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Rotating cylinder

  1. Jul 5, 2009 #1
    1. The problem statement, all variables and given/known data
    Cylinder with mass of 2 kg can be rotated around fixed horizontal geometrical axis. Around cylinder's circumference there is wounded rope, end of which is pulled by force equal 2,5 N in a horizontal direction, so that cylinder's rotation is accelerated. In what time does end of the rope move for 1,2 m in horizontal direction?
    http://img113.imageshack.us/img113/613/cylinder.png [Broken]


    2. Relevant equations
    • [itex]m=2kg[/itex]
    • [itex]F=2,5N[/itex]
    • [itex]x=1,2m[/itex]
    [itex]a_{tangential}=r\alpha[/itex]
    [itex]\phi=\omega_0t+\frac{1}{2}\alpha t^2[/itex]

    3. The attempt at a solution
    1. First and unsuccessful attempt (solution is [itex]1,0s[/itex]):
      [itex]\frac{F}{m}=a_{tangential}=r\alpha=r\frac{2\phi}{t^2}=r\frac{2\frac{x}{r}}{t^2}=\frac{2x}{t^2}\Rightarrow t=\sqrt\frac{2mx}{F}=1,4s[/itex]
    2. While writing this thread, I gave it another shot, this time including inertia [itex]I[/itex] (and getting, what seems, correct solution):
      [itex]\frac{2\phi}{t^2}=\alpha=\frac{\tau}{I}=\frac{Fr}{\frac{1}{2}mr^2}=\frac{F}{\frac{1}{2}m\frac{x}{\phi}}=\frac{2F\phi}{mx}\Rightarrow t=\sqrt\frac{mx}{F}=1,0s[/itex]

    What is wrong with first attempt? Ignoring inertia ([itex]I[/itex]) doesn't seem right, but why does (seemingly) rigorous 1. attempt not lead to correct solution (i.e., why is there an extra [itex]\sqrt2[/itex])?

    Yours truly,
    courteous.

    PS.: Quite likely that I've made grammatical mistakes in 'problem statement'. Please correct me.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 5, 2009 #2

    LowlyPion

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    Solution 2 is correct because it's not F = m*a in 1) is it?

    Isn't it really

    T = I * α

    Where I = 1/2*m*r2 and α is rotational acceleration.
     
  4. Jul 5, 2009 #3

    tiny-tim

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    Hi courteous! :wink:

    Yes, grammar is important! :smile:

    i] you keep leaving out "a" and "the" (but they're not needed for "with mass")

    ii] the present tense is "wind" (rhymes with "mind"), the past tense is "wound" (rhymes with "sound") (but if you stab someone with a sword, the present tense is "wound" (rhymes with "tuned"), and the past tense is "wounded")

    iii] "equal to"

    iv] never "move for" …

    v] why "geometrical"? :confused:

    A cylinder with mass (or "a mass") of 2 kg can be rotated around a fixed horizontal axis. Around the cylinder's circumference there is wound a rope (or "a rope is wound"), the end of which (or "whose end" or "and its end") is pulled by a force equal to 2,5 N (or "of 2,5 N ") in a horizontal direction (or "horizontally"), so that the cylinder's rotation is accelerated. How long does the end of the rope take to move 1,2 m in the horizontal direction?​
    Your first attempt is wrong because, in the formula F = ma, a is different for different parts of the cylinder …

    if the cylinder was only a cylindrical shell, so that all of it was at distance r from the axis, then I = mr, and your first attempt would be correct …

    but for a solid cylinder, a gets less as you get nearer the axis! :biggrin:
     
  5. Jul 5, 2009 #4
    Indeed, I wrongly thought that, for a point where the rope meets the cylinder, Newton's second law would hold.
    It is. Haven't I correctly used it in 2. attempt?

    Cylinder with mass of 2 kg can be rotated around a fixed horizontal geometrical axis. Around the cylinder's circumference there is a wounded rope, end of which is pulled by a force equal to 2,5 N in a horizontal direction, so that the cylinder's rotation is accelerated. In what time does the end of the rope move 1,2 m in a (or the:uhh:) horizontal direction?​
    Any missing or superfluous?

    Haven't thought of second "wound". Nice illustration with sword.

    Very gratious for pointing out my inveterate mistakes.:approve:

    You're right. It really is a tautology, "axis" already harbours "geometrical" ("axis" from latin axis:smile: or axle; better yet: sanskrit aksah "an axle, axis, beam of a balance").:blushing:

    tiny-tim, thank you!
     
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