Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A friend of mine fell while he was snowboarding which led to a discussion. He mentioned that, in general, if your center of mass is traveling at a certain velocity, and you fall forward, your head would contact the ground at twice the velocity that your center of mass was traveling. Is this true? And if so can you explain the physics behind it to me?

share|cite|improve this question
up vote 3 down vote accepted

If you assume that

  1. Your body is a uniform, thin rigid rod.

  2. One end of the rod is pivoted (aka your feet) during the fall.

Then one simply recalls that the angular velocity $\omega$ of rotation of your body is related to the tangential velocity $v$ of a point a distance $r$ from the pivot by \begin{align} v = \omega r \end{align} Now, if you have height $h$, then your center of mass will be a distance $h/2$ from your feet, and your head will be a distance $h$ from your feet, so the ratio $v_\mathrm{head}/v_\mathrm{COM}$ of the tangential velocity of your head to that of your center of mass is \begin{align} \frac{v_{\mathrm{head}}}{v_{\mathrm{COM}}} = \frac{\omega h}{\omega (h/2)} = 2. \end{align} In other words, the tangential velocity of your head will be twice as large as that of your center of mass.

The intuition behind this is simply that for a given change in the angular position of your body, your head moves twice as far as your center of mass.

Note that in the real world, the assumptions above aren't really correct, but they do give you an idea of the basic physics in a rough approximation.

share|cite|improve this answer
Thats pretty simple, I should have seen that. Thanks for the clear and timely response! – DanW Feb 10 '14 at 21:30
@DanW Sure thing. – joshphysics Feb 10 '14 at 21:30
But this also assumes that your feet are not moving relative to the ground. Since the question implies that "the friend" fell while moving (snowboarding). In this case the correlation will not hold, unless you are referring to the velocity normal to the surface (assuming the surface in flat). – fibonatic Feb 10 '14 at 23:24
@fibonatic Agreed, that's why I used the phrase "pivoted during the fall" when I enumerated the assumptions. In real life, a snowboard can easily be pivoted in the snow. – joshphysics Feb 11 '14 at 0:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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