Imagine a ladder leaning against a wall. All surfaces are smooth. Hence the ladder will slip and fall. While falling it rotates because there are external torques acting on it. My question is about which axis does the ladder rotate?


4 Answers 4


If the ladder is slipping on the floor as well as the wall, then the point of rotation is where the two normal forces intersect. This comes from the fact that reaction forces must pass through the instant center of motion, or they would do work.

In the diagram below forces are red and velocities blue. If the ladder rotated by any other point other than S then there would be a velocity component going through the wall, or the floor. S is the only point that keeps points A and B sliding.


This leads to the acceleration vector of the center of mass C to be

$$ \begin{aligned} \vec{a}_C &= \begin{pmatrix} \frac{\ell}{2} \omega^2 \sin\theta - \frac{\ell}{2} \dot{\omega} \cos\theta \\ -\frac{\ell}{2} \omega^2 \cos\theta -\frac{\ell}{2} \dot\theta \sin \theta \\ 0 \end{pmatrix} & \vec{\alpha} &= \begin{pmatrix} 0 \\ 0 \\ -\dot\omega \end{pmatrix} \end{aligned}$$

If only gravity is acting, then

$$\dot\omega = \frac{m\,g\,\frac{\ell}{2}\sin\theta}{I_C+m \left(\frac{\ell}{2}\right)^2} $$

  • $\begingroup$ How did you figure out the axis of rotation to lie outside the body? $\endgroup$ Dec 1, 2013 at 16:46
  • 1
    $\begingroup$ Like I said, the axis of rotation has to be where the reaction forces intersect. Also, it is the only point which makes the rigid body velocity vectors parallel to the slip directions. $\endgroup$ Dec 1, 2013 at 17:28
  • 1
    $\begingroup$ @shortstheory, You can arrive at it with the following logic. Since velocity at A has to be horizontal the rotation has to be somewhere along the vertical line through A. If the rotation center has a horizontal sepation with A then there will be component of the velocity up or down which is not allowed. For any sliding contact, the rotation is along the perpendicular line. The same for B leading to only one possible solution S. $\endgroup$ Dec 1, 2013 at 17:43
  • $\begingroup$ @ja72 -1: In my opinion, this answer is misleading. It's misleading to say that one is somehow forced to choose a particular point as the definition of the axis of rotation. Sure, if you demand that the point satisfy certain properties, then you might feel that a particular choice is natural, but that's a matter of preference. Look in books on analytic mechanics, and you'll see physicists will often choose an axis passing through the center of mass and for good reasons. $\endgroup$ Dec 1, 2013 at 20:30
  • 2
    $\begingroup$ @joshphysics The choice is done by the constraints and there is no other point that would work in this case. I did not simply choose point S, the constraints did. There is only one point of rotation in the plane. $\endgroup$ Dec 2, 2013 at 0:10

The ladder falls because it experiences unequal moments from the normal reactions at both its ends. That is to say that the surface pushes on the ladder from the bottom as well as the side. In the absence of tangential contact forces such as friction, the ladder rotates and falls.

To solve a problem with such a situation, you may choose any point as the origin. If you decide to choose the COM of the uniform ladder that is at the middle of the ladder, then you must also know that as the force $mg$ passes through the COM; it does not contribute to rotation of the ladder about the COM at all! But if you choose one of the ends of the origin, you will have to consider that $mg$ indeed exerts torque and contributes to rotation about that axis. There isn't strictly one axis which is correct to choose, you may choose the origin at your own convenience as the problem demands.

  • $\begingroup$ The choice is not arbitrary. Planar motion is always described as rotation about a single point, and kinematics dictate where this point is in this case. The velocity at the ends of the ladder must be parallel to the wall and floor, or else the ladder will penetrate through the wall. $\endgroup$ Dec 1, 2013 at 16:36
  • $\begingroup$ @ja72: Only if you insist on describing the (instantaneous) motion as a pure rotation, rather than as a rotation plus a translation. Admittedly, that's what the OP seems to be asking for (since otherwise their question doesn't make much sense), but in general it doesn't seem to me like a very useful way to model rigid body dynamics. $\endgroup$ Dec 1, 2013 at 17:26
  • $\begingroup$ @IlmariKaronen, actually it is. In 3D motion is a rotation about an instantaneous line (and a translation along the line) called a screw motion. When projected to 2D this becomes a point on the plane. It is useful because it lends to a decomposition of forces along reaction and active directions. The above problem becomes a 1 DOF problem with a virtual pin joint, instead of a 3 DOF + 2 constraints. $\endgroup$ Dec 1, 2013 at 17:33
  • 1
    $\begingroup$ @IlmariKaronen. Any rotation plus translation of a point A can be described as a pure rotation about a point P with coordinates $$x_P = x_A - \frac{\dot{y}_A}{\omega} \\ y_P = y_A + \frac{\dot{x}_A}{\omega}$$ where $\vec{v}_A = (\dot{x}_A,\dot{y}_A)$. Even in 3D the above becomes $$\vec{r}_P = \vec{r}_A + \frac{\vec{\omega}\times\vec{v}_A}{|\vec{\omega}|^2}$$ $\endgroup$ Dec 1, 2013 at 17:37
  • $\begingroup$ @ja72: Perhaps we're coming from different viewpoints here. I agree that the DoF reduction trick is neat here (and gave you a +1 for it), but coming from a numerics background, for general use I'd rather prefer a representation of motion that a) did not change in the absence of external forces, b) worked for all rotation rates, including zero, and c) was easy to (approximately) integrate over time. Translation + rotation around the CoM fits those criteria. $\endgroup$ Dec 1, 2013 at 17:43

This is a case of non-stationary axis rotation. The axis is generally taken as the COM of the ladder as about this axis, the mass distribution is equal on both sides.

  • $\begingroup$ I don't understand, what does symmetrical mass distribution have to do with the axis of rotation? $\endgroup$
    – user34304
    Dec 1, 2013 at 15:18
  • 1
    $\begingroup$ Although moments need to be summed about the center of mass, the CM is not motionless because the sum of the forces is not zero. $\endgroup$ Dec 1, 2013 at 16:37

from the point of any point on the rod , the rod is rotating about that point.refer to landau mechanics .

  • $\begingroup$ The OP asks about point of rotation from an inertial frame. From a local coordinate system, there is no motion because the ladder is a rigid body. $\endgroup$ Dec 16, 2013 at 13:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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