Rotating Cylinder Homework: 2kg Mass, 2.5N Force, 1.2m Horizontal Distance

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Homework Help Overview

The problem involves a cylinder with a mass of 2 kg that is rotated around a fixed horizontal axis by a rope wound around its circumference, which is pulled by a force of 2.5 N. The question seeks to determine the time it takes for the end of the rope to move 1.2 m horizontally.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to analyze the problem using Newton's second law and rotational dynamics, questioning the validity of their first attempt which ignored inertia. Some participants suggest that the relationship between force, mass, and acceleration may not apply uniformly across the cylinder.

Discussion Status

Participants are exploring different interpretations of the problem, particularly the implications of inertia in the calculations. There is a recognition that the first attempt may not account for the distribution of mass in the cylinder, leading to differing conclusions about the time calculation.

Contextual Notes

Some participants note grammatical issues in the problem statement, which may affect clarity but do not directly impact the physics involved. The discussion also touches on the nuances of language used in the problem description.

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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


  • m=2kg
  • F=2,5N
  • x=1,2m
a_{tangential}=r\alpha
\phi=\omega_0t+\frac{1}{2}\alpha t^2

The Attempt at a Solution


  1. First and unsuccessful attempt (solution is 1,0s):
    \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
  2. While writing this thread, I gave it another shot, this time including inertia I (and getting, what seems, correct solution):
    \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

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

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 (I) doesn't seem right, but why does (seemingly) rigorous 1. attempt not lead to correct solution (i.e., why is there an extra \sqrt2)?

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!
 

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