/*! 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 19 A playground merry-go-round has ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 19

A playground merry-go-round has a radius of 4.40 \(\mathrm{m}\) and a moment of inertia of 245 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) and turns with negligible friction about a vertical axle through its center. (a) A child applies a 25.0 \(\mathrm{N}\) force tangentially to the edge of the merry-go-round for 20.0 s. If the merry-go-round is initially at rest, what is its angular velocity after this 20.0 s interval? (b) How much work did the child do on the merry-go- round? (c) What is the average power supplied by the child?

Short Answer

Expert verified
(a) 8.98 rad/s, (b) 9,878 J, (c) 493.9 W.

Step by step solution

01

Understanding the Problem

In this problem, we need to find the angular velocity of a merry-go-round given a force applied by a child, the work done by the child, and the average power supplied.
02

Calculate Torque

The torque (蟿) applied about the center of the merry-go-round can be calculated using the formula: \( \tau = F \times r \)where \( F = 25.0 \ \mathrm{N} \) and \( r = 4.40 \ \mathrm{m} \). Plugging in these values gives:\( \tau = 25.0 \ \mathrm{N} \times 4.40 \ \mathrm{m} = 110.0 \ \mathrm{N} \cdot \mathrm{m} \).
03

Calculate Angular Acceleration

Using the moment of inertia \( I = 245 \ \mathrm{kg} \cdot \mathrm{m}^2 \), the angular acceleration \( \alpha \) can be found using the relation: \( \tau = I \times \alpha \) which rearranges to: \( \alpha = \frac{\tau}{I} = \frac{110.0}{245} \approx 0.449 \ \mathrm{rad/s}^2 \).
04

Calculate Angular Velocity

Using the angular acceleration \( \alpha \) and the time \( t = 20.0 \ \mathrm{s} \), the angular velocity \( \omega \) after 20 seconds can be found using: \( \omega = \omega_0 + \alpha \times t \)where \( \omega_0 = 0 \ \mathrm{rad/s} \) (initially at rest):\( \omega = 0 + 0.449 \times 20.0 = 8.98 \ \mathrm{rad/s} \).
05

Calculate Work Done

The work done by the child, \( W \), can be found using the formula:\( W = \tau \times \theta \)The angle \( \theta \) in radians is given by:\( \theta = \frac{1}{2} \times \alpha \times t^2 \)So, \( \theta = \frac{1}{2} \times 0.449 \times (20.0)^2 = 89.8 \ \mathrm{rad} \).Thus, \( W = 110.0 \times 89.8 = 9,878 \ \mathrm{J} \).
06

Calculate Average Power

The average power \( P \) is the work done divided by time:\( P = \frac{W}{t} = \frac{9,878}{20.0} = 493.9 \ \mathrm{W} \).

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.

