# Work analogy but with torque

1. Dec 13, 2007

### DaTario

Hi all,

If we take the integral of tau (the resultant torque) times the angular displacement theta, with respect to the angular variable theta, are we to expect the result to correspond to the variation in rotational kinetic energy?

Is is this the case, does this integral have any particular name (analogous to work) ? "WORQUE" :)

Thank you

DaTario

Last edited: Dec 13, 2007
2. Dec 13, 2007

### Staff: Mentor

Yes.
It's just called work.

3. Dec 13, 2007

### DaTario

It is also called work? But it reffers to different physical manifestation. Considering for example a disk rolling down an incline. The variation in kinetic energy is dou to the work done by the resutant force while the variation in rotational kinetic energy is produced by this "worque" ....

best wishes

DaTario

4. Dec 13, 2007

### Staff: Mentor

Realize that integrating torque over an angle is equivalent to integrating force over a distance. Both give a change in kinetic energy and can be called work. It requires work to increase translational KE and it requires work to increase rotational KE.

5. Dec 13, 2007

### DaTario

Ok, I know both energies are energies, but the numerical results in each calculation are not the same.
But if you are saying that both expressions are called work, it is ok to me.
In a certain way I was expecting something like rotational work or something like that.

Thank you

Best wishes

DaTario

Last edited: Dec 13, 2007
6. Dec 14, 2007

### robphy

Just to add to Doc Al's comment....
recall that a rigid body is a many-particle system.
The grouping of certain terms into a "rotational" part is just for convenience.

7. Dec 14, 2007

### stewartcs

I like the name "WORQUE", I think I'll start using it instead.

My boss is from NY, he tells me to get to "worque" all the time! :rofl:

8. Dec 14, 2007

### robphy

Should that be New Yorque?

9. Dec 14, 2007

### DaTario

I guess I understand this part.

I am a little bit confused, as in the case of a disk rolling down an incline 1/2 * M V^2 is the kinetic energy of the CM while 1/2* I Omega^2 is related to the rotational motion.

10. Dec 14, 2007

### Staff: Mentor

As robphy indicated, the real deal is that each piece of the rigid body--each tiny element of mass--is moving with some speed. So each mass element has some KE. Add them all up, and that's the total KE.

It's just easier to take advantage of the fact that it's a rigid body and separate out a component of kinetic energy associated with the translational motion of the center of mass and a component associated with rotation motion about the center of mass. But that sum is equal to what you'd get if you just added the KE of each mass element.

11. Dec 14, 2007

### robphy

Last edited by a moderator: Apr 23, 2017
12. Dec 14, 2007

### DaTario

Very nicely said. But now comes my question: the work of the resultant is the net variation of the CM component of the total KE.

Will the integral of Tau (resultant torque) with respect to Theta (angular displacement) correspond to the net variation of the rotational component of the total KE ?

best wishes

DaTario

13. Dec 14, 2007

### arildno

Can't you do the maths for once, DaTario?

For a point on a rigid body, acted upon by an external force $\vec{f}[/tex], with distance [itex]\vec{r}$ from the mass centre, with velocity $$\vec{v}=\vec{v}_{C.M}+\vec{\omega}\times\vec{r}$$ its contribution to the power of the rigis body is:
$$\vec{f}\cdot\vec{v}=\vec{f}\cdot\vec{v}_{C.M}+\vec{f}\cdot(\vec{\omega}\times\vec{r})=\vec{f}\cdot\vec{v}_{C.M}+(\vec{r}\times\vec{f})\cdot\vec{\omega}$$.

See if you can glean the result from this.

14. Dec 14, 2007

### DaTario

I am not asking you to make calculations. You misinterpreted me. I am asking if someone has this complementary information about the work-energy theorem.

If it is clear to you I would be pleased to know if the integral of Tau (resultant torque) with respect to Theta (angular displacement) corresponds to the net variation of the rotational component of the total KE?

Could you explain what is the purpose of this forum, if it isn't for people to interact, to exchange information and to debate?

Last edited: Dec 14, 2007
15. Dec 14, 2007

### arildno

Which is given by..the calculations.

I just gave you the information, I'm not into hand-wavy airiness, neither should you, if you actually want to learn some physics, which is the main point of this forum.

16. Dec 14, 2007

### Ben Niehoff

This is something you should be able to answer yourself. Take a simple, two-dimensional case, so you don't have to worry about vectors. Write down an equation relating the torque and angular displacement to the force, velocity and linear displacement of each small part of the object. Use this to set up two integrals, one for $F dr$, and one for $\tau d\theta$. Check whether the results are equal.

17. Dec 14, 2007

### Staff: Mentor

Yes, if by work you mean $\int \vec{F}_{net} \cdot d\vec{r}_{cm}$.

Yes, as long as the torque is taken with respect to the center of mass.

18. Dec 14, 2007

### DaTario

Doc Al, Thanks.
Ben Niehoff, welcome and thanks.
Aridno, you have talents.

best wishes

DaTario

19. Dec 14, 2007

### Shooting Star

(Groan) You are both dorques. Actually, I liked both, but now put a corque in it.