Calculating Net Torque on Counterweight System

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Homework Help Overview

The discussion revolves around calculating the net torque on a counterweight system involving a spool and a mass. The problem is situated within the context of rotational dynamics, specifically focusing on the forces and torques acting on a system comprising a counterweight and a spool.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the definition of "the system" in relation to torque calculations, questioning whether the net torque should include both the mass and the spool. There are attempts to clarify how external forces contribute to the net torque and discussions about the role of gravity in this context.

Discussion Status

The discussion is active, with participants providing hints and guidance on how to approach the problem. There is an exploration of different interpretations of torque and its application to the system, with some participants expressing confusion about the concepts involved.

Contextual Notes

Participants note the complexity of understanding torque in relation to linear forces and the specific conditions of the problem, such as the role of the center of mass and the definition of moment arm. There is an acknowledgment of the challenge in visualizing these concepts in the context of the problem set.

Antepolleo
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I can't for the life of me figure out what I'm doing wrong here. Here's the problem:

A 5.40 kg counterweight is attached to a light cord, which is wound around a spool (refer to Fig. 10.20). The spool is a uniform solid cylinder of radius 7.00 cm and mass 2.00 kg.

10-20.gif


Ok... the first question is:

What is the net torque on the system about the point O?

Here's my take on the situation... the only torque being applied to the system is from the tension T of the string.


T = tension of string
m = mass of counterweight
a = acceleration of counterweight
R = radius of wheel in meters
M = mass of pulley


[tex]T - mg = -ma[/tex]

[tex]T = m(g - a)[/tex]

[tex] \begin{equation*}<br /> \begin{split}<br /> \sum\tau &= TR \\<br /> &= mR(g - a)<br /> \end{split}<br /> \end{equation*}[/tex]

Also...

[tex] \begin{equation*}<br /> \begin{split}<br /> \tau &= I\alpha\\<br /> TR &= \frac{1}{2}MR^2\alpha\\<br /> \alpha &= \frac{2T}{MR}<br /> \end{split}<br /> \end{equation*}[/tex]

Now we know that:

[tex] \begin{equation*}<br /> \begin{split}<br /> a &= R\alpha\\<br /> <br /> a &= R(\frac{2T}{MR}) \\<br /> a &= \frac{2T}{M}<br /> <br /> \end{split}<br /> \end{equation*}[/tex]

So...

[tex] \begin{equation*}<br /> \begin{split}<br /> T &= m(g - \frac{2T}{M})\\<br /> T &= mg(\frac{M}{M + 2m})<br /> \end{split}<br /> \end{equation*}[/tex]

So am I right in saying that the net torque on the system is

[tex]\tau_{net} = [mg(\frac{M}{M + 2m})]R[/tex]
 
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Originally posted by Antepolleo

So am I right in saying that the net torque on the system is

[tex]\tau_{net} = [mg(\frac{M}{M + 2m})]R[/tex]
That depends on what you mean by "the system". You calculated (perfectly, as far as I can see) the net torque on the disk.

If the question really was find the net torque on "the system" (meaning mass plus disk), then no. (That's a bit of a trick question.)
 


Originally posted by Doc Al
That depends on what you mean by "the system". You calculated (perfectly, as far as I can see) the net torque on the disk.

If the question really was find the net torque on "the system" (meaning mass plus disk), then no. (That's a bit of a trick question.)

Hmm... if we call the mass plus the disk the system, how would I go about finding the torque on the system? I'm afraid I don't know where to start.
 


Originally posted by Antepolleo
Hmm... if we call the mass plus the disk the system, how would I go about finding the torque on the system? I'm afraid I don't know where to start.
Here's a hint: the only forces that can contribute to the net torque on the total system are external forces.
 


Originally posted by Doc Al
Here's a hint: the only forces that can contribute to the net torque on the total system are external forces.

The only external force acting the system is gravity... so perhaps its just mg! But why would that be a "torque"?

I'm afraid these angular concepts aren't as intuitive for me as the linear concepts. Am I way off?
 


Originally posted by Antepolleo
The only external force acting the system is gravity... so perhaps its just mg! But why would that be a "torque"?

I'm afraid these angular concepts aren't as intuitive for me as the linear concepts. Am I way off?
Yes, the only (unbalanced) external force is mg; what torque does that force exert about the axis?

(You are not off at all!)
 


Originally posted by Doc Al
Yes, the only (unbalanced) external force is mg; what torque does that force exert about the axis?

(You are not off at all!)

The force of gravity acts through the center of mass of both objects, so I don't see how it could create any torque. Unless it has something to do with the center of mass of the entire system... but I don't think that would be something that would be included in this problem set.
 


Originally posted by Antepolleo
The force of gravity acts through the center of mass of both objects, so I don't see how it could create any torque. Unless it has something to do with the center of mass of the entire system... but I don't think that would be something that would be included in this problem set.
Yes, but what's important is: does the force exert a torque about the axis?

The weight of the disk (and the force that holds it up) act right through the axis. So they exert no torque.

