# Question regarding W-E theorems

1. Dec 16, 2003

### StephenPrivitera

Are these quantites:

$$W=\int_{\theta_0}^{\theta_f}{\tau}{d}\theta=\frac{1}{2}I(\omega_f^2-\omega_0^2)$$
$$W=\int_{r_0}^{r_f}Fdr=\frac{1}{2}m(v_f^2-v_0^2)$$
the same or different?

Is "v" in the second equation the speed of the center of mass? IOW, does the bottom integral give the change in translational kinetic energy and the top the change in rotational kinetic energy (independent of each other) or do they both represent the total change in kinetic energy?

2. Dec 16, 2003

### Staff: Mentor

It is possible to represent the motion of a rigid body as a combination of translation (of the center of mass) and rotation (about the center of mass). Used in this manner, those equations represent different kinetic energies that can be treated independently.

Yes, the "v" in the second equation is the speed of the center of mass. For non-rigid bodies, that second equation can be generalized if both the displacement and speed are of the center of mass.

Does this help?

3. Dec 16, 2003

### turin

It depends on the context. Either equation could represent the total work/change in kinetic energy. But, I would infer that, if you see both of them together, that this would be intended to contrast the different forms of the kinetic energy into translational and rotational. The total work/change in kinetic energy would be the sum of these two integrals in that case.

4. Dec 16, 2003

### StephenPrivitera

Can you give specific example and work it both ways? That is, first considering the quantities are representing fractions of the total work and then the quantities representing the whole work. I can see what you're saying, but I don't see how to apply it to a problem.

I asked this question with the following problem in mind:
A rope of length L is wrapped around the side of a cylinder. The rope is pulled off the cylinder with constant force F. What is the translational and rotational speeds of the cylinder (ignore friction acting against motion)?
It's easy to find the angle over which the force works and so the first integral (above) can be solved easily. But that integral represents only rotational energy, correct?
However, it's difficult (to me) to find the distance over which the force acts, and so the bottom integral is difficult.

5. Dec 17, 2003

### turin

I haven't really given it that much thought, but this seems like it could actually be a little complicated. You need both integrals in this problem.

F is the tension in the rope. But, in order to get &tau;, it seems like you should need the radius (?). Also, in order to find the distance, it seems like you should need the mass (or angular moment of inertia) (?). Do you have these other two parameters?

6. Dec 17, 2003

### StephenPrivitera

I apologize severely. Mass and radius are given.

7. Dec 17, 2003

### NateTG

The two quantities are different. Both are fractions of the work that is done on an object.

In the scenario with the free spinning cylinder that you describe, you will note that for a distance $$x$$ and angle $$\theta$$ that the cylinder has traveled, there will have been $$x+\theta R$$ rope pulled away, so force times distance for tension would actually be $$F(x+\theta R)$$

The equations of for the motion of the cylinder are actually relatively simple.

8. Dec 17, 2003

### StephenPrivitera

Re: Re: question regarding W-E theorems

I figured it as such:
The rope has length L and the disc radius R. So the angle over which the torque acts will be (L/(2piR))(2pi)=L/R. So taking &omega;0=0,
$$\omega_f=\sqrt{\frac{2}{I}\int_{0}^{L/R}FRd\theta}=\sqrt{\frac{2FL}{I}}$$
I did this by imagining the disc as a pulley. I then wanted to consider translational motion, but I got stuck.
I don't quite understand your reasoning.

9. Dec 17, 2003

### Staff: Mentor

Here's how I would approach this problem. The dynamical equations are:

$$F=ma_{cm}$$

$$\tau =I\alpha$$

The "work"-energy equations are (assuming the cylinder starts from rest, for simplicity):

$$F\Delta x_{cm}=\frac{1}{2}m v_{cm}^2$$

$$\tau \Delta\theta=FR \Delta\theta=\frac{1}{2}I\omega^2$$

Of course, $$\Delta\theta=\frac{L}{R}$$ (in radians) is given, but $$\Delta x_{cm}$$ must be calculated from the dynamics.

