A cylinder on an inclined plane with rope

  • Thread starter darxsys
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Homework Statement



A cylinder, with a rope wrapped around it, is placed on top of an inclined plane.If the speed of the cylinder in the bottom of the plane, when the angle of the plane is 30 degrees, is 1m/s, find what speed will the cylinder have when it is placed on a plane with an angle of 60 degrees. The length of the plane is 3m, radius of the cilinder is 0,5m, and the coefficient of friction is 0.2. Find the time needed for the cylinder to get to the bottom of the plane in case of a 60 degrees angle and find angular velocity of the cylinder when it is exactly on half of the height of the plane.

I hope I translated this correctly and that you'll be able to get it. Any help appreciated.

picture: http://img822.imageshack.us/img822/3240/screenshot3wb.png [Broken]

Homework Equations


I dont know how to put the equations right. Both friction and tension are needed in translational part. I thought of something like this:

mgsin[tex]\beta[/tex] - T - Fr = ma

for the rotational part:

M = Mt - Mfr
I dont know what to do next.


The Attempt at a Solution



I tried something using conservation of energy to find speed at the bottom of the plane, but I dont think I got it right cause I never used the coefficient of friction in my calculations nor tension of the rope.
 
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Answers and Replies

  • #2
tiny-tim
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welcome to pf!

hi darxsys! welcome to pf! :smile:
… I tried something using conservation of energy to find speed at the bottom of the plane, but I dont think I got it right cause I never used the coefficient of friction in my calculations nor tension of the rope.
he he :biggrin:

from your picture, the rope is fixed to the top of the plane, and goes over the top of the cylinder …

that means that the cylinder "rolls along the rope", but slips down the plane :wink:
 
  • #3
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thanx :D


Well, if it slips down the plane, then my first equation becomes mgsin[tex]\beta[/tex] - [tex]\mu[/tex] mgcos[tex]\beta[/tex]- T?
But now, what to do with the angular momentum? It should stay the same as I wrote it?

I mean, I dont have an exact idea of what should I calculate in order to have enough data to find velocity at the bottom of the plane? are both rolling and slipping important for that? I can get the tension from the two equations above now that i know friction, so now i can just ignore rolling and only consider slipping?
 
  • #4
tiny-tim
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hi darxsys! :smile:

as with all rotation problems, you need a linear equation, an angular equation, and an equation relating v and ω to connect them :wink:
 
  • #5
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well thats the problem. i know that v = r[tex]\omega[/tex] and i wrote the equations for both rolling and slipping.


ma = mgsin[tex]\beta[/tex] - [tex]\mu[/tex]mgcos[tex]\beta[/tex] - T

I[tex]\alpha[/tex]=rT - rmg[tex]\mu[/tex]cos[tex]\beta[/tex]

from which I got T = [tex]\frac{mg}{2}[/tex]sin[tex]\beta[/tex]

and got the acceleration 0,753 m/s^2. but the problem is that now i get that velocity in beta case is 2,12m/s when it reaches bottom of the plane and not 1 m/s
 
  • #6
tiny-tim
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you've used I = mr2

it isn't :redface:
 
  • #7
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Why not? its a hollow cylinder. i forgot to mention that. sorry. but that doesnt really change a lot does it?

http://en.wikipedia.org/wiki/List_of_moments_of_inertia

first one?


edit: i assume that the friction in beta case is not the same as in alfa case, but then again, i dont even need the beta case to calculate everything. im totally confused :D
 

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