Why does ignoring cylinder inertia give wrong time for rope acceleration?

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Homework Statement


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

Homework 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]

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.
 
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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.
 
courteous said:

Homework Statement


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?

PS.: Quite likely that I've made grammatical mistakes in 'problem statement'. Please correct me.

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?​
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])?

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:
 
LowlyPion said:
Solution 2 is correct because it's not F = m*a in 1) is it?
Indeed, I wrongly thought that, for a point where the rope meets the cylinder, Newton's second law would hold.
LowlyPion said:
Isn't it really
T = I * α
Where I = 1/2*m*r2 and α is rotational acceleration.
It is. Haven't I correctly used it in 2. attempt?

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

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:rolleyes:) horizontal direction?​
Any missing or superfluous?

tiny-tim said:
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")
Haven't thought of second "wound". Nice illustration with sword.

tiny-tim said:
iii] "equal to"
tiny-tim said:
iv] never "move for" …
Very gratious for pointing out my inveterate mistakes.:approve:

tiny-tim said:
v] why "geometrical"? :confused:
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!