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Chapter 7: Problem 49

radial and tangential components) just before it is released.One method of pitching a softball is called the "windmill" delivery method, in which the pitcher's arm rotates through approximately \(360^{\circ}\) in a vertical plane before the 198-gram ball is released at the lowest point of the circular motion. An experienced pitcher can throw a ball with a speed of \(98.0 \mathrm{mi} / \mathrm{h}\). Assume the angular acceleration is uniform throughout the pitching motion and take the distance between the softball and the shoulder joint to be \(74.2 \mathrm{~cm}\). (a) Determine the angular speed of the arm in rev/s at the instant of release. (b) Find the value of the angular acceleration in \(\mathrm{rev} / \mathrm{s}^{2}\) and the radial and tangential acceleration of the ball just before it is released. (c) Determine the force exerted on the ball by the pitcher's hand (both

Short Answer

Expert verified
The estimated angular speed is 59 rad/s, the angular acceleration depends on the time it takes for one revolution (t), the radial acceleration is calculated to be \( (59)^2 * 0.742 \ m/s^2\) and the tangential acceleration is \( \frac{59}{t} * 0.742 \ m/s^2\). The force exerted on the ball also depends on the time (t).

Step by step solution

01

Convert the ball's speed

First, let's convert the speed from miles per hour (mi/h) to meters per second (m/s), the unit more commonly used in physics. 1 mile is approximately 1609 meters and 1 hour is 3600 seconds, therefore, \(98.0 \ \mathrm{mi/h} \approx 43.7 \ \mathrm{m/s}\).
02

Calculate the angular speed

Now, calculate the angular speed, which is the speed divided by the radius of the circle. In this case, the radius is the distance from the shoulder joint to the ball, which is given as 74.2 cm or 0.742 m. Hence, the angular speed \( \omega \) is \( \omega = \frac{43.7}{0.742} \ \mathrm{rad/s} \approx 59 \ \mathrm{rad/s}\). Remember that the angular speed is usually expressed in rad/s for SI consistency.
03

Calculate the angular acceleration

The angular acceleration is the change in angular speed per unit time. Since the arm starts at rest and takes a whole rotation (\(2\pi \ rad\)) to reach the final angular speed, and given that the angular acceleration is assumed to be uniform, the time of one revolution \(t\) can be expressed as \( t = \frac{2 \pi}{\omega} \). Thus, the angular acceleration \( \alpha \) is \( \alpha = \frac{\omega}{t} = \frac{59}{t} \ rad/s^2\).
04

Calculate the radial and tangential accelerations

The radial (centripetal) acceleration \(a_r\) is given by \(a_r = \omega^2 * r = {(59)^2 * 0.742} \ m/s^2\). The tangential acceleration \(a_t\) is \( \alpha * r = \frac{59}{t} * 0.742 \ m/s^2\).
05

Calculate the force exerted on the ball

The last step is to calculate the force exerted on the ball by the pitcher. This can be expressed as the sum of the radial and tangential forces (\(F = F_r + F_t\)), which are the product of the mass of the ball and the respective acceleration (remember that \(F = m * a\)). Hence, \(F = m * a_r + m * a_t\).

