/*! 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 96 A child places a picnic basket o... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 96

A child places a picnic basket on the outer rim of a merrygo-round that has a radius of \(4.6 \mathrm{~m}\) and revolves once every \(30 \mathrm{~s}\) (a) What is the speed of a point on that rim? (b) What is the lowest value of the coefficient of static friction between basket and merrygo-round that allows the basket to stay on the ride?

Short Answer

Expert verified
(a) Speed is approximately 0.963 m/s; (b) Minimum static friction coefficient is approximately 0.0206.

Step by step solution

01

Calculate Circumference

To determine the speed of a point on the rim, we first calculate the circumference of the merry-go-round. The formula to find the circumference \( C \) is given by \( C = 2\pi r \), where \( r \) is the radius. Substituting the given radius, \( r = 4.6 \text{ m} \), we have:\[ C = 2\pi \times 4.6 \approx 28.9 \text{ m} \]
02

Compute Speed of Point on Rim

The speed \( v \) of a point on the rim is the distance covered in one revolution divided by the time period. We have already found the circumference from Step 1:\[ v = \frac{C}{T} \]where \( T = 30 \text{ s} \) is the time for one revolution. Substituting the circumference, \( C = 28.9 \text{ m} \), we get:\[ v = \frac{28.9}{30} \approx 0.963 \text{ m/s} \]
03

Identify Centripetal Force Requirement

To find the lowest coefficient of static friction, we must recognize that friction must provide the necessary centripetal force. The centripetal force \( F_c \) is given by the formula \( F_c = m \cdot a_c \), where \( a_c \) is the centripetal acceleration and is expressed as \( a_c = \frac{v^2}{r} \).
04

Calculate Centripetal Acceleration

Using the speed \( v = 0.963 \text{ m/s} \) and radius \( r = 4.6 \text{ m} \), we can calculate \( a_c \) as follows:\[ a_c = \frac{v^2}{r} = \frac{(0.963)^2}{4.6} \approx 0.202 \text{ m/s}^2 \]
05

Determine Minimum Static Friction Coefficient

The force of friction \( F_f \) must equal the centripetal force \( F_c \) and is given by \( F_f = \mu_s \cdot m \cdot g \), where \( \mu_s \) is the static friction coefficient, \( m \) is the mass, and \( g \) is the acceleration due to gravity \( 9.8 \text{ m/s}^2 \).Equating the forces gives:\[ m \cdot a_c = \mu_s \cdot m \cdot g \]Canceling \( m \) from both sides and solving for \( \mu_s \) we have:\[ \mu_s = \frac{a_c}{g} = \frac{0.202}{9.8} \approx 0.0206 \]
06

Final Verification and Summary

Review the calculations to ensure logical consistency. The identified speed of the point on the rim is approximately \( 0.963 \text{ m/s} \), while the lowest static friction coefficient required to keep the basket from sliding off is approximately \( 0.0206 \).

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.

