# Torque and such

1. Nov 23, 2003

### Antepolleo

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.

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

$$T - mg = -ma$$

$$T = m(g - a)$$

$$\begin{equation*} \begin{split} \sum\tau &= TR \\ &= mR(g - a) \end{split} \end{equation*}$$

Also...

$$\begin{equation*} \begin{split} \tau &= I\alpha\\ TR &= \frac{1}{2}MR^2\alpha\\ \alpha &= \frac{2T}{MR} \end{split} \end{equation*}$$

Now we know that:

$$\begin{equation*} \begin{split} a &= R\alpha\\ a &= R(\frac{2T}{MR}) \\ a &= \frac{2T}{M} \end{split} \end{equation*}$$

So...

$$\begin{equation*} \begin{split} T &= m(g - \frac{2T}{M})\\ T &= mg(\frac{M}{M + 2m}) \end{split} \end{equation*}$$

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

$$\tau_{net} = [mg(\frac{M}{M + 2m})]R$$

2. Nov 23, 2003

### Staff: Mentor

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.)

3. Nov 23, 2003

### Antepolleo

Re: Re: Torque and such...

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.

4. Nov 23, 2003

### Staff: Mentor

Re: Re: Re: Torque and such...

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

5. Nov 23, 2003

### Antepolleo

Re: Re: Re: Re: Torque and such...

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?

6. Nov 23, 2003

### Staff: Mentor

Re: Re: Re: Re: Re: Torque and such...

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

(You are not off at all!)

7. Nov 23, 2003

### Antepolleo

Re: Re: Re: Re: Re: Re: Torque and such...

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.

8. Nov 23, 2003

### Staff: Mentor

Re: Re: Re: Re: Re: Re: Re: Torque and such...

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 !

9. Nov 23, 2003

### Antepolleo

Re: Re: Re: Re: Re: Re: Re: Re: Torque and such...

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. Nov 23, 2003

### Staff: Mentor

Re: Re: Re: Re: Re: Re: Re: Re: Re: Torque and such...

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. Nov 23, 2003

### Antepolleo

Re: Re: Re: Re: Re: Re: Re: Re: Re: Re: Torque and such...

*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. Nov 23, 2003

### Staff: Mentor

Re: Re: Re: Re: Re: Re: Re: Re: Re: Re: Re: Torque and such...

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. Nov 23, 2003

### Antepolleo

Re: Re: Re: Re: Re: Re: Re: Re: Re: Re: Re: Re: Torque and such...

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. Nov 23, 2003

### Staff: Mentor

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. Nov 23, 2003

### Antepolleo

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. Nov 23, 2003

### Staff: Mentor

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! )

17. Nov 23, 2003

### Antepolleo

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. Nov 23, 2003

### Staff: Mentor

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. Nov 15, 2004

### swatikiss

from where do you know the angular momentum of the disk?

how do you do the cross product?