/*! 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 55 (II) A merry-go-round has a mass... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 55

(II) A merry-go-round has a mass of 1440 kg and a radius of 7.50 m. How much net work is required to accelerate it from rest to a rotation rate of 1.00 revolution per 7.00 s? Assume it is a solid cylinder.

Short Answer

Expert verified
The work required is 5133.43 J.

Step by step solution

01

Understanding the Problem

We need to calculate the net work required to accelerate the merry-go-round from rest to a specified rotational speed. The system is a solid cylinder with a given mass and radius.
02

Convert Rotational Speed to Angular Velocity

The rotation rate is given in revolutions per second. First, convert this to angular velocity in radians per second. The angular velocity \( \omega \) can be calculated using the formula: \[ \omega = \frac{\text{revolutions}}{\text{seconds}} \times 2\pi \] Given 1 revolution per 7 seconds, \[ \omega = \frac{1}{7} \times 2\pi = \frac{2\pi}{7} \text{ rad/s} \]
03

Calculate Moment of Inertia

The moment of inertia \( I \) for a solid cylinder is given by the formula: \[ I = \frac{1}{2} m r^2 \] where \( m = 1440 \text{ kg} \) and \( r = 7.50 \text{ m} \). Substitute the values: \[ I = \frac{1}{2} \times 1440 \times (7.50)^2 = 40500 \text{ kg} \cdot \text{m}^2 \]
04

Use Rotational Kinetic Energy Formula

The formula for rotational kinetic energy \( KE \) is: \[ KE = \frac{1}{2} I \omega^2 \] Substituting for \( I \) and \( \omega \) from Steps 2 and 3: \[ KE = \frac{1}{2} \times 40500 \times \left(\frac{2\pi}{7}\right)^2 \]
05

Calculate the Net Work Done

Calculate the kinetic energy using the values found: \[ KE = \frac{1}{2} \times 40500 \times \left(\frac{2\pi}{7}\right)^2 = 5133.43 \text{ J} \] The net work done \( W \) is equal to the change in kinetic energy, which is \( 5133.43 \text{ J} \) since it starts from rest.

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.

Moment of Inertia
The moment of inertia is a fundamental concept in rotational dynamics. It describes an object's resistance to any change in its state of rotation.
Imagine it as the rotational equivalent of mass in linear motion. Just as a larger mass is harder to push, a larger moment of inertia is harder to spin.
For different shapes, the moment of inertia varies. In our exercise, we had a solid cylinder, and its moment of inertia is given by:
  • \( I = \frac{1}{2} m r^2 \),
where
  • \( m \) represents the mass of the cylinder,
  • and \( r \) is its radius.
A higher moment of inertia implies that more work is needed to spin the object to a certain angular velocity.
This is crucial in calculating how much energy is required to achieve a desired rotational speed.
Angular Velocity
Angular velocity is how we describe the speed of rotation. Unlike linear velocity, which measures straight-line speed, angular velocity measures rotation along a circular path.
When calculating angular velocity, we use radians per second. This may seem unusual at first because we're accustomed to thinking in terms of revolutions or degrees per second.
The conversion from revolutions to radians is straightforward:
  • 1 revolution = \( 2\pi \) radians.
In this particular exercise, the merry-go-round rotates at 1 revolution every 7 seconds.
So, its angular velocity \( \omega \) becomes:
  • \( \omega = \frac{1}{7} \times 2\pi = \frac{2\pi}{7} \text{ rad/s} \).
Understanding angular velocity helps us to explore deeply how quickly an object like our merry-go-round spins around its axis.
Rotational Kinetic Energy
Rotational kinetic energy shows the energy an object possesses due to its rotation. Just like kinetic energy in linear motion \( (KE = \frac{1}{2} mv^2) \), rotational kinetic energy has its own formula:
  • \( KE = \frac{1}{2} I \omega^2 \).
It's directly proportional to two key elements:
  • the moment of inertia \( I \)
  • and the square of the angular velocity \( \omega^2 \).
In our exercise, to calculate how much work is needed to get the merry-go-round spinning, we first figured out its rotational kinetic energy. Since it starts from rest, the work done is equal to the energy change which, in this case, is \( 5133.43 \text{ J} \).
This change in rotational kinetic energy tells us how much effort is required to get it up to speed.

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

(III) A hammer thrower accelerates the hammer \((mass = 7.30 kg)\) from rest within four full turns (revolutions) and releases it at a speed of 26.5 m/s. Assuming a uniform rate of increase in angular velocity and a horizontal circular path of radius 1.20 m, calculate \((a)\) the angular acceleration, \((b)\) the (linear) tangential acceleration, \((c)\) the centripetal acceleration just before release, \((d)\) the net force being exerted on the hammer by the athlete just before release, and \((e)\) the angle of this force with respect to the radius of the circular motion. Ignore gravity.

A spherical asteroid with radius \(r = 123 m\) and mass \(M = 2.25 \times 10^{10} kg\) rotates about an axis at four revolutions per day. A "tug" spaceship attaches itself to the asteroid's south pole (as defined by the axis of rotation) and fires its engine, applying a force \(F\) tangentially to the asteroid's surface as shown in Fig. 8-65. If \(F = 285 N\), how long will it take the tug to rotate the asteroid's axis of rotation through an angle of 5.0\(^{\circ}\) by this method?

(II) How fast (in rpm) must a centrifuge rotate if a particle 8.0 cm from the axis of rotation is to experience an acceleration of 100,000 g's?

(II) A grinding wheel is a uniform cylinder with a radius of 8.50 cm and a mass of 0.380 kg. Calculate \((a)\) its moment of inertia about its center, and \((b)\) the applied torque needed to accelerate it from rest to 1750 rpm in 5.00 s. Take into account a frictional torque that has been measured to slow down the wheel from 1500 rpm to rest in 55.0 s.

(II) A merry-go-round accelerates from rest to 0.68 rad/s in 34 s. Assuming the merry-go-round is a uniform disk of radius 7.0 m and mass 31,000 kg, calculate the net torque required to accelerate it.
See all solutions

Recommended explanations on Physics Textbooks

Physics of Motion

Read Explanation

Fundamentals of Physics

Read Explanation

Energy Physics

Read Explanation

Turning Points in Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Radiation

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.