# How is work-energy theorem explained for rigid bodies?

We were first introduced to the work-energy theorem assuming point masses.

$$\mathbf F\cdot\,\text d\mathbf s=\text d(\text{KE})$$

This was explained using basic kinematic and some calculus and therefore introduced the fact that "A force that moves an object along the direction of force does work".

But then rigid bodies were introduced. That's when things started becoming fuzzy.

Now let's say a ball is rolling, friction acting on it (Assuming any other necessary forces act too). Now we say that as the bottommost point is always at rest(point of contact), work done by friction is $$0$$ (pure rolling). But why exactly do we see only the point of contact and not the centre of mass?

The same should apply for normal forces too. But then why does it require a greater work to stand up on a moving swing when it is at the bottommost point, than just getting up similarly from the ground? In both cases I see only gravity doing work besides the man. The normal force shouldn't do work as the legs are always in contact with the surface. Then why the difference here?

Now let's say a ball is rolling, friction acting on it (Assuming any other necessary forces act too). Now we say that as the bottommost point is always at rest(point of contact), work done by friction is 0 (pure rolling). But why exactly do we see only the point of contact and not the centre of mass ?

The friction force acts on the contact patch, not on the center of mass. If you had a force that acted on the center of mass then you would use that to determine the mechanical power (work) of that force. This distinction is important in many different confusing scenarios. The mechanical power delivered by a given force, $$\vec F$$ is always given by $$\vec F \cdot \vec v$$ where $$\vec v$$ is the velocity of the material at the point of application of the force.

There is also a separate concept known as "center of mass work" or sometimes confusingly "net work" which is given by $$\vec F_{net} \cdot \vec v_{com}$$ where $$\vec F_{net}$$ is the net force (the vector sum of all forces) anc $$v_{com}$$ is the velocity of the center of mass. This quantity is equal to the change in kinetic energy, but it does not account for any change in total energy. It also does not identify the mechanical power provided by any specific individual force even in the situation where there is only a single external force.

The same should apply for normal forces too. But then why does it require a greater work to stand up on a moving swing when it is at the bottommost point, than just getting up similarly from the ground? In both cases I see only gravity doing work besides the man. The normal force shouldn't as the legs are always in contact with the surface. Then why the difference here?

At the moment that the swing is at the bottommost point the velocity is horizontal and the normal force is vertical so $$\vec F \cdot \vec v = 0$$. So no power is being delivered by the swing. Instead, the person's legs are extending converting chemical energy into potential energy and kinetic energy. Extra effort is required because on the ground the person is converting the chemical energy into only potential energy, but on the swing it is converted into kinetic energy also. Therefore more chemical energy must be converted and the person feels the difference.

Indeed, if the normal force were doing the work then the person wouldn't notice any additional effort. For example taking an escalator up the stairs, the normal force does the work and the person is not fatigued. So the fact that the person is more fatigued is actually an indication that the normal force is not doing any work.

• "But on the swing it is converted into kinetic energy also"-Assuming the height raised is very small, there is no significant change in velocity right? So isn't only potential energy changing either on the swing or the ground? Also quantifying the work done by man in the swing scenario it comes out to be (mg+m $\omega^2$r)×$\delta$x. Is it just a coincidence that it it is the same as Normal×$\delta$x ? Apr 12 at 15:29
• @VamsiKrishna assuming that the height raised is very small there is also no significant change in the potential energy. They are of the same order so you can't really get rid of just one and not the other. It is not coincidence, but you can easily make scenarios where they are not equal, so they are not generally the same.
– Dale
Apr 12 at 16:47
• I get it now! Kinetic energy changes due to the change in moment of inertia right? My math works out then! Thanks! Apr 12 at 16:57
• @VamsiKrishna yes, that is correct. Conservation of angular momentum with a changing moment of inertia gives an angular acceleration. That increases the KE as well as the PE. I am glad I could help!
– Dale
Apr 12 at 16:59

Well I am going to try and make this as simple as possible.See the definition of work is :

Work is defined as the product of a Force and displacement of the point of application of force in the direction of displacement.

As you can see in the bolded section Work done on a body has no relation to the displacement of its centre of mass.It merely depends on the displacement of the point where the force is acting.In case of pure rolling the point where friction acts lies at rest with respect to the ground.So displacement of this point is zero.Since friction acts only at this point the work done by friction depends only on the displacement of this point not on the displacement of centre of mass.So since displacement of this point is zero Work done by friction is also zero.

Now for the swing example: Here the normal force merely serves to cancel out gravity.When we stand up of the ground there is always a net force acting on your centre of mass that displaces it by h and causes a difference in potential energy of mgh.

Change in Mechanical energy = Work done by you

So net work done equals net change in mechanical energy =mgh. But when you are on a swing there is an additional downward force acting at you centre of mass which is the centrifugal force.When you raise your centre off mass by h height this force does a negative work since it acts opposite to the direction of displacement of your centre of mass so you have to do more work to create same potential energy difference.

Change in Mechanical energy = Work done by you + Work done by centrifugal force.

If work done by centrifugal force = -x Then,

Change in Mechanical energy = Work done by you + (-x)

or, Change in Mechanical energy + x = Work done by you

So naturally you have to do more work to stand up on a swing than on the ground.

Basically the difference between these two cases is that Friction acts on the bottommost point in case of pure rolling which is at rest so friction does zero work.

But The centrifugal force on the swing acts at your centre of mass so since you raise your centre of mass by height h the centrifugal force ends up doing negative work so you must do some extra work to overcome that negative work done by centrifugal force.

Please let me know if you have understood or do I need to elaborate ?

• Thanks! But what if I am looking from the ground frame? Then there is no centripetal force right? Then I do not find any forces besides gravity and normal and the normal does no work. How can i explain the extra work there? Apr 12 at 16:34
• If you are looking from the ground frame then your displacement of centre of mass won't be vertically upward but it will be a complex spiral type of motion which would be very complicated to process.So change in energy won't be mgh either. Apr 12 at 16:36
• But what if I stand up almost instantaneously? My horizontal displacement is negligible in that time interval but with finite vertical displacement Apr 12 at 16:52
• Look when you say "I have to do more work " it intrinsically means that you are observing from your reference frame i.e the swings reference frame .If you want to analyse it from ground basically you would have to be on ground while you friend is on the swing so it might appear to you that you friend is standing up almost effortlessly but for your friend it will be very difficult to stand since your friend will be feeling everything from the swing's reference frame. Apr 12 at 16:58
• Thank you very much. Apr 12 at 16:59