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Key Concepts

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

Angular Speed
Angular speed is a key concept in understanding how fast an object rotates around a fixed point. In the case of a softball pitcher using the "windmill" technique, the arm moves in a circular path. The angular speed, denoted by \( \omega \), is calculated by dividing the linear speed of the ball by the radius of the circular path. This provides a measure of how quickly the arm is rotating in radians per second (rad/s).
  • To calculate angular speed, use the formula \( \omega = \frac{v}{r} \), where \( v \) is the linear speed and \( r \) is the radius.
Angular speed helps in predicting the velocity at which the ball will be released, crucial for knowing how fast it will travel once it leaves the pitcher's hand.
Angular Acceleration
Angular acceleration, represented by \( \alpha \), is the rate at which the angular speed changes with respect to time. In our exercise, the pitcher's arm starts from rest, meaning the initial angular speed is zero. As the arm completes the circular path, it reaches its final angular speed at the point of release.
  • To find angular acceleration, you need to know the final angular speed \( \omega_f \) and the time taken to reach that speed \( t \): \( \alpha = \frac{\omega_f}{t} \).
  • The time \( t \) for one full revolution can be calculated by \( t = \frac{2\pi}{\omega_f} \), given uniform angular acceleration.
Understanding angular acceleration is essential for determining how the pitcher's arm increases its speed uniformly as it approaches the release point.
Radial and Tangential Acceleration
In rotational motion, like the pitcher's arm, we examine two types of accelerations: radial and tangential. These accelerations provide a complete picture of the forces at play just before the ball is released.
  • Radial acceleration \( a_r \) is also known as centripetal acceleration, aimed at keeping the object in a circular path. It's given by \( a_r = \omega^2 \times r \), where \( \omega \) is the angular speed and \( r \) is the radius.
  • Tangential acceleration \( a_t \) is responsible for the change in speed along the path and is calculated with \( a_t = \alpha \times r \), where \( \alpha \) is the angular acceleration.
Both types are crucial for determining how the ball achieves the necessary speed and direction when it leaves the pitcher's hand.
Centripetal Force
Centripetal force is vital for maintaining an object in a circular path and is directly related to radial acceleration. For a softball undergoing a windmill pitch, centripetal force ensures that it travels in the intended circular motion before release.
  • Calculate centripetal force \( F_c \) using the formula \( F_c = m \times a_r \), where \( m \) is the mass of the ball and \( a_r \) is the radial acceleration.
Centripetal force is exerted by the pitcher's hand on the ball, allowing it to travel smoothly in a circle. Once the ball is released, this force ceases to act, and the ball follows a linear path determined by its velocity at the point of release.

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Most popular questions from this chapter

A coin rests \(15.0 \mathrm{~cm}\) from the center of a turntable. The coefficient of static friction between the coin and turntable surface is \(0.350\). The turntable starts from rest at \(t=0\) and rotates with a constant angular acceleration of \(0.730 \mathrm{rad} / \mathrm{s}^{2}\). (a) Once the turntable starts to rotate, what force causes the centripetal acceleration when the coin is stationary relative to the turntable? Under what condition does the coin begin to move relative to the turntable? (b) After what period of time will the coin start to slip on the turntable?

Because of Earth's rotation about its axis, a point on the equator has a centripetal acceleration of \(0.0340 \mathrm{~m} / \mathrm{s}^{2}\), whereas a point at the poles has no centripetal acceleration. (a) Show that, at the equator,the gravitational force on an object (the object's true weight) must exceed the object's apparent weight. (b) What are the apparent weights of a \(75.0-\mathrm{kg}\) person at the equator and at the poles? (Assume Earth is a uniform sphere and take \(g=9.800 \mathrm{~m} / \mathrm{s}^{2}\).)

A digital audio compact disc carries data along a continuous spiral track from the inner circumference of the disc to the outside edge. Each bit occupies \(0.6 \mu \mathrm{m}\) of the track. A CD player turns the disc to carry the track counterclockwise above a lens at a constant speed of \(1.30 \mathrm{~m} / \mathrm{s}\). Find the required angular speed (a) at the beginning of the recording, where the spiral has a radius of \(2.30 \mathrm{~cm}\), and (b) at the end of the recording, where the spiral has a radius of \(5.80 \mathrm{~cm}\). (c) A full-length recording lasts for \(74 \mathrm{~min}, 33 \mathrm{~s}\). Find the average angular acceleration of the disc. (d) Assuming the acceleration is constant, find the total angular displacement of the disc as it plays. (e) Find the total length of the track.

A dentist's drill starts from rest. After \(3.20 \mathrm{~s}\) of constant angular acceleration, it turns at a rate of \(2.51 \times\) \(10^{4} \mathrm{rev} / \mathrm{min}\). (a) Find the drill's angular acceleration. (b) Determine the angle (in radians) through which the drill rotates during this period.

An air puck of mass \(m_{1}=0.25 \mathrm{~kg}\) is tied to a string and allowed to revolve in a circle of radius \(R=1.0 \mathrm{~m}\) on a frictionless horizontal table. The other end of the string passes through a hole in the center of the table, and a mass of \(m_{2}=1.0 \mathrm{~kg}\) is tied to it (Fig. P7.27). The suspended mass remains in equilibrium while the puck on the tabletop revolves. (a) What is the tension in the string? (b) What is the horizontal force acting on the puck? (c) What is the speed of the puck? the horizontal force acting on the puck? (c) What is the the horizontal force acting on the puck? (c) What is the speed of the puck?
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