Friction on a casing on a metal ring

  • Thread starter Victarion
  • Start date
  • #1
2
0

Homework Statement



A casing of mass m can glide along a thin metal ring lying horizontally. It has an initial speed of v0. How long a distance s will it glide before stopping if its friction coefficient is μ?

The answer is given as

s = r/2μ * ln( v02 + √(v02 + g2r2)/ g * r )

The task is to provide the deductive path to reach that answer.

Homework Equations



Centripetal force: mv2/r

Gravitational force: mg

Kinetic energy: mv2/2

The Attempt at a Solution



The casing will stop once all the kinetic energy has turned into friction work. The friction is a sum of that caused by the gravitational force and the centripetal force. The friction work Wf is thus

Wf = s * F = s * μ * ( mv2/r + mg )

Since Wf is equal to the kinetic energy we can write the following equation:

s = Kinetic energy / F = (mv2/2) / (μ * ( mv2/r + mg ))

This can be quite cleanly simplified into something that looks similar to the answer I am supposed to reach:

s = r/2μ * ( v2/ (gr + v2) )

But this is where I run into a brick wall. I can't figure out how to integrate it properly, all the similar equations I have seen integrated seem to include arctan somewhere in the answer.

I apologise any mistakes I made in the translation of the problem and in laying them out clearly, and thank you for your attention!
 
Last edited:

Answers and Replies

  • #2
Delphi51
Homework Helper
3,407
10
I haven't solved this, just starting now, but already I am thinking the two forces must be combined as vectors not simply added. We have mg vertically and mv²/R horizontally so the combined force magnitude will be the common m times the square root of g² + (v²/R)². Also, the velocity is a variable, and it has a complicated non-linear acceleration μF so we are in for some trouble - can we even find the velocity function?

This problem looks tough for me; I have long since forgotten my differential equations class 40 years ago so I hope others will pitch in!
 
  • #3
2
0
I am pretty sure that friction of this kind means that we do not have to look on the forces as vectors. What the friction coefficient is, as far as I am aware, is the portion of a perpendicular force that is turned into friction, and friction is by definition a force that is directly opposed to the movement of the object.

If that made any sense.

The velocity function sounds tricky to find. The way I figured we would get the value for s is by integrating the function with regards to v, with v going from v0 to 0. Since we know the start and end points of the interval I was hoping we would not need any closer information on how quickly the casing actually decelerates, but that depends on actually getting an analytical answer to the integral.

I really appreciate your input by the way!
 
  • #4
Delphi51
Homework Helper
3,407
10
Friction depends on the normal force, pressing the slider against the ring. That is the vector sum of the horizontal and vertical forces so I'm pretty sure the normal force magnitude is square root of {g² + (v²/R)²} all right.

I'm playing with this, writing F = ma, μFn = ma,
a = dv/dt = μF/m where F = the square root above.
This looks like it might be possible to integrate with a substitution. If you could get it to say
dw/sqrt(1 + w²) = constants * dt
could you integrate it? If so, try to come up with a w substitution that incorporates the R and a g so you can factor out the g from the square root and leave just (1 + w²) under the square root. If you need a hint, try letting w = v*sqrt(R) and see how it simplifies.
 
Last edited:
  • #5
Delphi51
Homework Helper
3,407
10
I did the integration and was able to solve it for v and integrate again to get x.
But I didn't get anything like the answer given for the question!
It is probably worth you working it through, though - I'm having a lot of trouble with calculations these days.
Your turn to help me! Hopefully somebody else will pitch in.
 

Related Threads on Friction on a casing on a metal ring

Replies
2
Views
688
  • Last Post
Replies
5
Views
19K
  • Last Post
3
Replies
51
Views
4K
  • Last Post
Replies
2
Views
2K
Replies
16
Views
930
  • Last Post
Replies
5
Views
5K
Replies
0
Views
2K
Replies
3
Views
6K
  • Last Post
Replies
5
Views
10K
Top