# The physics of tripping on a stone [closed]

You know, you are walking or running at a certain velocity, your foot gets stuck on an obstacle and you end up flat-faced to the ground.

What is the physics behind it? How does the linear motion change into rotational motion? Probably the formula for an ideal body tripping is simple; is there any energy subtracted from linear motion and given to the stone? Or is it just $\omega = \sqrt{2E_k/I}$?

• There is no need for any friction between the stone and the foot in order to trip. – CuriousOne Jun 10 '15 at 7:40
• You should revise your idea of the dynamics of extended bodies. – CuriousOne Jun 10 '15 at 7:44
• @CuriousOne, write an answer and give a detailed description, if you can – user77434 Jun 10 '15 at 7:49
• I did. Your question doesn't make any sense because it expects something that physically isn't necessary and plays no role in practical tripping. – CuriousOne Jun 10 '15 at 7:52
• @CuriousOne "There is no need for any friction between the stone and the foot in order to trip" Well in fact you DO need static friction, or else the stone will just slide along with you. The point is that static friction acts no work on either body. – Joshua Lin Jun 10 '15 at 8:25