How does the calculation of work change for a rotating or non-rigid object?

[nav888]

Homework Statement

The steam ejected from the nozzles provides a couple. The force at each nozzle is 0.12N. The
perpendicular distance between the nozzles is 8.2 × 10–2m.
What is the work done by the forces as the steam generator completes one revolution?

Relevant Equations

W = Fx cos theta

Here is the question, I am struggling with it as in all the past questions I have done it didnt matter if you worked out the work done by each force and then added it all together or if you added the forces together and then worked out the work done by the total force, but here the net force is 0 and hence the work done is 0 but that is not the correct answer and I would like some help/guidance please.

 

[kuruman]

The net force does no work as you already noted. What about the net torque?

 

[Lnewqban]

Here is the question, I am struggling with it as in all the past questions I have done it didnt matter if you worked out the work done by each force and then added it all together or if you added the forces together and then worked out the work done by the total force, but here the net force is 0 and hence the work done is 0 but that is not the correct answer and I would like some help/guidance please.

When both forces are opposite and equal, but not aligned, like in this case, we need to consider the distances between their line of action and the center of rotation.
The problem mentions a couple of forces, which mean, each force has a distance to be considered, which results in a net torque or moment.
In the rotational world, that net moment is equivalent to the net force, and distance is equivalent to angle of rotation (in radians).

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#rlin

Note that if both nozles were pointing in the same direction, no rotation would be induced (zero work).
That is very different to the linear world, where both forces would be added to estimate the net force producing a linear movement.

 

[BvU]

don't use temporary links to pictures....

 

[jbriggs444]

in all the past questions I have done it didnt matter if you worked out the work done by each force and then added it all together or if you added the forces together and then worked out the work done by the total force

This is because in all past problems you were dealing with either point-like objects or rigid objects that did not rotate. Pointlike objects and non-rotating rigid objects share an important feature. Every point on the object always moves at the same speed and in the same direction as all of the other points.

So if you multiply one external force by the displacement of its point of application and another external force by the displacement of its point of application, then those two displacements are guaranteed to be equal.$$\vec{F_1} \cdot \vec{s_1} + \vec{F_2} \cdot \vec{s_2} = (\vec{F_1} + \vec{F_2}) \cdot \vec{s_1} = (\vec{F_1} + \vec{F_2}) \cdot \vec{s_2} = (\vec{F_1} + \vec{F_2}) \cdot \vec{s_\text{COM}}$$because $\vec{s_1} = \vec{s_2} = \vec{s_\text{COM}}$.

For a rotating or non-rigid object, the displacements of different parts of the same object can be different from each other. So you can't just add up all of the forces and multiply by the displacement of the center of mass.

There are at least two notions of "work" that can be used. One is "center of mass work", aka "net work" where you take the net force and multiply by the displacement of the center of mass. The other is what I call "real work" where you take each force individually and multiply by the displacement of its point of application.

Because there are two notions of work, there are two versions of the work energy theorem.

For center of mass work, you get that the work done is equal to the change in the bulk kinetic energy of the object as a whole. $W = \Delta \frac{1}{2}mv_\text{com}^2$. Any kinetic energy associated with rotation, sloshing or any other sort of internal motion is ignored.

In the case at hand, the net force is zero. Center of mass work is zero and the object does not gain any bulk kinetic energy. It rotates faster, but its center of mass remains stationary.

For real work, you get that the work done is equal to the change in energy of the object as a whole. This time you get to count the energy of rotation or of moving parts within the object.

In the case at hand, the real work is non-zero. The object gains rotational kinetic energy.