/*! 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 33 Calculate the rotational inertia... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Calculate the rotational inertia of a wheel that has a kinetic energy of \(24400 \mathrm{~J}\) when rotating at 602 rev \(/ \mathrm{min}\).

Short Answer

Expert verified
The rotational inertia is approximately 12.27 kg m².

Step by step solution

01

Convert Units

First, we need to convert the rotational speed from revolutions per minute (rev/min) to radians per second (rad/s). Remember, there are \(2\pi\) radians in one revolution, and 60 seconds in a minute. So the conversion is: \[602 \ \text{rev/min} \times \frac{2\pi \ \text{rad}}{1 \ \text{rev}} \times \frac{1 \ \text{min}}{60 \ \text{s}} = \frac{602 \times 2\pi}{60} \ \text{rad/s}\approx 63.062 \ \text{rad/s}.\]
02

Use the Kinetic Energy Formula for Rotation

We know the kinetic energy \(K\) of a rotating object is given by: \[K = \frac{1}{2} I \omega^2,\] where \(I\) is the rotational inertia, and \(\omega\) is the angular velocity. We are given \(K = 24400\ \text{J}\) and \(\omega = 63.062 \ \text{rad/s}\).
03

Solve for Rotational Inertia (I)

Rearrange the kinetic energy formula to solve for \(I\): \[I = \frac{2K}{\omega^2}.\] Plug in the known values: \[I = \frac{2 \times 24400}{(63.062)^2}.\]Calculating this gives: \[I \approx \frac{48800}{3976.81} \approx 12.27 \ \text{kg m}^2.\]

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.

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. For rotating objects, kinetic energy is a bit different from linear kinetic energy. Here, the object’s mass distribution around its axis, also called rotational inertia or moment of inertia, plays a crucial role.
To calculate the kinetic energy of a rotating object, we use the formula given by: \[ K = \frac{1}{2} I \omega^2, \]where:
  • K is the kinetic energy in joules.
  • I represents the rotational inertia in kilogram-meter square (kg m2).
  • \( \omega \) is the angular velocity in radians per second (rad/s).

This formula tells us that the kinetic energy depends not only on how fast the object is spinning (angular velocity) but also on how its mass is distributed in relation to the axis of rotation.
Understanding how these factors contribute to kinetic energy helps in calculating energy for systems involving rotational motion, like wheels or gears.
Angular Velocity
Angular velocity describes how quickly an object rotates or revolves around an axis. It is the rotational equivalent of linear velocity. Angular velocity is measured in radians per second (rad/s).
To change rotational speed from one unit to another, such as revolutions per minute (rev/min) to radians per second, conversions are necessary. This is important because many physical formulas, such as those dealing with kinetic energy, require angular velocity in rad/s.
Here's how you convert revolutions per minute to radians per second:
  • Multiply the value in rev/min by \(2\pi\) to convert revolutions to radians.
  • Divide the result by 60 to convert minutes to seconds.
So, for 602 rev/min, the conversion would be:\[602 \ \text{rev/min} \times \frac{2\pi \ \text{rad}}{1 \ \text{rev}} \times \frac{1 \ \text{min}}{60 \ \text{s}} \approx 63.062 \ \text{rad/s}.\]
This calculation helps in accurately determining other physical quantities, like rotational inertia.
Conversion of Units
Conversions in physics are essential for calculation accuracy. Often different units are used in measurements, and understanding how to correctly convert between them is crucial.
One of the most common conversions in rotational dynamics is between revolutions per minute (rev/min) and radians per second (rad/s). Here's why it's important:
  • Consistency in Calculations: Calculations involving kinetic energy, power, or torque often require consistent units (SI units), such as radians and seconds.
  • Precision: Correct conversions ensure precision and reduce errors in finding quantities like angular velocity or inertia.
The conversion involves recognizing:
  • There are \(2\pi\) radians in one complete revolution.
  • There are 60 seconds in a minute.
So, always remember to adjust values to appropriate units before plugging them into formulas to ensure accurate and meaningful results.
Physical Formulas
Physical formulas are fundamental tools used in physics to relate different physical quantities. Understanding each component and how they interact within the formula is vital.
For instance, the kinetic energy formula for rotating objects \( K = \frac{1}{2} I \omega^2 \) has individual components that must be understood:
  • Rotational Inertia (I): This measures an object’s resistance to change in its rotational motion.
  • Angular Velocity (\( \omega \)): The rate of change of the angular position of an object, expressed in rad/s.
  • K (Kinetic Energy): Represents the energy an object holds because of its rotational motion.
With the given problem:
  • You start by calculating angular velocity through unit conversion, which is essential before applying the formula.
  • Then, rearrange the formula to solve for the unknown, which, in this case, is rotational inertia.
Finally, plug values into the rearranged formula based on known values, and solve. This structured approach to using physical formulas ensures accurate outcomes.

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

Starting from rest at \(t=0,\) a wheel undergoes a constant angular acceleration. When \(t=2.0 \mathrm{~s},\) the angular velocity of the wheel is \(5.0 \mathrm{rad} / \mathrm{s} .\) The acceleration continues until \(t=20 \mathrm{~s},\) when it abruptly ceases. Through what angle does the wheel rotate in the interval \(t=0\) to \(t=40 \mathrm{~s} ?\)

A bicyclist of mass \(70 \mathrm{~kg}\) puts all his mass on each downwardmoving pedal as he pedals up a steep road. Take the diameter of the circle in which the pedals rotate to be \(0.40 \mathrm{~m},\) and determine the magnitude of the maximum torque he exerts about the rotation axis of the pedals.

Trucks can be run on energy stored in a rotating flywheel, with an electric motor getting the flywheel up to its top speed of \(200 \pi \mathrm{rad} / \mathrm{s} .\) Suppose that one such flywheel is a solid, uniform cylinder with a mass of \(500 \mathrm{~kg}\) and a radius of \(1.0 \mathrm{~m}\). (a) What is the kinetic energy of the flywheel after charging? (b) If the truck uses an average power of \(8.0 \mathrm{~kW},\) for how many minutes can it operate between chargings?

A uniform spherical shell of mass \(M=4.5 \mathrm{~kg}\) and radius \(R=8.5 \mathrm{~cm}\) can rotate about a vertical axis on frictionless bearings (Fig. \(10-47) .\) A massless cord passes around the equator of the shell, over a pulley of rotational inertia \(I=3.0 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) and radius \(r=5.0 \mathrm{~cm},\) and is attached to a small object of mass \(m=0.60 \mathrm{~kg}\) There is no friction on the pulley's axle; the cord does not slip on the pulley. What is the speed of the object when it has fallen \(82 \mathrm{~cm}\) after being released from rest? Use energy considerations.

A golf ball is launched at an angle of \(20^{\circ}\) to the horizontal, with a speed of \(60 \mathrm{~m} / \mathrm{s}\) and a rotation rate of \(90 \mathrm{rad} / \mathrm{s} .\) Neglecting air drag, determine the number of revolutions the ball makes by the time it reaches maximum height.

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.