8
$\begingroup$

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:

enter image description here

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?

$\endgroup$
1
  • 5
    $\begingroup$ 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. $\endgroup$ Aug 9 at 4:48
12
$\begingroup$

The bottom of the stick, the part in contact with the table, is moving only horizontally. The normal force is applied vertically to the bottom of the stick. Therefore, the dot product of the velocity of the bottom of the stick and the normal force is 0, so no work is done.

$\endgroup$
2
  • $\begingroup$ you have too much reason! $\endgroup$
    – 3435
    Aug 9 at 4:15
  • $\begingroup$ Say we remove gravity from this problem, and instead make the surface provide a force equal to the opposite of gravity going upwards. Wouldn't the problem in this new reference frame remain unchanged, but the normal force now provide work in the form of torque on the stick? $\endgroup$ Aug 9 at 16:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.