/*! 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 4 A typical hard disk in a compute... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A typical hard disk in a computer spins at 5400 rpm. a. What is the frequency, in rev/s? b. What is the period, in seconds?

Short Answer

Expert verified
The frequency of the hard disk is 90 rev/s and the period is 0.01111 s/rev.

Step by step solution

01

Conversion of revolutions per minute to revolutions per second

Firstly, convert the provided speed of the hard disk from revolutions per minute (rpm) to revolutions per second (rev/s). There are 60 seconds in one minute, so, divide the given value by 60: \[ \text{rate (rev/s)} = \frac{5400 \, \text{rpm}}{60 \, \text{s/min}} = 90 \, \text{rev/s} \]
02

Calculation of the period

Next, calculate the period, which is the reciprocal of the frequency (or rate in rev/s). \[ \text{Period (s/rev)} = \frac{1}{\text{rate (rev/s)}} = \frac{1}{90 \, \text{rev/s}} = 0.01111 \, \text{s/rev} \]

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 91Ó°ÊÓ!

Key Concepts

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

Frequency
Frequency is a key concept in understanding rotational motion. It describes how often an event repeats over a given period of time. In the context of a rotating object like a computer hard disk, frequency measures how many complete rotations occur in one second. It's important to know that frequency is measured in hertz (Hz) when expressed in cycles per second. For practical calculations concerning disks and similar rotating devices, revolutions per second (rev/s) is often used.
  • To find the frequency in revolutions per second, you might need to convert from revolutions per minute (rpm).
  • This is done by dividing the rpm value by 60, because there are 60 seconds in one minute.
In the example given, a hard disk spins at 5400 rpm, which is converted into 90 rev/s after performing this division. Hence, the frequency of the hard disk is 90 revolutions per second. This step is essential in understanding the speed and performance of rotating machinery.
Period
The period is essentially the time it takes for one complete cycle of a rotating object. It is the inverse of frequency and is expressed in seconds. To make sense of its importance, consider it as the time one rotation of the disk takes, which helps in analyzing slow or fast spinning objects.
  • Calculate the period by taking the reciprocal of the frequency.
  • The formula for the period (T) is given by: \[ T = \frac{1}{f} \] where \( f \) is the frequency in cycles per second.
In our problem, with a frequency of 90 rev/s, the period of the disk is calculated as \( \frac{1}{90} \approx 0.01111 \) seconds. Knowing the period can be critical for synchronizing relevant components that depend on precise timing.
Revolutions per Minute
Revolutions per minute, or rpm, is a unit of rotational speed which indicates how many full rotations occur in one minute. It is a commonplace measurement in settings involving machinery and mechanical engineering. Unlike frequency, rpm provides a broader sense of an object's speed over longer intervals of time.
  • Rpm is straightforward and widely used because it conveniently matches the common minute-based time scale for many engineering calculations.
  • When converting rpm to other units like revolutions per second, you simply divide by 60 since there are 60 seconds per minute.
In the initial exercise, the hard disk's speed of 5400 rpm was effectively converted into revolutions per second to find frequency and understand motion dynamics better. Understanding and working with rpm helps in the broader planning and analysis of device performance.

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

You are driving your car through a roundabout that has a radius of \(9.0 \mathrm{m} .\) Your physics textbook is lying on the seat next to you. What is the fastest speed at which you can go around the curve without the book sliding? The coefficient of static friction between the book and the seat is 0.30 .

The spin cycle of a clothes washer extracts the water in clothing by greatly increasing the water's apparent weight so that it is efficiently squeezed through the clothes and out the holes in the drum. In a top loader's spin cycle, the 45-cm-diameter drum spins at 1200 rpm around a vertical axis. What is the apparent weight of a \(1.0 \mathrm{g}\) drop of water?

In addition to their remarkable top speeds of almost \(60 \mathrm{mph}\), cheetahs have impressive cornering abilities. In one study, the maximum centripetal acceleration of a cheetah was measured to be \(18 \mathrm{m} / \mathrm{s}^{2} .\) What minimum value of the coefficient of static friction between the ground and the cheetah's feet is necessary to provide this acceleration?

The gravitational force of a star on an orbiting planet 1 is \(F_{1} .\) Planet \(2,\) which is twice as massive as planet 1 and orbits at twice the distance from the star, experiences gravitational force \(F_{2} .\) What is the ratio \(F_{2} / F_{1} ?\) You can ignore the gravitational force between the two planets.

The radius of the earth's very nearly circular orbit around the sun is \(1.50 \times 10^{11} \mathrm{m} .\) Find the magnitude of the earth's (a) velocity and (b) centripetal acceleration as it travels around the sun. Assume a year of 365 days.

See all solutions

Recommended explanations on Physics Textbooks

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.