Work Energy Theorem: particle, system of particles, rigid body

[Pushoam]

Question: A) Derive the work - energy theorem for one particle.
B) Check whether it is applicable for a system of particles and a rigid bodyWork - energy theorem for one particle system,
total sum of work done by individual forces = work done by total force
To show the above equality,
let's say that there acts n forces on a particle and the net displacement of the particle is from $\vec s_i$ to $\vec s_f$.
Then , total sum of work done by individual forces = $Σ_{i =1} ^n W_i = Σ_{i =1} ^n \int _ {\vec s_i } ^ {\vec s_f} \vec F_i . d\, \vec s$
$= \int _ {\vec s_i } ^ {\vec s_f} \{Σ_{i =1} ^n \vec F_i\} . d\, \vec s$
$\vec F_{net} = Σ_{i =1} ^n \vec F_i$
So, total sum of work done by individual forces = $\int _ {\vec s_i } ^ {\vec s_f} \vec F_{net} . d\, \vec s$ = work done by total forceNow, the work energy theorem says ,
The total work done on a particle by net force is equal to change in its kinetic energy provided that the mass of the particle remains constant.
$W = \int _ {\vec s_i } ^ {\vec s_f} \vec F_{net} . d\, \vec s = \int _ {t_i } ^ {t_f} m \frac {d \vec v } {dt} . \vec v d\, t$
$= \int _ {v_i } ^ {v_f} m \frac 1 2 d\, v^2 =$ $\frac 1 2 m v_f ^2 - \frac 1 2 m v_i ^2$ = change in kinetic energyFor a system of particles,
Let's say that the work - energy theorem holds for tha j_th particle,
$W_j = ( K_f)_j - (K_i) _j$
For n - particles ,
$Σ_{ j = 1} ^n W_j = Σ_ {j = 1} ^n ( K_f)_j - Σ_{ j = 1} ^n (K_i) _j$

So, the sum of work done on all particles = sum of change in kinetic energy of all particles
⇒ work done on a system of particles = change in kinetic energy of the system of particles
where, work done on a system of particles ≡ sum of work done on each individual particle
kinetic energy of the system of particles ≡ sum of kinetic energy of each individual particle

Now, a rigid body is nothing but a special case of a system of particles, so what is valid for a system of particles is valid for a rigid body.

Here, I feel tempted to use the same analogy for a system of particles, too.
Can I say that what is valid for a single particle is valid for a system of particles and so, is valid for a rigid body, but not the vice - versa?

So, the answer to B is " yes".
Are the above arguments correct?

 

[vela]

You need to make a distinction between internal and external forces for a system of particles and for a rigid body.

Consider a one-dimensional system with two masses connected by a spring, and you apply force F to one mass and -F to the other. What's $F_\text{net}$ for the system? Will there be work done on the system? What's the change in kinetic energy of the system?

 

[Pushoam]

What's $F_\text{net}$for the system?

Both the net external force and the net force (including both external and internal forces ) is 0.

Will there be work done on the system?

Yes. work done by the internal forces is 0 as the work done by the spring force on the two masses gets canceled by the work done on the spring due to the forces by the two masses.
Work done by the external forces is nonzero.

This means that in case of a system of particles, even if the total force or total external force acting on the system is 0, the work done on the system may not be 0.

What's the change in kinetic energy of the system?

Sum of work done on the system by each force
So, work - kinetic energy theorem is valid here.
I didn't get the need of the following advice:

You need to make a distinction between internal and external forces for a system of particles and for a rigid body.

 

[vela]

So you apply the pair of forces for a short time, allowing the system to get stretched out, and then turn it off. There was no net force on the system, so $\int \vec{F}_\text{net}\cdot d\vec{s} = 0$, yet the masses will subsequently move so $\Delta K_1 + \Delta K_2 \ne 0$.

 

[Pushoam]

So you apply the pair of forces for a short time, allowing the system to get stretched out, and then turn it off. There was no net force on the system, so ∫⃗Fnet⋅d⃗s=0∫F→net⋅ds→=0\int \vec{F}_\text{net}\cdot d\vec{s} = 0, yet the masses will subsequently move so ΔK1+ΔK2≠0ΔK1+ΔK2≠0\Delta K_1 + \Delta K_2 \ne 0.

