# Work done by constraints on rotating rigid bodies

I am trying to understand why constraint forces do no work on extended, rotating bodies. For instance, consider the problem of a rigid rod falling on a frictionless surface (K&K 7.17)

There are no horizontal forces, so the work-energy theorem says that

$$\Delta K=\int_{y_0}^{y_1}\mathbf{F}\cdot dR+\int_{\theta_0}^{\theta_1}\tau\cdot d\theta$$

Now in non-rotating bodies, it's obvious that the normal force does no work, but clearly here the normal force is applying a force upward while the center of mass is moving in the same direction, doing negative work. At the same time, the contact point is exerting a torque on the rod, increasing the angular kinetic energy. It seems likely that these two contributions cancel each other, but I don't know how to prove or even intuit that this is true as a general rule. Why are we permitted to assume such a force is conservative?

• Just a suggestion: Think about static equilibrium, a condition that is met if the net torque on a object is zero. Apply a torque and it will become rotating. Rotating bodies can be constrained by forces e.g catching a cricket ball. Newton's Laws still apply. It's just our hands supply a non-conservative barrier. A force is conservative if its contour integral is equal to zero. That is there isn't a net energy expenditure throughout it's path.
– user97261
Nov 24, 2016 at 17:23

I think you're 'overthinking' a bit.

Your formula for $\Delta K$ counts the work twice.

As the CoG of the object has been lowered by $\frac{\ell}{2}$, the reduction in potential energy is simply:

$$W=\Delta K=\Delta U=mg\frac{\ell}{2}$$

We can also calculate it by the work done by the torque: $$W=\int_{path}\tau \mathrm{d}\theta$$ $$W=\int_0^{\pi/2}mg\frac{\ell}{2}\sin\theta \mathrm{d}\theta$$ $$W=mg\frac{\ell}{2}\Big[-\cos\theta\Big]_0^{\pi/2}=mg\frac{\ell}{2}(0-(-1))$$ $$W=mg\frac{\ell}{2}$$

These are equivalent ways of calculating $W=\Delta K=\Delta U$. If the object was both free falling and rotating we would have to add these energies but here the object is rotating only, not translating too.

The normal force (in the right hand 'resting point') as such performs no work because the point doesn't move in the direction of the force.

• Isn't the total external work zero, though, since $$\int\mathbf{F}\cdot dR=-mg\frac{l}{2}$$ and $$\int\mathbf{\tau}\cdot d\theta = mg\frac{l}{2}$$ as you showed? So $\Delta K_l+\Delta K_r=0$, and the only work done is by gravity, as desired? I was just wondering if there was some way of showing that this holds generally (for instance, if you have a rod sliding down a wall). Nov 24, 2016 at 22:27
• You're still starting from the wrong assumption that the two works have to be added up. Also, $\int\mathbf{F}\cdot dR=-mg\frac{l}{2}$ is actually wrong. Start from the definition of work: $\mathrm{d}W=\vec{F}.\mathrm{d}\vec{R}$. As the force vector and the displacement vector point in the same direction: $\mathrm{d}W>0$ and so also $W>0$!!
– Gert
Nov 24, 2016 at 23:26
• The force vector is in the positive $\hat{j}$ direction, and the displacement of the center of mass is in the negative $\hat{j}$ direction, isn't it? Nov 24, 2016 at 23:31
• No. The weight vector points downward and the object moves downward too: work done by the force is positive! Look at it from an energy point of view: when it was falling, did $K$ increase? Yes, of course: $W=\Delta K$.
– Gert
Nov 24, 2016 at 23:37
• Aren't we talking about the work done by the normal force on the body? Clearly gravity is doing positive work, but the question asked whether the normal force would do any net work, which it doesn't (by my first comment). Nov 24, 2016 at 23:38

I'm not exactly sure what you are aiming at with your question. We can strictly prove that the work done by constraint forces vanishes only for the case of particles (point systems), in which case the conclusion is near-trivial: These forces eliminate movement in the direction of the constraint, so they cannot perform work on the system. For extended systems, on the other hand, it is not always true that constraint forces do not perform any work. Wikipedia links to Goldstein's Classical Mechanics, 3rd Edition, page 16, for an example.

• The question was a bit ambiguous and over-general. I suppose I'm asking as a practical matter why we are justifying in solving a problem like this using energy methods (i.e. $mgy_0=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2$). K&K just says "because there are no dissipative forces", but while entirely reasonable, it's not clear why the normal force would do no work. Nov 24, 2016 at 17:59
• Ahh, that's a very different question. In this case what you are really asking is if there is work done on the system that does not result in a corresponding increase (or decrease) of the system's energy. In a purely mechanical system, the only way this could happen is through friction, or internal dissipation of energy in a deforming object. Since you have neither, the mechanical energy conservation principle you cite holds. So, the normal force in this case may or may not do work, but either way this work must result in a change of the rod's energy.
– Pirx
Nov 24, 2016 at 18:04
• There is still something very fundamental I'm not understanding. The normal force is external to the system, so it can easily increase the kinetic energy of the system if it does work. For particles, it's obvious that non-dissipative constraint forces do no work, but here the center of mass is displaced, in the direction of the force, so I'm not sure why no work is done as a matter of fact. Thanks for your help! Nov 24, 2016 at 21:59
• In this case, for instance, the 'linear' work is $-mgl/2$, and the rotational work is $mgl/2$, so no external work is done by the ground. But how can we show that this is always the case - if it even is? Nov 24, 2016 at 22:30
• Like I said, there are cases where work is being done by constraint forces, see the Goldstein reference. Also, the statement you quote from K&K does not make such a claim. All it says is that you can use the quoted conservation of energy principle if there are no dissipative forces.
– Pirx
Nov 25, 2016 at 6:40