/*! 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 51 An astronaut is rotated in a hor... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 51

An astronaut is rotated in a horizontal centrifuge at a radius of \(5.0 \mathrm{~m}\). (a) What is the astronaut's speed if the centripetal acceleration has a magnitude of \(7.0 \mathrm{~g}\) ? (b) How many revolutions per minute are required to produce this acceleration? (c) What is the period of the motion?

Short Answer

Expert verified
Speed: 18.52 m/s, RPM: 35.3, Period: 1.70 s.

Step by step solution

01

Understand the given values and formulas

Given radius: \(r = 5.0 \text{ m} \)Centripetal acceleration: \(a_c = 7.0 \times 9.8 \text{ m/s}^2 = 68.6 \text{ m/s}^2\)The relation between centripetal acceleration \(a_c\) and speed \(v\) is given by the formula: \(a_c = \frac{v^2}{r}\)
02

Calculate the astronaut's speed

Re-arrange the formula to solve for \(v\): \(v^2 = a_c \times r\)Substitute the known values: \(v^2 = 68.6 \text{ m/s}^2 \times 5.0 \text{ m}\)\(v^2 = 343 \text{ m}^2/\text{s}^2\)Take the square root of both sides: \(v = \sqrt{343} \text{ m/s} \approx 18.52 \text{ m/s}\)
03

Calculate the angular velocity

Angular velocity \(\omega\) is related to linear speed \(v\) by the formula: \(v = \omega r\)Substitute the known values: \(18.52 \text{ m/s} = \omega \times 5.0 \text{ m}\)Solve for \(\omega\): \(\omega = \frac{18.52}{5.0} \text{ rad/s} \approx 3.70 \text{ rad/s}\)
04

Convert angular velocity to revolutions per minute

1 revolution = \(2\pi\) radiansConvert \(\omega\) from rad/s to revolutions per second: \(\frac{3.70}{2\pi} \approx 0.589 \text{ rev/s}\)Convert rev/s to rev/min: \(0.589 \text{ rev/s} \times 60 \approx 35.3 \text{ rev/min}\)
05

Calculate the period of motion

The period \(T\) is the reciprocal of the frequency (revolutions per second):\(T = \frac{1}{f} \)Substitute the frequency: \(f = 0.589 \text{ rev/s}\)\(T = \frac{1}{0.589} \text{ s} \approx 1.70 \text{ s}\)

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.

Centripetal Acceleration
Centripetal acceleration is what keeps an object moving in a circular path. It's directed towards the center of the circle. This force is essential for any rotating system.
The formula to find centripetal acceleration is: \(a_c = \frac{v^2}{r}\)
  • \(a_c\): Centripetal acceleration
  • \(v\): Speed or linear velocity of the object
  • \(r\): Radius of the circular path
In the given problem, the astronaut's centripetal acceleration is \(a_c = 68.6 \text{ m/s}^2\), calculated from \(7.0 \times 9.8\). By rearranging the formula to solve for speed (\(v\)), we can determine the astronaut's velocity along the circular path.
Angular Velocity
Angular velocity signifies how fast an object is rotating. It's different from speed because it doesn't care about the path length, just the angle it sweeps out.
The formula linking angular velocity (\(\omega\)) and linear speed (\(v\)) is: \(v = \omega r\)
  • \(\omega\): Angular velocity
  • \(v\): Linear speed
  • \(r\): Radius
By knowing the speed (\(v\)) and the radius (\(r\)), you can solve for \(\omega\) as 3.70 rad/s in our problem. This helps us understand how quickly the astronaut completes one circular path part of their motion.
Period of Motion
The period of motion is the time it takes for one complete revolution. It's inversely related to the frequency.
The formula is: \(T = \frac{1}{f}\)
  • \(T\): Period
  • \(f\): Frequency (revolutions per second)
In the exercise, the frequency was found by converting the angular velocity to revolutions per second. The astronaut's frequency was around 0.589 rev/s, giving a period (\(T\)) of about 1.70 seconds per complete revolution.
Revolutions per Minute
Revolutions per minute (RPM) tell us how many complete spins or cycles happen in one minute. It's a practical way to measure rotating systems.
To convert angular velocity (rad/s) to RPM, use these steps:

  • Convert to revolutions per second: \(\text{\(\omega\)} \div 2\pi\)
  • Multiply by 60 to get RPM
In our example, we transformed 3.70 rad/s to RPM, resulting in approximately 35.3 RPM. This means the astronaut completes about 35.3 full spins every minute.

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

An Earth Satellite An Earth satellite moves in a circular orbit \(640 \mathrm{~km}\) above Earth's surface with a period of \(98.0 \mathrm{~min}\). What are (a) the speed and (b) the magnitude of the centripetal acceleration of the satellite?

You are to ride a jet-cycle over a lake, starting from rest at point \(1:\) First, moving at \(30^{\circ}\) north of due east: 1\. Increase your speed at \(0.400 \mathrm{~m} / \mathrm{s}^{2}\) for \(6.00 \mathrm{~s}\). 2. With whatever speed you then have, move for \(8.00 \mathrm{~s}\). 3\. Then slow at \(0.400 \mathrm{~m} / \mathrm{s}^{2}\) for \(6.00 \mathrm{~s}\). Immediately next, moving due west: 4\. Increase your speed at \(0.400 \mathrm{~m} / \mathrm{s}^{2}\) for \(5.00 \mathrm{~s}\). 5\. With whatever speed you then have, move for \(10.0 \mathrm{~s}\). 6\. Then slow at \(0.400 \mathrm{~m} / \mathrm{s}^{2}\) until you stop. In magnitude-angle notation, what then is your average velocity for the trip from point 1 ?

. Fast Bullets A rifle that shoots bullets at \(460 \mathrm{~m} / \mathrm{s}\) is to be aimed at a target \(45.7 \mathrm{~m}\) away and level with the rifle. How high above the target must the rifle barrel be pointed so that the bullet hits the target?

Volleyball For women's volleyball the top of the net is \(2.24 \mathrm{~m}\) above the floor and the court measures \(9.0 \mathrm{~m}\) by \(9.0 \mathrm{~m}\) on each side of the net. Using a jump serve, a player strikes the ball at a point that is \(3.0 \mathrm{~m}\) above the floor and a horizontal distance of \(8.0 \mathrm{~m}\) from the net. If the initial velocity of the ball is horizontal, (a) what minimum magnitude must it have if the ball is to clear the net and (b) what maximum magnitude can it have if the ball is to strike the floor inside the back line on the other side of the net?

A particle starts from the origin at \(t=0\) with a velocity of \(\vec{v}_{1}=(8.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) and moves in the \(x y\) plane with a constant acceleration of \(\vec{a}=\left(4.0 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(2.0 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). At the instant the particle's \(x\) coordinate is \(29 \mathrm{~m}\), what are (a) its \(y\) coordinate and (b) its speed?
See all solutions

Recommended explanations on Physics Textbooks

Mechanics and Materials

Read Explanation

Energy Physics

Read Explanation

Turning Points in Physics

Read Explanation

Thermodynamics

Read Explanation

Torque and Rotational Motion

Read Explanation

Wave Optics

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.