How Does a Potter Stop a Rotating Wheel Using Friction?

  • Thread starter Thread starter mirandasatterley
  • Start date Start date
  • Tags Tags
    Rotational
Click For Summary

Homework Help Overview

The problem involves a potter's wheel, specifically examining the dynamics of stopping a rotating wheel using friction. The wheel has a defined radius and mass, and the scenario includes the application of a force to halt its motion over a specified time period.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to analyze the forces acting on the wheel, including friction and normal forces, while expressing confusion about the relationship between these forces and torque. Some participants suggest calculating angular acceleration and using it to find the frictional force, leading to the coefficient of friction.

Discussion Status

Participants are exploring the relationships between torque, angular acceleration, and frictional forces. Guidance has been provided regarding the steps to take, but there is no explicit consensus on the approach as multiple interpretations of the problem are being discussed.

Contextual Notes

The original poster expresses uncertainty about the forces affecting torque and the nature of the forces involved in the problem, indicating a need for clarification on these points.

mirandasatterley
Messages
61
Reaction score
0

Homework Statement



A potter's wheel - A thick stone disk of radius 0.5m and mass 100kg is freely rotating at 50rev/min. The potter can stop the wheel in 6s by pressing a rag against the rim and exerting a radially inward force of 70N. Find the effective coefficient of kinetic friction between the wheel and the rag.

Homework Equations



torque = RF
Torque = I(moment of inertia) x radial acceleration

The Attempt at a Solution



Sum of forces on the x axis: F(applied) - F(normal) = 0
F(applied) = F(normal)
The froce of friction is on the y-axis and is opposite to the direction the wheel turns, and F(friction) = (coefficient of friction)F(normal)

Sum of forces on the y axis: I'm not sure what other force opposes the frictional force. Normally I would have thought F(friction) = ma, but it is not a particle that we are looking at, so I'm a little confused.

And for the sum of the forces for torque, i know Torque = I(angular acceleration). Fir this i think I= 1/2 MR^2, and I can find angular acceleration with angular speed(final) = angular speed(initial) + angular acceleration X t.

I'm also not sure what forces affect torque.

Any help is appreciated.
 
Physics news on Phys.org
Your nearly there. Basically, the friction from the rag is the torque, you know how long it takes the wheel to stop and its initial angular velocity and therefore you can work out the angular acceleration. From this you can work out the applied torque, and hence the frictional force; from which you can calculate the coefficient of friction.
 
So, i find the angular acceleration, using:
final angular speed = initial angular speed + (angular acceleration)t

Then, I use the angular acceleration to find the frictional force by:
torque = RF(friction) and torque = I x (angular acceleration), therefore,
RF(friction) = I x angular acceleration.

Finally I use F(friction) = coefficient x F(normal), where F(normal) = F(applied), and I solve for the coefficient of friction.

Is this what I need to do?
Thanks a lot for all your help.
 
mirandasatterley said:
So, i find the angular acceleration, using:
final angular speed = initial angular speed + (angular acceleration)t

Then, I use the angular acceleration to find the frictional force by:
torque = RF(friction) and torque = I x (angular acceleration), therefore,
RF(friction) = I x angular acceleration.

Finally I use F(friction) = coefficient x F(normal), where F(normal) = F(applied), and I solve for the coefficient of friction.

Is this what I need to do?
That's exactly what you need to do :approve:
mirandasatterley said:
Thanks a lot for all your help.
Twas a pleasure
 

Similar threads

  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
13
Views
2K
Replies
19
Views
3K
  • · Replies 33 ·
2
Replies
33
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
4
Views
2K