Picture a rigid body such as a sledge hammer. Imagine that the base of the handle is attached to a fixed point such that it can rotate but not translate. I give the hammer a good push to get it rotating laterally about the pivot point, with its angular velocity vector pointed straight up. (Note that it's NOT rotating about its center of mass.) Assume there's no friction, but gravity is present. Note also that it's not quite in the horizontal plane because of the force of gravity. The path of the body describes a shallow cone.

Here's my question: It seems to me that the body would rotate forever with its angular velocity vector pointed straight up. So the rate of change of angular velocity is zero. But there is a net torque on the body about the pivot point because of the force of gravity. This net torque implies that there should be a rate of change of angular velocity. What am I missing?


There is a torque, but it points sideways, perpendicular to the orientation of the hammer at any time. Because of this, as the hammer rotates, the direction of the torque also rotates around with it. The torque only acts to rotate the system horizontally around in space, not to change the direction of its angular velocity.

Let's see this with a calculation. Suppose I model the hammer as a rod of length $L$ and mass $m_r$ with a point mass $m_p$ on the end.

enter image description here

The moment of inertia of the hammer at this moment can be computed by taking the moment of inertia of a similar configuration aligned along the $x$ axis and rotating it by an angle $-\theta$ in the $y$ direction:

$$\begin{align}I &= R_y^{-1}\begin{pmatrix}0 & 0 & 0 \\ 0 & \frac{mL^2}{3} + m_pL^2 & 0 \\ 0 & 0 & \frac{mL^2}{3} + m_pL^2\end{pmatrix}R_y \\ &= \begin{pmatrix}ML^2\sin^2\theta & 0 & -ML^2\sin\theta\cos\theta \\ 0 & ML^2 & 0 \\ -ML^2\sin\theta\cos\theta & 0 & ML^2\cos^2\theta\end{pmatrix}\end{align}$$

where $R_y$ is the rotation matrix around the $y$ axis and $M = \frac{m_r}{3} + m_p$. Computing the angular momentum using $\vec{L} = I\vec\omega$, where $\vec\omega = \omega\hat{z}$, I get

$$\vec{L} = ML^2\cos\theta(\hat{z}\cos\theta - \hat{x}\sin\theta)$$

The torque, on the other hand, is

$$\vec\tau = \vec{r}\times\vec{F} = (\hat{x}\cos\theta - \hat{z}\sin\theta)\times (-mg\hat{z}) = mg\hat{y}\cos\theta$$

So the torque actually pushes the angular momentum in a direction perpendicular to both its direction and the orientation of the hammer. This means there will be no change in the amount of angular momentum. It also means that the hammer rotates along with everything else, so that the angular momentum and the momenta of inertia preserve their relative orientation and thus there will be no change in its angular velocity.

  • $\begingroup$ This is correct, but it doesn't explain the confusion--- how can you have constant angular velocity when the angular momentum is nonconstant? $\endgroup$ – Ron Maimon May 1 '12 at 13:16
  • $\begingroup$ @RonMaimon The confusion for the op is how can you have a torque, yet no change in angular velocity? $\endgroup$ – John McVirgooo May 1 '12 at 16:24
  • 1
    $\begingroup$ @JohnMcVirgo: Yes, the idea is that if the angular velocity is constant, the angular momentum is constant, and the most obvious example of where this is not true is this problem (r cross p precesses around the axis of symmetry). This is one of the falsehoods that is caused by the avoidance of tensor of inertia in elementary physics classes. $\endgroup$ – Ron Maimon May 1 '12 at 18:13
  • 1
    $\begingroup$ +1: nice answer! but a little peeve--- while what you did is correct, it is correct because for rotations, there is no distinction between covariant and contravariant indices, so the tensor of inertia is the same as a matrix of inertia. It works just fine, you did it right, but I prefer to not use matrix notation in this context, because the generalization to relativity, or to nonorthonormal frames requires that one be comfortable with objects that transform covariantly in both indices. $\endgroup$ – Ron Maimon May 1 '12 at 21:18
  • 1
    $\begingroup$ @Ron true, though I don't think any of that is worth worrying about that in this case. It's pretty common to ignore it in practical usages of classical mechanics. I mainly wanted to show the structure of $I$ and $\omega$ in terms of components, and I figured matrix notation is the best (most straightforward) way to do that. $\endgroup$ – David Z May 1 '12 at 21:41

What you are missing is that the angular velocity and the angular momentum don't point in the same direction. The angular velocity is the angular velocity, and the angular momentum is the angular momentum. The two are only proportional for an object like a sphere, where the three principal moments of inertia are equal. In order to have a constant angular velocity for a rigidly rotating hammer (which is only rigidly rotating for a one parameter set of initial condition, which does not include the one you described, starting horizontal), the angular momentum has to precess around the axis, and this precession requires the steadily moving torque.

  • 1
    $\begingroup$ Good answer. Even the sphere only has angular momentum and angular velocity proportional to each other when you calculate about the center of mass. $\endgroup$ – Mark Eichenlaub May 1 '12 at 14:03

Thanks very much everyone for your input. Ater reading all your responses, my mistake was presuming that the direction of angular momentum was the same as the direction of angular velocity. In fact it's not. So even though the magnitude and direction of the angular velocity don't change, the direction (but not the magnitude) of angular momentum does change as the body rotates around the pivot. This change in angular momentum is consistent with the torque applied by the force of gravity.

Me = smarter now. Thanks!

  • $\begingroup$ Um, you could say that without words, by just accepting my answer. You can also accept David's much more detailed answer. $\endgroup$ – Ron Maimon May 3 '12 at 5:31
  • $\begingroup$ Gee, sorry for wasting electrons. I'm a newbie to the site -- didn't know there was an "accept" button. $\endgroup$ – Joe May 3 '12 at 23:24
  • $\begingroup$ It's fine, just delete this answer, and accept the one you like best, and that's it. $\endgroup$ – Ron Maimon May 4 '12 at 0:09

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.