Finding Tension and Mass in a Rotational Motion Problem

In summary: Force of the torque being perpendicular: so\sin(\frac{\pi}{2}) = 1leaving me with r*m*a.That doesn't make sense. The tension force is not perpendicular to the radius... And why did you just write "rma" when you actually evaluated the sin for the angle between them?In summary, the problem involves a 50 kg mass tied to a massless rope wrapped around a solid cylindrical drum mounted on a frictionless horizontal axle. The mass is released and has a final downward acceleration of 3.7 m/s^2. The tension
  • #1
xaer04
38
0

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 canceled 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?
 
Last edited:
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  • #2
xaer04 said:
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
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.
 

Attachments

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  • #4
xaer04 said:
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
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
xaer04 said:
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.

xaer04 said:
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
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
xaer04 said:
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
Doc Al said:
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
e(ho0n3 said:
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
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.
 

What is rotational motion?

Rotational motion is the movement of an object around an axis or center point. This type of motion is often seen in objects such as wheels, planets, and rotating machinery.

What is angular velocity?

Angular velocity is the rate at which an object rotates around an axis. It is measured in radians per second and is denoted by the symbol ω.

What is moment of inertia?

Moment of inertia is a measure of an object's resistance to rotational motion. It depends on the mass, shape, and distribution of mass of the object. It is denoted by the symbol I and has units of kg m².

What is torque?

Torque is a measure of the force that causes an object to rotate. It is calculated by multiplying the force applied to an object by the distance from the axis of rotation to the point where the force is applied.

How do I solve rotational motion problems?

To solve rotational motion problems, you can use equations such as Newton's second law for rotational motion, which states that the net torque on an object is equal to its moment of inertia times its angular acceleration. It is also important to draw a diagram and clearly label all forces and distances involved in the problem.

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