I've just come across an exercise that consists of a rigid, ideal rod that can move around a fixed axis: enter image description here

After letting it move , the rod's angular velocity at time t (when it is vertical) is asked.

The most obvious thing is to use conservation of energy, combining the rod's rotational energy and its potential energy. In that case, we get $$ w=\sqrt{\frac{3g}l}$$ However, I wanted to try out other methods to see if all were equivalent. While calculating the work done by gravity's torque yielded the same answer, I also tried to do it with a point particle. My reasoning was that I could apply the work done by gravity to a particle at the bottom of the rod and get a speed. Since all particles have the same angular speed, the rod as a whole should have the same one. Knowing $v$, I could just do $w$=$vr$. However, this gives $\sqrt{\frac{2g}l}$ instead.

Why is this? I thought that maybe the rest of the particles exerted a force in the rod's direction to the one at the bottom that would add to its speed, but since that force would be perpendicular to the particle's line of motion (The red one), its total work would be 0 and thus it wouldn't give it any speed.

In addition, I wonder if the fact that the rod's center of mass is moving means that it has got a linear momentum. We were explained back then that it's just $mv$, but if the rod had linear momentum, then it would have a translational kinetic energy that simply doesn't appear here. I have tried to interpret that $v$ as the translational component of all the speed, so the rod would have 0 translational component according to a frame fixed in the rotation point. However, at any other point, this wouldn't be the case, and so the rod would also have a linear momentum and translational kinetic energy to consider. It looks weird to me that we are forced to choose the centre of rotation as the only frame to do some calculations.

Is my interpretation right? That is, does linear momentum depend on a specific type of velocity? And why do we seemingly have to choose this particular frame for our calculations?

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    $\begingroup$ If you model this as a single particle process, then this particle must reside in the COM of rod, not at the bottom of rod. As all forces acts on body COM in free body diagram. This may be one source of error. Btw, correct equation would be $\omega = v/r$, (not $\omega = vr$), if you would do a dimensional analysis, you would see that units in your equation does not match. $\endgroup$ Commented Nov 8, 2022 at 19:33
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    $\begingroup$ Reducing a rigid body to single particle at the center of mass is only sufficient for translation type motion. For any sort of motion with rotation you can replace a planar rigid body with two half masses placed at $\ell / \sqrt{3} $ apart symmetric about the center of mass. $\endgroup$ Commented Nov 8, 2022 at 19:38

2 Answers 2


No, your intuition it's not correct because is not a point-like mass. In this case you must use the cardinal equations, for example:

$ I_0 \alpha= mg( \frac{L}{2}) , \quad I_0=\frac{1}{3} m L^2 $

Note that in $I_0$ is contained the rigid body information, in particular, as you can check, the angular velocity depends crucially from that:

$ \omega= \sqrt{\frac{mg L}{I_0}} $


If the particles in the rod could only exert radial forces on one another then a particle at a distance $r$ from the pivot would be moving at a speed $\sqrt{2gr}$ when it reached the lowest point of its swing. But because the rod is rigid we know that the speed of each particle at the lowest point of its swing must be proportional to $r$ (so that the rod has a uniform angular speed), not proportional to $\sqrt r$.

Therefore the particles in a rigid rod must exert tangential forces on one another in order to redistribute kinetic energy along the rod and give it a uniform angular speed.

  • $\begingroup$ I thought it could make sense, but I'm trying to find a general way to think about it. For example, in another exercise I have a cylinder rolling on a larger cylindrical shell, and now it has translational k.e as well as rotational. If the distance of the cylinder's c.o.m is constant w.r.t the rotation point, why would it be considered to be translating? I'm kinda confused. $\endgroup$ Commented Nov 9, 2022 at 10:40
  • $\begingroup$ @JaimeYepesdePaz The centre of mass of the cylinder is moving, which is where the "translation" part comes from. The cylinder has one total k.e. but sometimes it is useful to split this into translational k.e. which is the same for all particles in the object and other types of k.e. (rotational k.e., vibrational k.e. etc.) which can vary from one particle to another. It is just terminology. $\endgroup$
    – gandalf61
    Commented Nov 9, 2022 at 14:39
  • $\begingroup$ Sure, the c.o.m is moving and that gives it a kinetic energy, but in the rod, whose c.o.m is moving as well, we don't consider it. I don't get why... $\endgroup$ Commented Nov 9, 2022 at 19:20

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