/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 108 A wheel is rotating at\(200 \mat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 108

A wheel is rotating at\(200 \mathrm{rev} / \mathrm{sec}\). Find the angular velocity in radians per minute (to the nearest tenth).

Short Answer

Expert verified
75398.4 rad/min

Step by step solution

01

- Convert Revolutions per Second to Revolutions per Minute

First, convert the angular velocity from revolutions per second to revolutions per minute. Since there are 60 seconds in a minute, multiply the given revolutions per second by 60. \[ 200 \text{ rev/sec} \times 60 = 12000 \text{ rev/min} \]
02

- Convert Revolutions per Minute to Radians per Minute

Next, convert the angular velocity from revolutions per minute (rev/min) to radians per minute (rad/min). Since one revolution is equivalent to \(2\pi\) radians, multiply the revolutions per minute by \(2\pi\). \[ 12000 \text{ rev/min} \times 2\pi \text{ rad/rev} = 24000\pi \text{ rad/min} \]
03

- Approximate the Value

Finally, approximate the product \(24000\pi\) to the nearest tenth. Using the approximation \(\pi \approx 3.1416\): \[ 24000 \times 3.1416 \approx 75398.4 \text{ rad/min} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Revolutions per Minute
Revolutions per minute, often abbreviated as RPM, is a unit of angular velocity. It indicates how many times an object rotates completely (360 degrees) in one minute. For example, if a wheel spins at 60 RPM, it completes 60 full turns every minute. Understanding RPM is fundamental when dealing with rotating objects like wheels, motors, and gears. It provides a simple way to quantify and compare their speeds. To solve the exercise, the first step involves converting revolutions per second to revolutions per minute by using the relationship between seconds and minutes. Since there are 60 seconds in a minute, simply multiply the given value in rev/sec by 60.
Radians per Minute
Radians per minute is another way to express angular velocity. A radian is a measure of the angle formed when the radius of a circle is wrapped around its circumference. Since there are \(2\pi\) radians in one full circle (which is one revolution), converting from revolutions per minute to radians per minute involves multiplying by \(2\pi\). This conversion helps when dealing with mathematical equations and functions involving angles, since radians are the preferred unit in mathematics due to their direct relationship with the radius and arc length of a circle. To complete the exercise, the given RPM value needs to be multiplied by \(2\pi\) to obtain the angular velocity in radians per minute.
Unit Conversion
Unit conversion is a crucial skill in physics and engineering. It involves converting a given quantity from one unit to another by using conversion factors. These factors are determined based on known relationships between units. In the exercise, we performed two key unit conversions:
  • Converting from revolutions per second to revolutions per minute using the factor of 60 (since 1 minute = 60 seconds).
  • Converting from revolutions per minute to radians per minute using the factor of \(2\pi\) (since 1 revolution = \(2\pi\) radians).
The final angular velocity was approximated to the nearest tenth for practical usage. Mastering unit conversions allows you to switch between different measurement systems and understand diverse scientific data accurately.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Area of a Slice of Pizza A slice of pizza with a central angle of \(\pi / 7\) is cut from a pizza with a radius of 10 in. What is the area of the slice to the nearest tenth of a square inch?

Inscription Rock rises almost straight upward from the valley floor. From one point the angle of elevation of the top of the rock is \(16.7^{\circ} .\) From a point \(168 \mathrm{~m}\) closer to the rock, the angle of elevation of the top of the rock is \(24.1^{\circ} .\) How high is Inscription Rock? Round to the nearest tenth of a meter.

Accurately draw a small \(45-45-90\) triangle on paper or cardboard and cut it out. Label the angles \(45^{\circ}, 45^{\circ},\) and \(90^{\circ} .\) Label the sides \(1,1,\) and \(\sqrt{2}\). Do the same for a \(30-60-90\) triangle. Label its angles \(30^{\circ}, 60^{\circ}\) and \(90^{\circ} .\) Label its sides \(1,2,\) and \(\sqrt{3}\).

Find \(w\) if \(f(w)=5\) and \(f(x)=-3 x-7\)

Solve each problem. Too much trail will cause the front wheel of a motorcycle to turn more than expected (flop over) when the handlebars are rotated away from the straight ahead position. This is why choppers can be difficult to control. The formula $$F=T \sin \theta \cos \theta$$ gives the flop factor \(F\) as a function of the trail \(T\) and the head angle \(\theta\). Find the flop factor for a motorcycle with a head angle of \(39^{\circ}\) and a trail of 10 inches.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.