But this doesn't tell me why I should distinct between internal and external forces. I want to know how this distinction helps us.
By the way, if there is no damping force acting on the system, the system will continue oscillating forever. So, the fact that we can't generate perpetual motion is due to the existence of damping force in nature. Right?
If atoms oscillate perpetually, does this mean that the net force acting on it is not damping?

 

[vela]

But this doesn't tell me why I should distinct between internal and external forces. I want to know how this distinction helps us.

What I was trying to get at, though not very clearly, is that the net force on a system of particles is a force that's external to the system, but in your analysis, you looked at the forces on individual particles, which is a combination of internal and external forces.

By the way, if there is no damping force acting on the system, the system will continue oscillating forever. So, the fact that we can't generate perpetual motion is due to the existence of damping force in nature.Right?

Yup.

If atoms oscillate perpetually, does this mean that the net force acting on it is not damping?

Yup.

 

[Pushoam]

What I was trying to get at, though not very clearly, is that the net force on a system of particles is a force that's external to the system, but in your analysis, you looked at the forces on individual particles, which is a combination of internal and external forces.

So, you mean that in standard convention, by " net force ", "net external force " is meant. So, I should distinct the forces into internal and external forces. I will consider this point from next time.

I think in work -energy theorem, we need to consider both external and internal forces. Right?

 

[BvU]

Yes. In particular, there is something that has to do with Newton's third for internal forces.
In post #2, Vela introduced a nasty kind of external forces; other kinds are forces that work on the system as a whole (all the same F), or a force like gravity (all the same g).

 

[Pushoam]

In particular, there is something that has to do with Newton's third for internal forces.

Because of the Newton's 3rd Law, the net internal forces acting on the system of particles is 0. But the net work-done by the internal forces i.e. the sum of the work done by each internal force may not be 0 .
Hence, work - kinetic energy theorem for a system off particles: net work done on the system = sum of work done by each of both internal and external forces = change in the total kinetic energy of the system

Is there any easy way to figure out whether the work done by internal forces is zero?
I think if the internal forces is electromagnetic or gravitational , the work done by the internal force is not zero.
Generalizing it , if the internal forces are central forces, work-done by them may not be 0.
Since, the forces in action - reaction pair act on different bodies, and the different bodies could have different mass and so different displacement, the work-done may not be 0.
In case of rigid bodies the problem gets simplified, in the sense that the work done by the internal forces is 0.

 

[vela]

Hence, work - kinetic energy theorem for a system off particles: net work done on the system = sum of work done by each of both internal and external forces = change in the total kinetic energy of the system

How are you defining the kinetic energy of the system? When you talk about the kinetic energy of, say, a ball, you don't typically include the thermal energy.

 

[Pushoam]

How are you defining the kinetic energy of the system?

I am talking here about a system of particles. So, K. E. of the system here will be sum of kinetic energy of all particles.

When you talk about the kinetic energy of, say, a ball, you don't typically include the thermal energy.

That depends whether you consider a ball as a system of particles or a rigid- body.
Considering it as a system of particles, you have to include thermal energy.
Considering it as a rigid body, thermal energy is 0 as each constituent particle of the system is fixed under rigid body approximation.

 

[BvU]

Focus on the center of mass of a system of particles, see if at least that gives you a work/energy relationship

 

[Pushoam]

Focus on the center of mass of a system of particles, see if at least that gives you a work/energy relationship

The net force acting on the center of mass is net external force acting on the system of particles .With this COM is essentially a one particle system and so, work kinetic energy theorem will be applicable here.
But, we can reduce a system of particles to a COM - one particle system for applying work- kinetic energy theorem only if the change in kinetic energy of the system of particles due to internal interactions is 0. Isn't it so?

 

[vela]

The net force acting on the center of mass is net external force acting on the system of particles .With this COM is essentially a one particle system and so, work kinetic energy theorem will be applicable here.

I think the point of the problem is for you to show whether or not you can treat the COM of a system as a single particle.