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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 $\omega = \sqrt{2E_k/I}$?

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    $\begingroup$ There is no need for any friction between the stone and the foot in order to trip. $\endgroup$
    – CuriousOne
    Jun 10, 2015 at 7:40
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    $\begingroup$ You should revise your idea of the dynamics of extended bodies. $\endgroup$
    – CuriousOne
    Jun 10, 2015 at 7:44
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    $\begingroup$ @CuriousOne, write an answer and give a detailed description, if you can $\endgroup$
    – user77434
    Jun 10, 2015 at 7:49
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    $\begingroup$ 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. $\endgroup$
    – CuriousOne
    Jun 10, 2015 at 7:52
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    $\begingroup$ @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. $\endgroup$
    – QCD_IS_GOOD
    Jun 10, 2015 at 8:25

2 Answers 2

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If you think of the runner as being stationary and the ground as being the thing that moved, you can think of a rock hitting a rigid object off center. A torque is imposed on the body based on the force perpendicular to the body and the distance from it's axis of rotation.

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When you move with a certain velocity, you possess what is known as inertia of motion which is the measure of the difficulty to avert the motion. Since you have velocity, you have an associated momentum.

Now, when your foot gets struck to the stone, by Newton's third law, the stone exerts a reactive force due to its deformation by the force applied by you. This reactive force provides sufficient impulse to stop your feet by destroying the momentum of the foot. However, as your body has still the momentum left, by Newton's first law, your body moves & eventually leans forward ultimately in a position of unstable equilibrium. This makes the great THUD!

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  • $\begingroup$ @user77434: Have I written that torque is not acting??? Have you read the answer completely? If so, haven't you seen "unstable equlibrium"?? Do you what happens when a body reaches this state? Do you know that the linear terms will change by then? Gravity is providing the torque when the body reaches this position, do you know that? Isn't it implied here?? If you aren't satisfied, The answer-box is here; write your own answer! $\endgroup$
    – user36790
    Jun 10, 2015 at 7:50
  • $\begingroup$ @Daffy: Hey, can you explain me why a person leans & sometimes falls forward when a moving bus applies brakes???? $\endgroup$
    – user36790
    Jun 10, 2015 at 9:17
  • $\begingroup$ @Daffy: You are skipping my question wittily. I do know the explanation & it is the same as above. BTW, thanks for downvoting:) $\endgroup$
    – user36790
    Jun 10, 2015 at 13:36
  • $\begingroup$ @Daffy: Momentum is associated with the whole body. And answer yourself the question I have asked & read any valid explanation in the net. Then come again. $\endgroup$
    – user36790
    Jun 10, 2015 at 13:42