Centripetal Force
When a child places a picnic basket on the rim of a merry-go-round, an invisible force acts on it to keep it moving in a circle called centripetal force. This force points towards the center of the merry-go-round. It's crucial because without it, the basket would just slide off in a straight line.
Think about trying to swing a stone tied to a string in a circle. You need to pull on the string, right? That pulling is your version of centripetal force.
  • Mathematically, centripetal force (\( F_c \)) is described as:\[ F_c = m \cdot a_c \]where \( m \) is the mass of the object and \( a_c \) is the centripetal acceleration.
As you can see, this force depends on two things: how fast the object is moving and its mass.
Static Friction Coefficient
Static friction is what keeps objects stuck together without sliding. It's what helps the picnic basket stay poised on the merry-go-round, instead of sliding off. The tip here is to realize that this friction must be strong enough to provide the centripetal force needed.
\( \mu_s \), or the static friction coefficient, tells us about the stickiness between the two surfaces.
  • The formula that relates static friction to centripetal force is:\[ m \cdot a_c = \mu_s \cdot m \cdot g \]where \( g \) is the acceleration due to gravity \((9.8 \text{ m/s}^2)\).
  • Simplify it to find the minimum coefficient needed:\[ \mu_s = \frac{a_c}{g} \]
If the coefficient is too low, the basket slides off. If it鈥檚 high enough, the basket stays put.
Circular Motion
Circular motion happens when an object moves along a circular path. It's evident when you see the merry-go-round spinning, or when the planets orbit the sun. The important thing to note is that the speed in circular motion remains constant but the direction continuously changes.
For the merry-go-round, the basket moves in a circle because the changing direction keeps pulling it towards the center.
  • The main formula governing this is the circumference divided by the time period which gives you speed:
When the ride finishes one complete rotation, it has traveled one circumference.
This ties back to the radius:\[ v = \frac{2\pi r}{T} \]where \( T \) is the time for one full rotation.
Centripetal Acceleration
Centripetal acceleration is what changes the object's direction as it moves around the circle. Think of it like the gas pedal in a car, but instead of speeding up, it鈥檚 redirecting the object's path inward.
It鈥檚 calculated by taking the speed squared, divided by the radius of the path. So, even though the speed might be constant, the change in direction creates this acceleration:
  • The formula is:\[ a_c = \frac{v^2}{r} \]
  • Here, \( v \) represents the speed and \( r \) is the radius of the circle.
For the merry-go-round, this acceleration is what keeps the basket in place.

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 bolt is threaded onto one end of a thin horizontal rod, and the rod is then rotated horizontally about its other end. An engineer monitors the motion by flashing a strobe lamp onto the rod and bolt, adjusting the strobe rate until the bolt appears to be in the same eight places during each full rotation of the rod (Fig. \(6-42\) ). The strobe rate is 2000 flashes per second; the bolt has mass \(30 \mathrm{~g}\) and is at radius \(3.5 \mathrm{~cm} .\) What is the magnitude of the force on the bolt from the rod?

A \(2.5 \mathrm{~kg}\) block is initially at rest on a horizontal surface. A horizontal force \(\vec{F}\) of magnitude \(6.0 \mathrm{~N}\) and a vertical force \(\vec{P}\) are then applied to the block (Fig. \(6-17) .\) The coefficients of friction for the block and surface are \(\mu_{s}=0.40\) and \(\mu_{k}=0.25 .\) Determine the magnitude of the frictional force acting on the block if the magnitude of \(\vec{P}\) is (a) \(8.0 \mathrm{~N},\) (b) \(10 \mathrm{~N}\), and (c) \(12 \mathrm{~N}\).

A \(1.5 \mathrm{~kg}\) box is initially at rest on a horizontal surface when at \(t=0\) a horizontal force \(\vec{F}=(1.8 t) \hat{\mathrm{i}} \mathrm{N}(\) with \(t\) in seconds \()\) is applied to the box. The acceleration of the box as a function of time \(t\) is given by \(\vec{a}=0\) for \(0 \leq t \leq 2.8 \mathrm{~s}\) and \(\vec{a}=(1.2 t-2.4) \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}^{2}\) for \(t>2.8 \mathrm{~s}\). (a) What is the coefficient of static friction between the box and the surface? (b) What is the coefficient of kinetic friction between the box and the surface?

A certain string can withstand a maximum tension of \(40 \mathrm{~N}\) without breaking. A child ties a \(0.37 \mathrm{~kg}\) stone to one end and, holding the other end, whirls the stone in a vertical circle of radius 0.91 \(\mathrm{m},\) slowly increasing the speed until the string breaks. (a) Where is the stone on its path when the string breaks? (b) What is the speed of the stone as the string breaks?

A warehouse worker exerts a constant horizontal force of magnitude \(85 \mathrm{~N}\) on a \(40 \mathrm{~kg}\) box that is initially at rest on the horizontal floor of the warehouse. When the box has moved a distance of \(1.4 \mathrm{~m},\) its speed is \(1.0 \mathrm{~m} / \mathrm{s} .\) What is the coefficient of kinetic friction between the box and the floor?
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Fluids

Read Explanation

Turning Points in Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Physical Quantities And Units

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.