In this simple case we can find the real work on the system (of cylinder + string) since the force F is applied to the string for a distance of $$\Delta s=\Delta x_{cm}+L$$. This force times distance represents real work on the system:
$$F\Delta s={KE}_{total}=\frac{1}{2}m v_{cm}^2+\frac{1}{2}I\omega^2$$

10. Dec 17, 2003

### StephenPrivitera

So two questions:
How would I find delta s?

What is the real (general) work-energy theorem?

11. Dec 17, 2003

### Staff: Mentor

By adding &Delta;x and L. When I said we could find the real work, I meant we could identify the actual applied force at a point on the system and its displacement. I know of no shortcut to find &Delta;x.
By real work I mean "work" as used in conservation of energy, or in the first law of thermo. You have to integrate forces and displacement around the boundary of a system: not pretty, except in the simplest of cases.

The "pseudowork"-energy theorem is general: that's the "center of mass" version I gave above. Sorry if I'm just adding to the confusion.

It even works for wacky non-rigid body problems. Like when you jump in the air. The ground pushes on your feet, and your center of mass accelerates. You can use F&Delta;x to calculate the KE (of center of mass)--- yet the floor does no real work, since displacement (at the point of application) is zero.

12. Dec 17, 2003

### NateTG

I should correct myself, the two are the same, but it may be convenient to seperate forces that act in rotation and linearly.

Let's take a look at your cylinder/disk.

If I'm tugging on the rope with force F, then the total work done is F * x where x is the distance that the rope moves -- not the distance that the cylinder moves.

The linear motion is governed by F=ma.
The angular motion is &Tau;=I&omega;

What confused me is that it's easy to underestimate the amount of work that's being done.

13. Dec 18, 2003

### StephenPrivitera

Ok, so I'm still kinda confused. What is the cylinder was only able to rotate. Then
$$W=\int_{0}^{L/R}FRd\theta=FL=\DeltaKE$$
In this case, the change in KE is only due to rotational motion.
What if the cylinder could only move translationally.
Then if the force acts over a distance L
$$W=\int_{0}^{L}Fdx=FL=\DeltaKE$$
but in this case delta KE is change in energy of the CM.

Actually, I'm really confused now. I can understand how these equations work if only one of the processes (i.e., either rotation or translation but not both) is occurring. But what about when both occur? Doc Al said the general WE theorem is the one involving the CM. But not all the KE is associated with the motion of the CM if there is also rotational motion involved...

14. Dec 18, 2003

### Staff: Mentor

All true! But if constrained to only rotate, the net force must be zero, right? And if constrained to not rotate, the net torque must be zero.
What I meant was that Fnet*&Delta;Xcm will always give you the correct answer for 1/2 m V2 where V is the speed of the cm. This is not (necessarily) the actual work done by an external force! (Some folks object to calling this the work-energy theorem for just this reason.) It may turn out to be, in simple cases. To find the real work (which equals the change in energy of the system) you have to identity the external forces and their displacements at the boundary of the system. For cases like this cylinder problem, that involve only mechanical energy (not chemical, thermal, deformation energy changes, etc. or heat flow), the real work always equals the change in total KE of the object. Sometimes the total KE is purely translational or purely rotational, as you point out.

Last edited: Dec 18, 2003
15. Dec 18, 2003

### StephenPrivitera

Ok, so in conclusion,
$$W_{net,real}=\frac{1}{2}\left(I\omega_f^2+mV_f^2\right)-\frac{1}{2}\left(I\omega_0^2+mV_0^2\right)$$

And

$$W_{net,real}=W_{trans}+W_{rot}$$
where
$$W_{trans}=\int_{\vec{x}_0}^{\vec{x}_f}\vec{F}(x)\cdot d\vec{x}$$and the limits of integration are the final and initial positions of the CM
and
$$W_{rot}=\int_{\theta_0}^{\theta_f}F(\theta)Rsin\theta d\theta$$ and the limits of integration are the final and initial positions some arbitrary point not lying on the axis of rotation (for rigid bodies, anyway).

I want to think about this some more. If what I have above correctly summarizes your point, then I understand. But I still am uncomfortable with it. Would it ever be practical to apply this general WE theorem? For instance, in the cylinder problem, I still can't figure out a way to find the displacement of the CM and so I can't find its final speed.

