/*! 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 10 A computer is reading data from ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A computer is reading data from a rotating CD-ROM. At a point that is \(0.030 \mathrm{m}\) from the center of the disc, the centripetal acceleration is \(120 \mathrm{m} / \mathrm{s}^{2}\). What is the centripetal acceleration at a point that is \(0.050 \mathrm{m}\) from the center of the disc?

Short Answer

Expert verified
The centripetal acceleration at 0.050 m is 200 m/s².

Step by step solution

01

Define Centripetal Acceleration Formula

The centripetal acceleration \( a_c \) is given by the formula \( a_c = r \cdot \omega^2 \), where \( r \) is the distance from the center, and \( \omega \) is the angular velocity. Our task is to find the centripetal acceleration \( a_c \) at a different radius given the same angular velocity.
02

Find Angular Velocity from Initial Data

Using the given data at \( r = 0.030 \) m with \( a_c = 120 \) m/s², we can solve for \( \omega \) using \( 120 = 0.030 \cdot \omega^2 \). Solving for \( \omega \), we find \( \omega = \sqrt{\frac{120}{0.030}} = \sqrt{4000} = 63.245 \) rad/s.
03

Calculate New Centripetal Acceleration

Use the angular velocity found in Step 2 to calculate the centripetal acceleration at \( r = 0.050 \) m. Use \( a_c = 0.050 \cdot \omega^2 \). Substitute \( \omega = 63.245 \) rad/s to get \( a_c = 0.050 \times 63.245^2 = 0.050 \times 4000 = 200 \) m/s².

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.

Angular Velocity
Angular velocity, denoted by the symbol \( \omega \), plays a crucial role in understanding rotational motion. It describes how fast an object is rotating around a particular axis. For any rotating object, like a CD-ROM in a computer, the angular velocity allows us to relate linear quantities, such as the speed of data reading at different radii, to the motion of spinning.
  • Angular velocity is measured in radians per second (rad/s).
  • It relates to the physical rotation directly, providing a measure of how quick a rotation is happening.
In the exercise, \( \omega \) was calculated using the centripetal acceleration at a known radius. By understanding that the angular velocity remains constant for a rigid, spinning disc like the CD-ROM, we can use this same \( \omega \) to find answers at any other radius on the disc.The calculation \( \omega = \sqrt{\frac{120}{0.030}} = 63.245 \) rad/s shows how rapid the rotation is without changing its value for different points on the same disc.
Rotational Motion
Rotational motion is everywhere around us and is described as the movement of an object in a circular path around a central point. In systems like rotating CD-ROMs or planets orbiting in space, rotation leads to fascinating phenomena such as centripetal acceleration.
  • Rotational motion involves circular paths with continuous changing directions.
  • Key components include angular velocity, radius, and centripetal acceleration.
  • Centripetal acceleration \( a_c \) ensures that an object stays on its circular path, always pointing towards the center.
The CD-ROM example illustrates rotational motion well, where calculating different centripetal accelerations can reveal velocities and speeds at various points. Rotational motion analysis gives a deeper understanding of the forces and speeds involved when objects are in circular motion, allowing predictions about behavior at different points.
Physics Problem Solving
Tackling physics problems like the centripetal acceleration of a spinning CD-ROM requires a strategic approach to break down and solve each part systematically.
  • Start by identifying known and unknown values. In the problem, you know the radius and centripetal acceleration.
  • Use relevant formulas. The centripetal acceleration formula \( a_c = r \cdot \omega^2 \) ties the variations in acceleration to angular velocity and radius.
  • Solving for intermediate variables, such as \( \omega \) in this example, helps apply known values to new conditions.
By calculating \( \omega \) first, you establish a constant that transforms the problem from one radius to another seamlessly, enabling an answer to a whole set of similar problems. Adopting an organized method ensures accuracy and confidence in finding solutions to physical scenarios involving rotational motion.

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

A jet flying at \(123 \mathrm{m} / \mathrm{s}\) banks to make a horizontal circular turn. The radius of the turn is \(3810 \mathrm{m},\) and the mass of the jet is \(2.00 \times 10^{5} \mathrm{kg}\) Calculate the magnitude of the necessary lifting force.

A car travels at a constant speed around a circular track whose radius is \(2.6 \mathrm{km}\). The car goes once around the track in \(360 \mathrm{s}\). What is the magnitude of the centripetal acceleration of the car?

Two cars are traveling at the same speed of \(27 \mathrm{m} / \mathrm{s}\) on a curve that has a radius of \(120 \mathrm{m}\). Car A has a mass of \(1100 \mathrm{kg}\), and car \(\mathrm{B}\) has a mass of \(1600 \mathrm{kg}\). Find the magnitude of the centripetal acceleration and the magnitude of the centripetal force for each car.

A car is safely negotiating an unbanked circular turn at a speed of \(21 \mathrm{m} / \mathrm{s} .\) The road is dry, and the maximum static frictional force acts on the tires. Suddenly a long wet patch in the road decreases the maximum static frictional force to one-third of its dry-road value. If the car is to continue safely around the curve, to what speed must the driver slow the car?

Two newly discovered planets follow circular orbits around a star in a distant part of the galaxy. The orbital speeds of the planets are determined to be \(43.3 \mathrm{km} / \mathrm{s}\) and \(58.6 \mathrm{km} / \mathrm{s} .\) The slower planet's orbital period is 7.60 years. (a) What is the mass of the star? (b) What is the orbital period of the faster planet, in years?

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.