But the weight of the mass (mg) does exert a torque (about O) = mgR !
 


Originally posted by Doc Al
Yes, but what's important is: does the force exert a torque about the axis?

The weight of the disk (and the force that holds it up) act right through the axis. So they exert no torque.

But the weight of the mass (mg) does exert a torque (about O) = mgR !

Oh of course! Because the counterweight is connected to the pulley, it transfers the force exterted on it due to gravity into a torque about the axis.

I see it now. Thanks for your help!
 
  • #10


Originally posted by Antepolleo
Oh of course! Because the counterweight is connected to the pulley, it transfers the force exterted on it due to gravity into a torque about the axis.
Careful about thinking about anything "transfering" because it's connected to the pulley. If I cut the rope, the torque about 0 is still mgR ! (by definition of torque) No, you're not going crazy.

Just for fun: Treating the system as a whole, find the acceleration of the mass. Like this:

Torquenet = mgR = Inet&alpha;

Then compare to the answer you get using your original method of treating each body (mass, disk) separately.
 
  • #11


Originally posted by Doc Al
Careful about thinking about anything "transfering" because it's connected to the pulley. If I cut the rope, the torque about 0 is still mgR ! (by definition of torque) No, you're not going crazy.

Just for fun: Treating the system as a whole, find the acceleration of the mass. Like this:

Torquenet = mgR = Inet&alpha;

Then compare to the answer you get using your original method of treating each body (mass, disk) separately.


*blinks twice*

*passes out*

I'm interested in understanding this instead of just memorizing the problems and equations, so if you'll bear with me I'd really appreciate it.

How is the torque of the system still mgR if there is force acting to rotate an object?
 
  • #12


Originally posted by Antepolleo
How is the torque of the system still mgR if there is force acting to rotate an object?
I don't understand the question. Torque = Force x R, where R is the moment arm.

(Keep going... I'll be back in a few minutes.)
 
  • #13


Originally posted by Doc Al
I don't understand the question. Torque = Force x R, where R is the moment arm.

(Keep going... I'll be back in a few minutes.)

Hmmm, maybe my definitions are mixed up. I like to think of torque as the part of a force that goes into rotating an object about a defined axis.

How would you define moment arm?
 
  • #14
How about this definition of torque. The torque about the origin of Force (F) acting at position (R, a vector measured from the origin):

&Tau; = RF sin&theta; , where &theta; is the angle between the position vector and the force vector. Make any sense at all?

"Moment arm" is an old fashioned term for the perpendicular distance from the line on which the force acts and the axis; it equals R sin&theta;
 
  • #15
Originally posted by Doc Al
How about this definition of torque. The torque about the origin of Force (F) acting at position (R, a vector measured from the origin):

&Tau; = RF sin&theta; , where &theta; is the angle between the position vector and the force vector. Make any sense at all?

"Moment arm" is an old fashioned term for the perpendicular distance from the line on which the force acts and the axis; it equals R sin&theta;

I'm getting that... I'm just trying to get a grasp on the significance of torque, as it applies to motion and force in general.

To do this I'm trying to equate torque with something that I can easily get a handle on. I can understand the component of a force that causes an object to rotate, and that appears to be what torque is in essence. As the radious increases, so does the torque, meaning that a greater force is required to angularly accelerate the object.
 
  • #16
Originally posted by Antepolleo
To do this I'm trying to equate torque with something that I can easily get a handle on. I can understand the component of a force that causes an object to rotate, and that appears to be what torque is in essence. As the radious increases, so does the torque, meaning that a greater force is required to angularly accelerate the object.
Let's not get lost in semantics. I'm sure you have the right idea, but I don't like that last sentence. As the radius increases you need less force to get the same angular acceleration (if the rotational inertia remains the same).

Think of a wrench. The longer it is, the easier it is to exert torque on a bolt. (I'm sure you know that! :smile: )
 
  • #17
Originally posted by Doc Al
Let's not get lost in semantics. I'm sure you have the right idea, but I don't like that last sentence. As the radius increases you need less force to get the same angular acceleration (if the rotational inertia remains the same).

Think of a wrench. The longer it is, the easier it is to exert torque on a bolt. (I'm sure you know that! :smile: )

Heh, I messed up a bit in my though process... you're right.

Here's the second part of the problem:

When the counterweight has a speed v, the pulley has an angular speed &omega; = v/R. Determine the total angular momentum of the system about O.

Now that we've changed my concept of what system we're working in, I'm not sure how to tackle this.
 
  • #18
Originally posted by Antepolleo
Now that we've changed my concept of what system we're working in, I'm not sure how to tackle this.
You know the angular momentum of the disk. For the angular momentum of the mass, go back to the basic definition:

Angular momentum = R x Linear Momentum x Sin&theta; (look familiar?)

Make sense? It's a bit strange to talk of the angular momentum of something going straight, but you certainly can do it!

The total angular momentum of the system is the sum of the angular momenta of the parts.
 
  • #19
from where do you know the angular momentum of the disk?

how do you do the cross product?
 

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