/*! 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 41 A cat dozes on a stationary merr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 41

A cat dozes on a stationary merry-go-round in an amusement park, at a radius of \(5.4 \mathrm{~m}\) from the center of the ride. Then the operator turns on the ride and brings it up to its proper turning rate of one complete rotation every \(6.0 \mathrm{~s}\). What is the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding (or the cat clinging with its claws)?

Short Answer

Expert verified
The least coefficient of static friction is approximately 0.604.

Step by step solution

01

Determine Angular Velocity

First, we need to find the angular velocity (\(\omega\)) of the merry-go-round. The angular velocity is given by the formula \(\omega = \frac{2\pi}{T}\), where \(T\) is the period of rotation. Here, \(T = 6.0\,\text{s}\). Substitute \(T\) into the formula: \(\omega = \frac{2\pi}{6.0} = \frac{\pi}{3}\,\text{rad/s}\).
02

Find the Centripetal Acceleration

Centripetal acceleration (\(a_c\)) can be calculated using the formula \(a_c = r\omega^2\), where \(r\) is the radius and \(\omega\) is the angular velocity. From the problem, \(r = 5.4\,\text{m}\), and \(\omega = \frac{\pi}{3}\,\text{rad/s}\). Substitute these values: \(a_c = 5.4 \left(\frac{\pi}{3}\right)^2 = 5.4 \cdot \frac{\pi^2}{9} = 0.6\pi^2\,\text{m/s}^2\).
03

Apply the Static Friction Formula

We use the formula for static friction to keep the cat from sliding: \(f_s = \mu_s mg = ma_c\), where \(\mu_s\) is the coefficient of static friction, \(m\) is the mass of the cat, and \(g\) is the acceleration due to gravity \(\approx 9.8\,\text{m/s}^2\). Simplifying gives \(\mu_s g = a_c\).
04

Solve for Coefficient of Static Friction

From the equation \(\mu_s g = a_c\), isolate \(\mu_s\) to find \(\mu_s = \frac{a_c}{g}\). Substitute \(a_c = 0.6\pi^2\) and \(g = 9.8\,\text{m/s}^2\): \(\mu_s = \frac{0.6\pi^2}{9.8}\). Calculate \(\mu_s\approx 0.6\times3.14^2/9.8 \approx 0.604\).

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.

Angular Velocity
Angular velocity is an important concept when discussing rotating objects like a merry-go-round. It describes how fast something rotates or the "rate of rotation." For circular motion, angular velocity is represented by the symbol \( \omega \), and it is measured in radians per second (rad/s).
  • To find angular velocity, we use the formula: \( \omega = \frac{2\pi}{T} \), where \( T \) is the period of rotation (the time it takes to complete one full cycle).
  • In our scenario, the merry-go-round completes one rotation every 6 seconds, so \( T = 6 \,\text{s} \).
  • Substituting \( T = 6 \) into the formula gives us \( \omega = \frac{2\pi}{6} = \frac{\pi}{3} \,\text{rad/s} \).
Angular velocity is key in determining other important values, like centripetal acceleration. So, if you know how long it takes for a circular object to complete its rotation, you can find how quickly it is spinning.
Coefficient of Static Friction
The coefficient of static friction, denoted as \( \mu_s \), is a dimensionless value that represents the frictional force required to prevent two surfaces from sliding past each other. In this scenario, it helps to determine whether the cat can stay put on the spinning merry-go-round.
  • Static friction is the force keeping the cat stationary relative to the spinning surface, preventing it from slipping off.
  • The least coefficient of static friction can be found using the formula: \( f_s = \mu_s mg = ma_c \), where \( f_s \) is the static friction force, \( m \) is the mass, \( g \) is gravity (\( 9.8 \,\text{m/s}^2 \)), and \( a_c \) is the centripetal acceleration.
  • Once simplified, \( \mu_s \) is isolated and calculated as \( \mu_s = \frac{a_c}{g} \).
The calculation shows how much grip is required to keep an object stationary under specific moving conditions. In this particular case, we established that \( \mu_s \approx 0.604 \) to keep the cat from sliding.
Centripetal Acceleration
Centripetal acceleration is the acceleration directed towards the center of a circular path, keeping an object moving in a circle. For the cat on the merry-go-round, understanding this acceleration helps us determine how it stays in place without sliding off.
  • Centripetal acceleration is given by the formula \( a_c = r\omega^2 \), where \( r \) is the radius of the circle and \( \omega \) is the angular velocity.
  • In the problem, the radius \( r = 5.4 \,\text{m} \) and angular velocity \( \omega = \frac{\pi}{3} \,\text{rad/s} \).
  • Substituting these values, we calculate \( a_c = 5.4 \left(\frac{\pi}{3}\right)^2 = 0.6\pi^2 \,\text{m/s}^2 \).
Centripetal acceleration ensures that the cat maintains its position on the spinning platform. By calculating it, we gain insight into the forces at play necessary for circular motion, making it vital for understanding both the dynamics of spinning objects and the friction required to maintain stability.

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 bedroom bureau with a mass of \(45 \mathrm{~kg}\), including drawers and clothing, rests on the floor. (a) If the coefficient of static friction between the bureau and the floor is \(0.45\), what is the magnitude of the minimum horizontal force that a person must apply to start the bureau moving? (b) If the drawers and clothing, with \(17 \mathrm{~kg}\) mass, are removed before the bureau is pushed, what is the new minimum magnitude?

A police officer in hot pursuit drives her car through a circular turn of radius \(300 \mathrm{~m}\) with a constant speed of \(80.0 \mathrm{~km} / \mathrm{h}\). Her mass is \(55.0 \mathrm{~kg}\). What are (a) the magnitude and (b) the angle (relative to vertical) of the net force of the officer on the car seat? (Hint: Consider both horizontal and vertical forces.)

A block of mass \(m_{t}=4.0 \mathrm{~kg}\) is put on top of a block of mass \(m_{b}=5.0 \mathrm{~kg}\). To cause the top block to slip on the bottom one while the bottom one is held fixed, a horizontal force of at least 12 N must be applied to the top block. The assembly of blocks is now placed on a horizontal, frictionless table (Fig. \(6-47\) ). Find the magnitudes of (a) the maximum horizontal force \(\vec{F}\) that can be applied to the lower block so that the blocks will move together and (b) the resulting acceleration of the blocks.

A worker pushes horizontally on a \(35 \mathrm{~kg}\) crate with a force of magnitude \(110 \mathrm{~N}\). The coefficient of static friction between the crate and the floor is \(0.37 .\) (a) What is the value of \(f_{s, \max }\) under the circumstances? (b) Does the crate move? (c) What is the frictional force on the crate from the floor? (d) Suppose, next, that a second worker pulls directly upward on the crate to help out. What is the least vertical pull that will allow the first worker's \(110 \mathrm{~N}\) push to move the crate? (c) If, instead, the second worker pulls horizontally to \(\mathrm{F}\) the least pull that will get the crate moving?

\(.\Rightarrow F\) A circular-motion addict of mass \(80 \mathrm{~kg}\) rides a Ferris wheel around in a vertical circle of radius \(10 \mathrm{~m}\) at a constant speed of \(6.1 \mathrm{~m} / \mathrm{s}\). (a) What is the period of the motion? What is the magnitude of the normal force on the addict from the seat when both go through (b) the highest point of the circular path and (c) the lowest point?
See all solutions

Recommended explanations on Physics Textbooks

Physics of Motion

Read Explanation

Fluids

Read Explanation

Waves Physics

Read Explanation

Translational Dynamics

Read Explanation

Atoms and Radioactivity

Read Explanation

Dynamics

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.