/*! 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 36 A potter's wheel having a radius... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 36

A potter's wheel having a radius of \(0.50 \mathrm{~m}\) and a moment of inertia of \(12 \mathrm{~kg} \cdot \mathrm{m}^{2}\) is rotating freely at \(50 \mathrm{rev} / \mathrm{min}\). The potter can stop the wheel in \(6.0 \mathrm{~s}\) by pressing a wet rag against the rim and exerting a radially inward force of \(70 \mathrm{~N}\). Find the effective coefficient of kinetic friction between the wheel and the wet rag.

Short Answer

Expert verified
The coefficient of kinetic friction between the wheel and the wet rag is the value obtained in Step 5.

Step by step solution

01

Convert revolution per minute to radian per second

First, convert the initial angular speed from revolutions per minute(rpm) to radian per second(rad/s) because the standard unit of angular speed in physics is rad/s. To do this, using conversion factor \(1 revolution = 2\pi rad\) and \(1 minute = 60 sec\), you multiply the value in rpm by \(2\pi / 60\).
02

Calculate the angular acceleration

The angular acceleration, \( \alpha \), can be calculated using the kinematic equation which relates final angular speed, initial angular speed, angular acceleration and time: \( \omega_{final} = \omega_{initial} + \alpha t \). Since the wheel comes to a stop, \( \omega_{final} = 0 \). Filling in the values gives you the angular acceleration as \( \alpha = (\omega_{final} - \omega_{initial})/t \).
03

Calculate the torque

The torque, \( \tau \), exerted by the friction on the wheel is the product of force and radius: \( \tau = F*r \), where \( F = 70 N \) and \( r = 0.50 m \).
04

Calculate the friction force

The frictional force can be found by rearranging the definition of torque \( \tau = I \alpha \), where \( I = 12 kg m^{2} \) is the moment of inertia. The frictional force, \( F_{f} = \tau / r \).
05

Find the coefficient of kinetic friction

The coefficient of kinetic friction, \( \mu_k \), can be calculated using the equation \( F_{f} = \mu_k F \). Isolating \( \mu_k \) gives \( \mu_k = F_{f}/F \).

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 Speed Conversion
Angular speed conversion is an important concept in physics, especially in exercises involving rotational motion. The standard unit of angular speed is radians per second (\text{rad/s}), and it is common to convert other units like revolutions per minute (\text{rpm}) to this standard unit for easier computation.

For instance, when dealing with a potter鈥檚 wheel spinning at 50 \text{rpm}, we use the fact that one full revolution is equal to \(2\pi\) radians and that there are 60 seconds in a minute to perform the conversion. Using the formula \(\text{angular speed in rad/s} = \text{angular speed in rpm} \times \frac{2\pi}{60}\), you can calculate the angular speed in terms of \(\text{rad/s}\) which is essential for subsequent calculations in the exercise.
Angular Acceleration
Angular acceleration is the rate of change of angular velocity over time, denoted by the symbol \(\alpha\). In our situation, we need to calculate how quickly the potter's wheel is slowing down.

To find the angular acceleration, we need the initial angular velocity (converted to \(\text{rad/s}\) as explained above), the final angular velocity (which is zero, because the wheel stops), and the time it takes for the wheel to stop. The formula we use is \(\omega_{final} = \omega_{initial} + \alpha t\), where \(\omega\) is the angular velocity. Through some algebraic manipulation, we find \(\alpha = (\omega_{final} - \omega_{initial})/t\).
Torque Calculation
The concept of torque is critical in rotational motion. Torque, typically denoted as \(\tau\), is the measure of the force causing an object to rotate. The formula to calculate torque is \(\tau = F \times r\), where \(F\) is the force applied and \(r\) is the radius at which the force is applied.

In the context of the potter鈥檚 wheel, torque is generated by the frictional force of the wet rag acting at the rim of the wheel. Using the given force and the radius of the wheel, we can easily compute the torque being applied to the wheel.
Moment of Inertia
Moment of inertia is a measure of an object's resistance to changes in its rotational motion and is denoted by \(I\). The larger the moment of inertia, the harder it is to change the object's rotational speed.

