Finding velocity given radius and coefficient of friction

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Homework Help Overview

The problem involves a box on the edge of a circular platform, with parameters including mass, radius, and coefficient of friction. The goal is to determine the speed at which the box slips off the edge, relating to concepts of circular motion and friction.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between centripetal force and friction, with some suggesting that the acceleration may be ignored. Others question how non-uniform circular motion affects the analysis and whether tangential acceleration can be disregarded.

Discussion Status

There is an ongoing exploration of the conditions under which the box will slip, with some guidance provided on focusing on the balance between centripetal force and friction. Multiple interpretations of the problem's dynamics are being considered, particularly regarding tangential acceleration.

Contextual Notes

Participants note that the problem may involve complexities due to non-uniform circular motion and the potential for kinetic friction to come into play, which adds layers to the analysis.

Mooshk
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Homework Statement


A box with mass 2kg is on the edge of a circular platform of radius 6.0m. The coefficient of friction between the platform and the box is 0.3. The platform accelerates. Determine the speed when the box slips off the edge.

Homework Equations


Fs = μsN
F⃗ net=ΣF⃗ =ma⃗
a(t) = d(vt)/dt
a(r) = v^2/r
a = [sqrt (ar^2) + (at^2)]

The Attempt at a Solution


I know that this is non-uniform circular motion.
I don't think that what I'm doing is right.

ΣF(t) = -Ff = m(at)
at = - μs(g)
a(t) = - (0.3)(9.8)
a(t) = -2.94 m/s^2

ΣF(r) = m(ar)
ΣF(r) = (mv^2) / r

ΣF(net) = m * [sqrt (ar^2) + (at^2)]
Ff = m * [sqrt (ar^2) + (at^2)]
μs(g) = [sqrt (μs^2(g^2)) + (m^2*v^4) / r^2]
(μg)^2 = (μs^2(g^2)) + (m^2v^4) / r^2
((μg)^2 - μs^2(g^2)) /r^2 = (m^2*v^4)

So if I solve for v, I would take the fourth root of the left side after bringing m^2 over, and that doesn't seem right to me.
 
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I think you are working too hard. The acceleration is not given, suggesting you ignore F = ma. To my mind, it sounds like the condition for beginning to slip is simply when friction can no longer provide the necessary centripetal force to hold it on:
centripetal force = force of friction
which easily solves for v.
 
Delphi51 said:
I think you are working too hard. The acceleration is not given, suggesting you ignore F = ma. To my mind, it sounds like the condition for beginning to slip is simply when friction can no longer provide the necessary centripetal force to hold it on:
centripetal force = force of friction
which easily solves for v.
How would that model non-uniform circular motion? Or does that not matter since the friction force is perpendicular?
 
Strictly speaking, it does not apply to non-uniform circular motion where it could be that high tangential acceleration will cause the thing to slip into kinetic friction mode before centrifugal force overcomes the grip. If you have been doing problems like that in class, then do it with this one, too. Note that you will have to consider the combined radial and tangential forces and calculate when that exceeds the friction force.

But the wording suggests to me that the tangential acceleration is so small it can be ignored and slipping will occur in the radial direction.
 

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