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Chapter 9: Problem 30

A typical Major League fastball is thrown at approximately \(88 \mathrm{mph}\) and with a spin rate of \(110 \mathrm{rpm} .\) If the distance between the pitcher's point of release and the catcher's glove is exactly \(60.5 \mathrm{ft},\) how many full turns does the ball make between release and catch? Neglect any effect of gravity or air resistance on the ball's flight.

Short Answer

Expert verified
Answer: 0 full turns

Step by step solution

01

Convert mph to ft/s

Convert the velocity of the fastball from 88 mph to feet per second (ft/s). 1 mile = 5280 feet and 1 hour = 3600 seconds. So, we have the following conversion formula: \(88\frac{\text{miles}}{\text{hour}} \times \frac{5280\text{ feet}}{1\text{ mile}} \times \frac{1\text{ hour}}{3600\text{ seconds}}.\)
02

Calculate the fastball's velocity in ft/s

Now, calculate the fastball's velocity in ft/s. \(88 \frac{\text{miles}}{\text{hour}} \times \frac{5280\text{ feet}}{1\text{ mile}} \times \frac{1\text{ hour}}{3600\text{ seconds}}\approx 128.96\text{ ft/s}.\)
03

Convert rpm to turns/s

Convert the spin rate of the fastball from 110 rpm to turns per second (turns/s). 1 minute = 60 seconds. So, we have the following conversion formula: \(110\frac{\text{turns}}{\text{minute}} \times \frac{1\text{ minute}}{60\text{ seconds}}.\)
04

Calculate the spin rate in turns/s

Now, calculate the spin rate of the fastball in turns/s. \(110\frac{\text{turns}}{\text{minute}} \times \frac{1\text{ minute}}{60\text{ seconds}}\approx 1.83\text{ turns/s}\)
05

Find the travel time

Calculate the time it takes for the fastball to travel the given distance of 60.5 ft. Speed = distance / time Re-arrange for time: time = distance / speed \(\text{time} = \frac{60.5\text{ ft}}{128.96\text{ ft/s}}\approx 0.47\text{ seconds}\)
06

Calculate the total turns

Multiply the time it takes for the ball to travel the distance by the spin rate in turns/s to find out how many total turns the ball makes between release and catch. Total turns = turns/s × time \(\text{Total turns} = 1.83\text{ turns/s} \times 0.47\text{ seconds} \approx 0.86\text{ turns}\). Since only a whole number of full turns can be made, the number of full turns that the ball makes between release and catch is 0.

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Most popular questions from this chapter

A race car is making a U-turn at constant speed. The coefficient of friction between the tires and the track is \(\mu_{\mathrm{s}}=1.20 .\) If the radius of the curve is \(10.0 \mathrm{~m},\) what is the maximum speed at which the car can turn without sliding? Assume that the car is performing uniform circular motion.

Assuming that the Earth is spherical and recalling that latitudes range from \(0^{\circ}\) at the Equator to \(90^{\circ} \mathrm{N}\) at the North Pole, how far apart, measured on the Earth's surface, are Dubuque, Iowa \(\left(42.50^{\circ} \mathrm{N}\right.\) latitude \()\), and Guatemala City \(\left(14.62^{\circ} \mathrm{N}\right.\) latitude \() ?\) The two cities lie on approximately the same longitude. Do not neglect the curvature of the Earth in determining this distance.

A speedway turn, with radius of curvature \(R\), is banked at an angle \(\theta\) above the horizontal. a) What is the optimal speed at which to take the turn if the track's surface is iced over (that is, if there is very little friction between the tires and the track)? b) If the track surface is ice-free and there is a coefficient of friction \(\mu_{s}\) between the tires and the track, what are the maximum and minimum speeds at which this turn can be taken? c) Evaluate the results of parts (a) and (b) for \(R=400 . \mathrm{m}\), \(\theta=45.0^{\circ},\) and \(\mu_{\mathrm{s}}=0.700 .\)

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