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

In summary, the end of the rope moves for 1,2 m in a horizontal direction when the cylinder's rotation is accelerated around its circumference.
  • #1
courteous
39
0

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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  • #2
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.
 
  • #3
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:
 
  • #4
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:uhh:) 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!
 

1. What is the formula for calculating the rotational force on the cylinder?

The formula for calculating the rotational force (torque) on the cylinder is T = Fd, where T is the torque, F is the applied force, and d is the distance from the axis of rotation to the point where the force is applied.

2. How do you convert the applied force of 2.5N to units of torque?

To convert the applied force of 2.5N to units of torque, you simply multiply it by the distance of 1.2m. This will give you a torque of 3Nm (newton-meters).

3. What is the moment of inertia for the 2kg mass?

The moment of inertia for a 2kg mass can be calculated by using the formula I = mr^2, where I is the moment of inertia, m is the mass, and r is the distance from the axis of rotation to the mass. In this case, the moment of inertia would be 2kg x (1.2m)^2 = 2.88kgm^2.

4. How would changing the distance of the applied force affect the rotational force?

Changing the distance of the applied force would affect the rotational force by changing the magnitude of the torque. As the distance increases, the torque also increases, and vice versa.

5. What is the relationship between torque and angular acceleration for a rotating cylinder?

The relationship between torque and angular acceleration for a rotating cylinder is given by the formula T = Iα, where T is the torque, I is the moment of inertia, and α is the angular acceleration. This means that the torque is directly proportional to the angular acceleration, and a greater torque will result in a greater angular acceleration.

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