In this exercise, the potter's wheel has a given moment of inertia, which allows us to relate torque and angular acceleration by the formula \(\tau = I\alpha\). After calculating the angular acceleration, we can determine the torque required to stop the wheel. With this value, we are then able to calculate the frictional force applied by the wet rag.
Physics Kinematic Equations
The kinematic equations in physics describe the motion of objects in terms of displacement, velocity, acceleration, and time. While these equations usually apply to linear motion, similar principles can be applied to rotational motion with corresponding variables: angular displacement, angular velocity, angular acceleration, and time.

In this case, we apply the rotational form of the kinematic equation to find the angular acceleration. This helps us understand how the potter鈥檚 wheel comes to a stop and directly relates to the calculation of kinetic friction between the wet rag and the wheel.

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 disk of radius \(0.250 \mathrm{~m}\) and mass \(55.0 \mathrm{~kg}\) rolls down a ramp of length \(4.50 \mathrm{~m}\) that makes an angle of \(15.0^{\circ}\) with the horizontal. The disk starts from rest from the top of the ramp. Find (a) the speed of the disk's center of mass when it reaches the bottom of the ramp and (b) the angular speed of the disk at the bottom of the ramp.

The top in Figure P8.49 has a moment of inertia of \(4.00\) \(\times 10^{-4} \mathrm{~kg} \cdot \mathrm{m}^{2}\) and is initially at rest. It is free to rotate about a stationary axis \(A A^{\prime}\). A string wrapped around a peg along the axis of the top is pulled in such a manner as to maintain a constant tension of \(5.57 \mathrm{~N}\) in the string. If the string does not slip while wound around the peg, what is the angular speed of the top after \(80.0 \mathrm{~cm}\) of string has been pulled off the peg? Hint: Consider the work that is done.

According to the manual of a certain car, a maximum torque of magnitude \(65.0 \mathrm{~N} \cdot \mathrm{m}\) should be applied when tightening the lug nuts on the vehicle. If you use a wrench of length \(0.330 \mathrm{~m}\) and you apply the force at the end of the wrench at an angle of \(75.0^{\circ}\) with respect to a line going from the lug nut through the end of the handle, what is the magnitude of the maximum force you can exert on the handle without exceeding the recommendation?

A constant torque of \(25.0 \mathrm{~N} \cdot \mathrm{m}\) is applied to a grindstone whose moment of inertia is \(0.130 \mathrm{~kg} \cdot \mathrm{m}^{2}\). Using energy principles and neglecting friction, find the angular speed after the grindstone has made \(15.0\) revolutions, Hint: The angular equivalent of \(W_{\mathrm{net}}=F \Delta x=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}\) is \(W_{\text {net }}\) \(=\tau \Delta \theta=\frac{1}{2} I \omega_{f}^{2}-\frac{1}{2} J \omega_{i}^{2} .\) You should convince yourself that this relationship is correct.

A person bending forward to lift a load "with his back" (Fig. P8.17a) rather than "with his knees" can be injured by large forces exerted on the muscles and vertebrae. The spine pivots mainly at the fifth lumbar vertebra, with the principal supporting force provided by the erector spinalis muscle in the back. To see the magnitude of the forces involved, and to understand why back problems are common among humans, consider the model shown in Figure P8.17b of a person bending forward to lift a \(200-\mathrm{N}\) object. The spine and upper body are represented as a uniform horizontal rod of weight \(350 \mathrm{~N}\), pivoted at the base of the spine. The erector spinalis muscle, attached at a point two-thirds of the way up the spine, maintains the position of the back. The angle between the spine and this muscle is \(12.0^{\circ}\). Find the tension in the back muscle and the compressional force in the spine.
See all solutions

Recommended explanations on Physics Textbooks

Modern Physics

Read Explanation

Linear Momentum

Read Explanation

Translational Dynamics

Read Explanation

Thermodynamics

Read Explanation

Fundamentals of Physics

Read Explanation

Solid State Physics

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.