Oh, and in answer to my original question (first post), these quantities are not at all the same? They each give a different part of the total work.

16. Dec 18, 2003

### NateTG

In a non-rigid body situation, something might either be considered rotational kinetic energy, or it might be linear kinetic energy depending on which is more convenient.

For example, if you have a planet orbitting a star in a circle, then you can either consider it's linear kinetic energy, or it's angular kinetic energy. You'll find that it's the same amount of energy.

What happens with spinning rigid bodies is that the linear movements of internal pieces cancel, so if we want to do a macroscopic energy calculation, we need to account for that in the form of rotational kinetic energy.

In other words, if you took $$KE=\int \frac{1}{2}{v(m)}^2 dm$$ over a rigid body you would also come up with the same value for kinetic energy. It's just that for linearly moving rigid bodies, $$v(m)$$ is constant, so the integral is trivial. For spinning rigid bodies, $$v(m)$$ can be divided into a angular component $$\omega$$ and a linear component $$v$$ which is constant over the entire body. Naturally, $$v$$ and $$\omega$$ are dependant on the frame of reference.

BTW, confusion is not always a bad sign. It usually indicates that you're learning.

So, let's take a look at the cylinder's linear motion.

There is a single force with magnitude $$F$$ acting on it, so we can apply the kinematic formulae:

$$a=\frac{F}{m}$$
so
$$x(t)=\frac{1}{2}at^2=\frac{F}{2m}t^2$$
and
$$v(t)=\frac{F}{m}t$$

17. Dec 18, 2003

### StephenPrivitera

Right, as long as the planet isn't also rotating. You can consider it rotating about an axis through the Sun. Then you have
$$K=\frac{1}{2}mv^2=\frac{1}{2}mR^2\left(\frac{v^2}{R^2}\right)=\frac{1}{2}I\omega^2$$
But if in addition the planet spins, it has an additional kinetic energy due to its spin. What if it spins with angular speed &omega; in addition to its orbit? Is its total KE then
$$K=\frac{1}{2}mv^2+\frac{1}{2}I_{diameter}\omega^2$$?

Ok, imagine the CM moves with constant velocity V. Imagine also that the body spins with angular speed &omega; about an axis through its CM and perpendicular to its direction of motion. A mass element dm a distance R (perpendiculat to both rot axis and CM's velocity) will have KE=(1/2)(dm)(V+&omega;R)2 and a mass element the same distance away but to the other side will have KE=(1/2)(dm)(V-&omega;R)2. These two average out to KE=(1/2)(dm)V2. So since we're talking about distance from the CM, the average KE should be (1/2)MV2. Is this then the total KE including rotation?

Actually, every time I think I got it, I don't. The more I think about it, the more unclear it is.

18. Dec 18, 2003

### NateTG

Um:
$$\frac{(V+\omega R)^2+(V-\omega R)^2}{2}=V^2+(\omega R)^2$$

only the middle terms cancel, and the extra R becomes part of $$I$$

Last edited: Dec 18, 2003
19. Dec 18, 2003

### turin

If you can calculate the rotational part (I think you were saying that you could do this), then you can infer a time duration from that. There is a constant torque (because F is constant and r is constant), so you should be able to figure a time duration from the rotational part. Then, realize that the total linear force on the system is a net imbalance of F (in the direction the rope is being pulled). Newton's second law will tell you the linear acceleration of the system, and since you know the time duration, you should be able to calculate the final velocity.

I am actually uncomfortable with this approach; it just seems too round-a-bout, but I can't think of anything else right now, and I'm too lazy to work through it.

20. Dec 18, 2003

### Staff: Mentor

Here's what I meant by figuring out &Delta;Xcm from the dynamics.
Considering rotation first (since we have L):

$$\tau =I\alpha$$

$$\alpha=\frac{FR}{I}$$

$$\frac{L}{R}=\frac{1}{2}\alpha T^2$$

So...
$$T^2=\frac{2LI}{FR^2}=\frac{LM}{F}$$

Combining this with translation:

$$a=\frac{F}{M}$$

$$\Delta X_{cm}=\frac{1}{2}a T^2=\frac{L}{2}$$