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A quarterback throws a pass that is a perfect spiral. In other words, the football does not wobble, but spins smoothly about an axis passing through each end of the ball. Suppose the ball spins at \(7.7 \mathrm{rev} / \mathrm{s}\). In addition, the ball is thrown with a linear speed of \(19 \mathrm{m} / \mathrm{s}\) at an angle of \(55^{\circ}\) with respect to the ground. If the ball is caught at the same height at which it left the quarterback's hand, how many revolutions has the ball made while in the air?

Short Answer

Expert verified
The ball makes approximately 24.49 revolutions while in the air.

Step by step solution

01

Calculate the Time of Flight

The time of flight for projectile motion when the projectile lands at the same height it was launched from can be calculated using the formula \( t = \frac{2v_0 \sin \theta}{g} \), where \( v_0 = 19 \, \text{m/s} \) is the linear speed, \( \theta = 55^{\circ} \) is the launch angle, and \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity. Convert the angle to radians: \( \theta = 55^{\circ} \times \frac{\pi}{180} = 0.9599 \, \text{radians} \). Substituting the values, \( t = \frac{2 \times 19 \times \sin(0.9599)}{9.8} = \frac{2 \times 19 \times 0.8192}{9.8} \approx 3.18 \, \text{seconds} \).
02

Calculate the Number of Revolutions During Flight

The number of revolutions is given by \( ext{Number of Revolutions} = ( ext{Angular Velocity}) \times ( ext{Time of Flight}) \). The angular velocity is \( 7.7 \, \text{rev/s} \). The time of flight is \( 3.18 \, \text{sec} \) (from Step 1). Therefore, \( ext{Number of Revolutions} = 7.7 \, \text{rev/s} \times 3.18 \, \text{s} \approx 24.486 \, \text{revolutions} \).

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Key Concepts

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

Angular Velocity
Angular velocity is a crucial concept when studying rotational motion. It describes how quickly an object spins around an axis. In our exercise, the football spins with an angular velocity of \(7.7 \text{ rev/s}\), which means it makes 7.7 complete turns around its central axis every second. This rotational speed remains constant throughout the ball's flight unless any external forces act on it to change its rate of spin.
Angular velocity is generally measured in radians per second (rad/s), but in this problem, it's given in revolutions per second (rev/s). Remember that one full revolution equals \(2\pi\) radians. Thus, if needed, you can convert between these units by multiplying the number of revolutions by \(2\pi\) to get radians.
Some key points to understand about angular velocity:
  • It is a vector quantity, meaning it has both magnitude and direction. The direction is determined by the right-hand rule.
  • For a perfectly spiraling football, the axis is along the length of the ball itself.
  • In our context, angular velocity helps us calculate how many spins the ball completes during its flight.
Time of Flight
Time of flight is a fundamental aspect of projectile motion, which explains how long an object remains airborne. When the projectile launches and lands at the same height, we can use a special formula to determine this time. The equation is \( t = \frac{2v_0 \sin \theta}{g} \). Here, \(v_0\) represents the initial speed, \(\theta\) is the launch angle, and \(g\) is the acceleration due to gravity.
In our football problem, we have:
  • Initial speed \(v_0 = 19 \text{ m/s}\)
  • Launch angle \(\theta = 55^\circ\), which is converted to radians for calculations.
  • Gravity \(g = 9.8 \text{ m/s}^2\).
Plugging these values into the formula gives a time of flight, \(t \approx 3.18 \text{ seconds}\). This helps us understand how long the ball stays in the air from the moment it leaves the quarterback's hand until it is caught. With time of flight, we can explore further calculations like horizontal distance or, as in this case, the number of revolutions.
Number of Revolutions
The number of revolutions gives us a sense of how many complete turns an object makes as it moves through the air. In the context of our spiraling football, it allows us to visualize its spinning motion from launch to catch.
To find the number of revolutions a projectile makes while in the air, multiply the angular velocity by the time of flight. Using the equation \( \text{Number of Revolutions} = \text{Angular Velocity} \times \text{Time of Flight} \), we have:
  • Angular velocity is \(7.7 \text{ rev/s}\)
  • Time of flight is \(3.18 \text{ seconds}\)
Multiplying these gives us approximately \(24.486\) revolutions.
Keep these points in mind:
  • The longer the time of flight, the more revolutions the object tends to make.
  • The higher the angular velocity, the more revolutions occur in a given period.
  • This calculation helps understand how angular motion relates to projectile motion in practical scenarios like sports or physics experiments.

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Most popular questions from this chapter

The wheels of a bicycle have an angular velocity of \(+20.0 \mathrm{rad} / \mathrm{s}\). Then, the brakes are applied. In coming to rest, each wheel makes an angular displacement of +15.92 revolutions. (a) How much time does it take for the bike to come to rest? (b) What is the angular acceleration (in \(\left.\mathrm{rad} / \mathrm{s}^{2}\right)\) of each wheel?

An automobile tire has a radius of \(0.330 \mathrm{m},\) and its center moves forward with a linear speed of \(v=15.0 \mathrm{m} / \mathrm{s}\) (a) Determine the angular speed of the wheel. (b) Relative to the axle, what is the tangential speed of a point located \(0.175 \mathrm{m}\) from the axle?

A racing car, starting from rest, travels around a circular turn of radius \(23.5 \mathrm{m} .\) At a certain instant, the car is still accelerating, and its angular speed is \(0.571 \mathrm{rad} / \mathrm{s}\). At this time, the total acceleration (centripetal plus tangential) makes an angle of \(35.0^{\circ}\) with respect to the radius. (The situation is similar to that in Figure \(8.12 b .\) ) What is the magnitude of the total acceleration?

Two Formula One racing cars are negotiating a circular turn, and they have the same centripetal acceleration. However, the path of car A has a radius of \(48 \mathrm{m},\) while that of \(\operatorname{car} \mathrm{B}\) is \(36 \mathrm{m} .\) Determine the ratio of the angular speed of car A to the angular speed of car B.

In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates in a horizontal circle. A trainee riding in the chamber of a centrifuge rotating with a constant angular speed of 2.5 rad/s experiences a centripetal acceleration of 3.2 times the acceleration due to gravity. In a second training exercise, the centrifuge speeds up from rest with a constant angular acceleration. When the centrifuge reaches an angular speed of \(2.5 \mathrm{rad} / \mathrm{s},\) the trainee experiences a total acceleration equal to 4.8 times the acceleration due to gravity. (a) How long is the arm of the centrifuge? (b) What is the angular acceleration of the centrifuge in the second training exercise?

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