Work done By Internal Forces Inside a hinged Rotating Rod

Question

Following the previous posted question , I got to know that Normal reaction is not doing any work as it does not displace any particles of the thin rod.

But,I got another concept dilemma.

There will be internal forces that help in driving all particles in circular motion around the hinged point, right ? If normal force is acting at only the free end and gravity is considered to be acting at the center of mass, then there must be internal forces acting at the other point inside the thin rod.

Now, if that's the case, then internal forces will do work too ? What Would Be its Magnitude ? Is Mechanical Energy still conserved ?

A good explanation would be appreciated; that would clear my ideas.

Answer 1

In general, internal forces can do non-zero work on a particle system, and thus change kinetic energy of particles in the system. This is taken into account in the work-energy theorem: work of all forces (including internal forces) equals change of kinetic energy of particles in the system. Work of internal forces is important when the system is made of particles interacting at a distance, e.g. a system of gravitationally interacting bodies, where kinetic energy can change as a result of action of internal gravitational forces.

However, in case the system is a rigid body, work of internal forces is zero.

Why?

In general, work of internal forces done during a small time can be expressed as sum of works done on each particle in this time. In this sum, we have many pairs of terms

$$ \mathbf{F}_{ba} \cdot \Delta \mathbf{r}_a + \mathbf{F}_{ab} \cdot \Delta \mathbf{r}_b $$

where $\mathbf F_{ba}$ is force of particle $b$ acting on particle $a$, and $\Delta \mathbf{r}_a$ is displacement of the particle $a$, and so on.

In a general system, there is no simple relation between the displacement $\Delta \mathbf{r}_a$ and $\Delta \mathbf{r}_b$ of two interacting particles, but in a rigid body, all interacting pairs consist of particles that are very near each other (no action at a distance), and displacements of both particles in the pair are therefore the same. This means the above expression is zero in a rigid body for all interacting pairs, and thus is the sum, equal to work of internal forces.

Answer 2

For these types of problems they are treated as a rigid body and as such if you broke down the rod into smaller pieces or particles the distance between adjacent particles within the rigid body would be fixed with respect to each other and if the body is moved the distances are all the same and fixed and the body movements can be described as a translation or rotation in some coordinate system. This is often good enough for engineering when deflections are small and the materials doesn’t change shape. In these cases the total energy in the system is assumed to be constant.

However, in detail you are correct when you bend or move something you can have friction, or in the case or bending movement of defects or grains of material that at the macroscopic level we usually consider as some sort of dissipation of energy or energy loss. If this energy loss is small it can sometimes be hard to understand where the energy loss is coming from in details.

But on the bright side since the energy loss is so all or the bending or strain is small, treating something as a rigid body gives good answers.