Torque
Torque is a measure of how much a force acting on an object causes that object to rotate. It's essential in understanding rotational motion. Torque (\( \tau \)) depends on two main factors:
  • Magnitude of the force (\( F \)).
  • Distance from the pivot point, which is called the lever arm (\( r \)).
The formula to calculate torque is:\[ \tau = F \times r \]where, in our merry-go-round example, a 25 \( \mathrm{N} \) force is applied at a 4.40 \( \mathrm{m} \) lever arm. Therefore, the torque generated is 110 \( \mathrm{N} \cdot \mathrm{m} \). This calculated torque helps determine how the merry-go-round spins in response to the force.
Moment of Inertia
Moment of inertia (\( I \)) is a property of a rotating body, which determines how much torque is needed for a desired angular acceleration. Think of it as the "rotational mass," representing resistance to change in rotational motion. The formula relating torque and moment of inertia is:\[ \tau = I \times \alpha \]where:
  • \( \tau \) is the torque.
  • \( I \) is the moment of inertia.
  • \( \alpha \) is the angular acceleration.
In our example, the merry-go-round has a moment of inertia of 245 \( \mathrm{kg} \cdot \mathrm{m}^2 \). This value is used with the torque to find the angular acceleration. As a result, we calculated an angular acceleration of approximately 0.449 \( \mathrm{rad/s}^2 \). This explains how quickly the merry-go-round picks up speed when the force is applied.
Work Done
Work done on a rotating object is similar to the linear work concept but involves torque and angular displacement. It represents the energy transferred when a force acts over a distance.The work done (\( W \)) in the context of rotation is given by:\[ W = \tau \times \theta \]where:
  • \( \tau \) is the torque.
  • \( \theta \) is the angular displacement, in radians.
In this exercise, the angular displacement is found using the formula:\[ \theta = \frac{1}{2} \times \alpha \times t^2 \]Using the known values, the work done by the child on the merry-go-round is approximately 9,878 \( \mathrm{J} \). This amount of energy has been transferred to the merry-go-round to get it spinning from rest.
Average Power
Average power (\( P \)) measures how quickly work is done or energy is transferred over time. It is an essential concept in understanding how efficient the energy transfer is.The formula to calculate average power is:\[ P = \frac{W}{t} \]where:
  • \( W \) is the work done.
  • \( t \) is the time over which the work is done.
With the work done by the child being 9,878 \( \mathrm{J} \) and the time being 20 seconds, the average power supplied by the child is 493.9 \( \mathrm{W} \). This value helps in understanding how much energy per second the child transfers to the merry-go-round, giving insight into the child's contribution to its 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 solid, uniform cylinder with mass 8.25 \(\mathrm{kg}\) and diameter 15.0 \(\mathrm{cm}\) is spinning at 220 \(\mathrm{rpm}\) on a thin, frictionless axle that passes along the cylinder axis. You design a simple friction brake to stop the cylinder by pressing the brake against the outer rim with a normal force. The coefficient of kinetic friction between the brake and rim is \(0.333 .\) What must the applied normal force be to bring the cylinder to rest after it has turned through 5.25 revolutions?

A certain type of propeller blade can be modeled as a thin uniform bar 2.50 m long and of mass 24.0 \(\mathrm{kg}\) that is free to rotate about a frictionless axle perpendicular to the bar at its midpoint. If a technician strikes this blade with a mallet 1.15 \(\mathrm{m}\) from the center with a 35.0 \(\mathrm{N}\) force perpendicular to the blade, find the maximum angular acceleration the blade could achieve.

Prior to being placed in its hole, a \(5700-\mathrm{N}, 9.0\) -m-long, uniform utility pole makes some nonzero angle with the vertical. A vertical cable attached 2.0 \(\mathrm{m}\) below its upper end holds it in place while its lower end rests on the ground. (a) Find the tension in the cable and the magnitude and direction of the force exerted by the ground on the pole. (b) Why don't we need to know the angle the pole makes with the vertical, as long as it is not zero?

You are trying to raise a bicycle wheel of mass \(m\) and radius \(R\) up over a curb of height \(h .\) To do this, you apply a horizontal force \(\vec{F}\) (Fig. \(10.81 ) .\) What is the smallest magnitude of the force \(\vec{F}\) that will succeed in raising the wheel onto the curb when the force is applied (a) at the center of the wheel, and (b) at the top of the wheel? (c) In which case is less force required?

A large wooden turntable in the shape of a flat disk has a radius of 2.00 \(\mathrm{m}\) and a total mass of 120 \(\mathrm{kg}\) . The turntable is initially rotating at 3.00 \(\mathrm{rad} / \mathrm{s}\) about a vertical axis through its center. Suddenly, a 70.0 kg parachutist makes a soft landing on the turntable at a point on its outer edge. Find the angular speed of the turntable after the parachutist lands. (Assume that you can treat the parachutist as a particle.)
See all solutions

Recommended explanations on Physics Textbooks

Radiation

Read Explanation

Physical Quantities And Units

Read Explanation

Circular Motion and Gravitation

Read Explanation

Rotational Dynamics

Read Explanation

Atoms and Radioactivity

Read Explanation

Oscillations

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.