Maximum radius given rps, static and kinetic friction coefficients

Click For Summary

Homework Help Overview

The problem involves a scenario where a pizza crumb is placed on a rotating magnetic disk, and the objective is to determine the maximum radius from the axis at which the crumb can remain without slipping. The context includes concepts of static and kinetic friction, centripetal force, and rotational motion.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to equate centripetal force with frictional force and questions the correct approach to the problem. Other participants discuss the role of static friction and the relationship between normal force and centripetal force, while also considering the implications of the kinetic friction coefficient.

Discussion Status

Participants are exploring the forces involved and clarifying the role of static versus kinetic friction. There is a recognition that static friction is the primary concern until motion occurs, and some guidance has been provided regarding the relationship between forces and acceleration.

Contextual Notes

There is uncertainty regarding the mass of the crumb and how it affects the calculations, as well as the specific role of the kinetic friction coefficient in this scenario.

tehnexus
Messages
2
Reaction score
0

Homework Statement



A computer scientist has dropped a tiny pizza crumb onto a magnetic disk. Later, when it's turned on, the disk is rotating at 88 revolutions per second about its axis, which is vertical. The coefficients of kinetic and static friction between pizza crumbs and the disk are 1.3 and 1.5 respectively. What is the maximum radius from the axis at which a pizza crumb will stay on the disk?

Homework Equations



F=ma, a=v^2/r, v=rω

The Attempt at a Solution



i really am not sure how to do this question, am i supposed to equate the centripetal force on the crumb with the frictional force? i have converted the rps to rad/s which is 552.93. please help me!
 
Physics news on Phys.org
There are a couple of things to consider here. The first thing is understanding the forces on the object.
F_M = \mu_s * F_N
F_M = Maximum force of friction
\mu_s = Static coefficient of friction
F_N = Normal force

Static friction doesn't always produce a force. It only produces a force up to a maximum in opposition to motion on a plane. Now you already mentioned centripetal force. This is the force that is required to make something rotate. If you don't have enough centripetal force, then you don't have a solid orbit. Normal force is perpendicular to the plane of motion. Can you think of any forces that are perpendicular to the plane of motion? Basically what we need to find is what horizontal centripetal force exceeds the perpendicular normal force times the coefficient.

F = ma

We don't know the mass, but we know the forces in question are applied to the same mass, so what matters is the acceleration. If you can figure out an acceleration acting perpendicular to the disc, then you should be able to find the maximum horizontal acceleration toward the center that can occur with the static coefficient of friction.
 
So the kinetic friction coefficient does not affect the question? but i understand what you are saying thanks.
 
Last edited:
Correct. Static friction is a ratio that represents how well two things are stuck together. It is all that applies until we have movement between the objects. Once we have movement you kind of get a skipping motion on the microscopic level. Think of it like two pieces of metal that have fitting triangular teeth on their surface. If you press against them and push, they will stay together. If you ever managed to push hard enough the teeth would start to skip against each other, and the amount of resistance in that skipping is less than when they fully fit together. That is why we need to numbers. We need one for when things are fully fit together and move as one, and one for when they start to slip and skip against each other.
 
Last edited:

Similar threads

  • · Replies 16 ·
Replies
16
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 33 ·
2
Replies
33
Views
3K
  • · Replies 28 ·
Replies
28
Views
6K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
5
Views
3K
  • · Replies 8 ·
Replies
8
Views
8K
Replies
48
Views
8K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
9
Views
4K