature can affect the path.
  • Gravity: It constantly pulls the bullet downwards.
  • Drag: The resistance experienced by the bullet as it moves through the air.
In the given exercise, we observe that the bullet speed reduces significantly over a distance. The initial and final speeds, along with the distance covered, help determine how drag affects the bullet's trajectory. As the bullet moves forward, drag causes it to lose speed, altering its path.
Aerodynamics
Aerodynamics plays a significant role in understanding how objects move through the air. It refers to the forces that impact an object as it flies. These forces are:
  • Lift: Acts perpendicular to the direction of motion.
  • Drag: Acts opposite to the direction of motion and is a primary focus here.
  • Weight: Due to gravity, pulls the object downwards.
  • Thrust: Works to propel the object forwards, typically in aviation, but in bullets, it's the initial firing action.
For a bullet, aerodynamics comes into play primarily through drag. The drag coefficient, which was calculated as approximately 0.295 in the given exercise, is a measure of aerodynamic resistance. A lower drag coefficient means the object can move more efficiently through the air by facing less resistance. Both the shape and speed of a bullet influence this coefficient, impacting its overall flight performance.
Newton's Second Law
Newton's Second Law of Motion is pivotal in understanding how forces affect motion. The law states that the force acting on an object is equal to the mass of that object multiplied by its acceleration:\[ F = m \cdot a \] Here's what each component means:
  • Force (F): A push or pull acting on an object. In this case, it's the drag force slowing down the bullet.
  • Mass (m): The amount of matter in an object. The bullet in the exercise has a mass of 0.02 kg.
  • Acceleration (a): The rate of change of velocity. For the bullet, it's the deceleration caused by drag.
This law helps us calculate the drag force as the bullet moves. Using the bullet's mass and the calculated deceleration, the exercise showed the drag force of approximately -1.8 N. Newton's Second law bridges the motion and the forces, allowing us to understand how drag influences speed reduction.
Kinematic Equations
Kinematic equations describe the motion of objects without considering the forces that cause them. They are essential for predicting future motion based on current velocity, acceleration, and time. In the context of the given problem, the kinematic equation used is:\[ v_f^2 = v_i^2 + 2a \cdot d \]This equation relates:
  • Final velocity \(v_f\): Speed of the bullet after covering the distance, 180 m/s.
  • Initial velocity \(v_i\): Speed of the bullet at the start, 300 m/s.
  • Acceleration \(a\): Change in velocity over time, found to be -90 m/s虏.
  • Distance \(d\): The length over which the bullet is analyzed, 200 m.
By rearranging this equation, we find the average deceleration of the bullet, which aids in further calculations. The use of kinematic equations simplifies understanding how the bullet's velocity changes over the specified distance.

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Most popular questions from this chapter

A small military helicopter has four blades with a rotor diameter of \(10.70 \mathrm{~m},\) and each blade has a width of \(0.73 \mathrm{~m}\). When the helicopter is flying at normal speed, the blades rotate at \(360 \mathrm{rpm} .\) The standard engine on the helicopter has a rated power of \(606 \mathrm{~kW}\). Assuming that the blades can be treated as flat plates for the purpose of estimating the frictional force that must be overcome in turning the blades, estimate the power required to turn the blades. What percentage of the engine power is used to turn the blades? Assume standard air in your analysis.

A balloon is filled with helium and is supported by a light string. The diameter of the balloon is \(650 \mathrm{~mm}\), and the tension in the string is measured as \(2.5 \mathrm{~N}\) when the balloon is in standard air at sea level. If a wind speed of \(5 \mathrm{~m} / \mathrm{s}\) is imposed on the balloon and the string deflects \(45^{\circ}\) from the vertical, estimate the drag coefficient of the balloon.

A blimp that is used for sightseeing has a length of \(58.5 \mathrm{~m}\), has a maximum diameter of \(15.2 \mathrm{~m},\) and is powered by two \(157 \mathrm{~kW}\) engines. The drag coefficients of blimps are generally estimated to be in the range of \(0.020-0.025 .\) If all of the engine power was used in overcoming hydrodynamic drag, what would be the maximum speed of the blimp? Assume standard air.

An aircraft has a wing planform area of \(180 \mathrm{~m}^{2}\), an aspect ratio of \(7.5,\) a zero-lift drag coefficient of \(0.0185,\) and a weight of \(800 \mathrm{kN}\) when fully loaded. Estimate the speed of the aircraft that will minimize the required engine thrust when the aircraft is flying under standard sea-level atmospheric conditions.

A prototype sports car has an engine that can deliver \(360 \mathrm{~kW}\) of power. The shape of the car is such that is has an estimated drag coefficient of 0.20 and a frontal area of \(2.50 \mathrm{~m}^{2}\). If the car is to be tested on a track at sea level under standard conditions, estimate the maximum possible speed the car can attain.
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