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Rotational Motion problem

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Homework Statement



From the text: "A [itex]50 kg[/itex] mass is tied to a massless rope wrapped around a solid cylindrical drum. The drum is mounted on a frictionless horizontal axle. When the mass is released, it falls with acceleration [itex]a = 3.7 \frac{m}{s^2}[/itex]. Find (a) the tension in the rope and (b) the mass of the drum.

Mass of the falling object = [itex]50 kg[/itex]
Acceleration of gravity = [itex]9.8 \frac{m}{s^2}[/itex]
Net acceleration of falling object = [itex]3.7 \frac{m}{s^2}[/itex]

Homework Equations



Newton's 2nd Law
[tex]\sum \vec{F} = m \vec{a}[/tex]

Definiton of torque in terms of radius and Force applied
[tex]\tau = r \vec{F} \sin (\theta)[/tex]

Definition of torque in terms of rotational intertia and rotational acceleration
[tex]\tau = I \alpha[/tex]

Definition of rotational acceleration in terms of radius and tangential acceleration
[tex]\alpha = \frac{a}{r}[/tex]

Rotational intertia for a cylinder
[tex]I = \frac{1}{2} MR^2[/tex]

The Attempt at a Solution



I know that [itex]a_\textit{gravity} - a_\textit{rope} = a_\textit{final}[/itex] for the falling block. this gave me an upward acceleration of [itex]6.1 \frac{m}{s^2}[/itex], which gave me my first answer:
[tex] T = -305N[/tex]

For my second answer, i set the definitons of torque in terms of quantities that i knew and the one quantity that i didn't know - mass of the falling object, mass of the drum, acceleration of the falling object, force of the falling object.

My equation looked like this before i started cancelling out things:

[tex] (\frac{1}{2}Mr^2) (\frac {a}{r}) = (rma)[/tex]

The radius values and accelerations cancelled out, leaving me with this:

[tex] M = 2m [/tex]

This can't be right because the mass of the larger object can't be entirely dependant on the mass of the smaller object. Could someone help me out?
 
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Answers and Replies

  • #2
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I know that [itex]a_\textit{gravity} - a_\textit{rope} = a_\textit{final}[/itex] for the falling block. this gave me an upward acceleration of [itex]6.1 \frac{m}{s^2}[/itex], which gave me my first answer:
[tex] T = -305N[/tex]
I have not checked the numbers. However, I don't understand how you conclude that the difference between the acceleration due to gravity and the rope is equal to the final acceleration. I suggest drawing a force-body diagram. You should quickly determine what the tension is from it.

[tex] M = 2m [/tex]

This can't be right because the mass of the larger object can't be entirely dependant on the mass of the smaller object. Could someone help me out?
Why can't that be? It is perfectly possible for the mass of the drum to be twice that of the falling mass.
 
  • #3
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Additional info

sorry about that. i attached a free body diagram. also, i posted the formula incorrectly, but the posted answer i obtained like this.

[tex]9.8 + a_\textit{rope} = 3.7[/tex]
[tex]a_\textit{rope} = -6.1[/tex]
[tex]T = -305N[/tex]

as for my question - I predicted that the mass of the drum would be larger, but i suppose I was looking for a more complicated equation. Maybe I'm having an issue grasping that the acceleration and radius cancels out completely in my formula, because it doesn't seem plausible.
 

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  • #4
Doc Al
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I know that [itex]a_\textit{gravity} - a_\textit{rope} = a_\textit{final}[/itex] for the falling block.
Where does this come from? Instead, write the force equation for the hanging mass per Newton's 2nd law.
 
  • #5
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ok, ok, lol... i'll write the force equation.

[tex] \sum \vec{F} = m \vec{a}[/tex]
[tex] \sum \vec{F} = \vec{F}_\textit{gravity} + \vec{F}_\textit{rope}[/tex]
[tex] \sum \vec{F} - \vec{F}_\textit{gravity} = \vec{F}_\textit{rope}[/tex]
[tex] 185N - 490N = -305N[/tex]

I opted to leave mass out and work with acceleration in the beginning for easier calculating... that didn't account for my sign error in my equation though, i just copied it wrong... unless i'm calculating for the tension incorrectly????:(
 
  • #6
Doc Al
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ok, ok, lol... i'll write the force equation.

[tex] \sum \vec{F} = m \vec{a}[/tex]
[tex] \sum \vec{F} = \vec{F}_\textit{gravity} + \vec{F}_\textit{rope}[/tex]
[tex] \sum \vec{F} - \vec{F}_\textit{gravity} = \vec{F}_\textit{rope}[/tex]
[tex] 185N - 490N = -305N[/tex]

I opted to leave mass out and work with acceleration in the beginning for easier calculating... that didn't account for my sign error in my equation though, i just copied it wrong... unless i'm calculating for the tension incorrectly????:(
Much better. (That acceleration equation made no sense on its own.)

You are messing up the directions of the forces (and thus their signs), that's why you get a negative answer. The forces are gravity and rope tension: which way do they act? What's the direction of the acceleration? Hint: Use a consistent sign convention.

For my second answer, i set the definitons of torque in terms of quantities that i knew and the one quantity that i didn't know - mass of the falling object, mass of the drum, acceleration of the falling object, force of the falling object.

My equation looked like this before i started cancelling out things:

[tex] (\frac{1}{2}Mr^2) (\frac {a}{r}) = (rma)[/tex]
Your expression for torque--the right hand side of the equation--is incorrect.
 
  • #7
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Much better. (That acceleration equation made no sense on its own.)

You are messing up the directions of the forces (and thus their signs), that's why you get a negative answer. The forces are gravity and rope tension: which way do they act? What's the direction of the acceleration? Hint: Use a consistent sign convention.
alright... i'll have upward be positive, giving me -185 + 490 = 305 N. I don't know why i had it backwards, to be honest.

Your expression for torque--the right hand side of the equation--is incorrect.
really? Are you talking about this: [itex]\tau = r \vec{F} \sin(\theta)[/itex]? I changed F to ma, and i see the point that the rope begins to hang freely as making a horizontal line with the radius, making the Force of the torque being perpendicular: so
[tex]\sin(\frac{\pi}{2}) = 1[/tex]
leaving me with r*m*a.
 
  • #8
Doc Al
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Are you talking about this: [itex]\tau = r \vec{F} \sin(\theta)[/itex]? I changed F to ma...
Why? The F you need is the force exerting the torque on the drum--what's that? "ma" would be the net force on the mass--not relevant.
 
  • #9
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Why? The F you need is the force exerting the torque on the drum--what's that? "ma" would be the net force on the mass--not relevant.
Would the tension on the rope be the force exerting the torque on the drum?
 
  • #10
Doc Al
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Would the tension on the rope be the force exerting the torque on the drum?
Of course. It's the rope that pulls on the drum.
 
  • #11
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ok... so the F on the right side of my equation is 305N, and then a on the left side is 3.7m/s^2, the net acceleration of the closed system. therefore, my final equation should be like this:

[tex]M = \frac{2F_\textit{rope}}{a_\textit{net}}[/tex]

*calculating*

about 165kg. that makes a LOT more sense, heh:) So my equation was almost right, i just misunderstood what variables meant what. thank you very much, Doc Al and e(ho0n3 :) Thanks to you... i have all week to do other homework.
 

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