# Why does the normal force do no work in the falling stick problem?

I'm studying the next problem what I found in a book.

A stick of length $$l$$ and mass $$M$$, initially upright on a frictionless table, starts falling. The problem is to find the speed of the center of mass as a function of the angle $$\theta$$ from the vertical.

I have analyzed the solution using energy methods, and I understand the problem almost completely, but I still do not understand why the normal force does not exert work.

Part of the solution that I found is below:

The key lies in realizing that because there are no horizontal forces, the center of mass must fall straight down. Since we must find velocity as a function of position, it is natural to apply energy methods.

The sketch shows the stick after it has rotated through angle $$\theta$$ and the center of mass has fallen distance $$y$$.

The initial energy is

\begin{aligned}E&=K_0+U_0\\&=\frac{Mgl}2\end{aligned}

The kinetic energy at a later time is

$$K=\frac12l_0\dot\theta^2+\frac12M\dot y^2$$

and the corresponding potential energy is

$$U=Mg\left(\frac l2-y\right)$$

Because there are no dissipative forces, mechanical energy is conserved and $$K+U=K_0+U_0=Mgl/2$$. Hence

$$\frac12M\dot y^2+\frac12l_0\dot\theta^2+Mg\left(\frac12-y\right)=Mg\frac l2$$

We can eliminate $$\dot\theta$$ by using the constraint equation. The sketch shows that

$$y=\frac l2(1-\cos\theta)$$

Hence

$$\dot y=\frac l2\sin\theta\;\dot\theta$$

Question: Why does the normal force not do any work?

• To close voters, note that this is not asking for a solution to the exercise. It's asking a conceptual question about work done by a force in a specific scenario. Aug 9 '21 at 4:48