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I'm studying a problem and I encountered a strange problem:

When a ball bounces how much time does the ball spend while touching the floor?

To be more clear I suppose that when a ball bounces the actual bounce can't start EXACTLY when the ball touches the floor but the ball touches the floor, the Energy from the fall is given to the floor then the floor gives the energy back and the ball bounce; but in what time?

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  • $\begingroup$ Hi Peterix. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic Jan 12 '15 at 9:23
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An possible (simplistic) answer would be the following: a simple model for the bouncing ball is a spring that shrinks to absorb all the initial kinetic energy and restores fully. To put it into equations, call $v_0$ the initial velocity of the ball, $m$ its mass and $K$ the spring stiffness. The initial kinetic energy is $\frac12mv_0^2$. If the spring shrinks by a length $x$, the elastic energy is $\frac12Kx^2$. The mechanic energy is conserved, so we have all along the move $$\frac12mv^2+\frac12Kx^2=\frac12mv_0^2.$$ Let us express the velocity $v$ as a function of $x$ during the first half of the move (when the spring shrinks and the ball slows down) $$v=\sqrt{v_0^2-\frac Kmx^2}.$$ Now remark that $v=\frac{\mathrm dx}{\mathrm dt}$, so we can separate the $x$ variable and get the differential $$\frac{\mathrm dx}{\sqrt{v_0^2-(K/m)x^2}}=\mathrm dt$$ and we can integrate this equation from the moment to ball hits the ground to the moment it stops (the $\frac12$ coefficient is there because it is only one half of the movement) $$ \frac12T=\int_0^{x_0}\frac{\mathrm dx}{\sqrt{v_0^2-(K/m)x^2}}=\frac\pi2\sqrt{\frac mK}$$ (with $x_0=v_0\sqrt{m/K}$). So the result is $$\boxed{T=\pi\sqrt{\frac mK}.}$$ Interestingly, it does not depend on the initial velocity !

A more refined model would take into account the spherical shape of the ball. If it has a Young modulus $E$, the elastic energy could be (very roughly) approximated by $\frac{2\pi}3 Ex^3$ (this means that the more the ball is shrinked the more it resists to an extra shrinking). The result is $$ T=2.19187\left(\frac m{E v_0}\right)^{1/3}.$$ Now it depends on the initial velocity. (The constant is equal to $6^{1/3}\pi^{1/6}\Gamma(4/3)/\Gamma(5/6)$).

Of course the total deformation of the ball should be taken into account, the preceding approach is valid only for hard balls (large $E$). Note also that dissipation of energy has not been taken into account.

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  • $\begingroup$ Look at highered.mcgraw-hill.com/sites/dl/free/0073529281/365764/… for how to treat a sphere on plane (or other sphere) problem. Note that the contact penetration is $$x=\left( \frac{1}{d_1} + \frac{1}{d_2}\right) a^2$$ where $a$ is the contact patch radius, and $d$'s are the diameters of the contacting surfaces. $\endgroup$ – ja72 Jun 4 '14 at 15:18
  • $\begingroup$ Why is $x_0=v_0\sqrt{m/K}$ ? $\endgroup$ – garyp Jun 4 '14 at 18:15
  • $\begingroup$ @garyp. Dunring the bounce, the ball slows down until it reaches a velocity of zero and then accelerates upwards. The length $x_0$ corresponds to the moment where the velocity reaches zero. At this moment, all the initial kinetic energy has been converted into elastic energy, hence $\frac12Kx_0^2=\frac12mv_0^2$. $\endgroup$ – Tom-Tom Jun 5 '14 at 8:32
  • $\begingroup$ @ja72. Thanks for your comment. In the case of a ball bouncing on a plane, we have $d_1=2R$, $d_2=\infty$. I did indeed use this formula to estimate the elastic energy in the following way: using Young's modulus standard formula, we have $F=E\times Ax/R$. From your formula, the contact area is $A=\pi a^2=2\pi R x$, therefore $F=2\pi E x^2$, therefore the elastic energy is $\int F\mathrm dx=\frac{2\pi}3Ex^3$. (There was an extra factor $2$, that is corrected now.) $\endgroup$ – Tom-Tom Jun 5 '14 at 8:59
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The time will depend on the elasticity of the material by which the ball is made.

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    $\begingroup$ It also depends on the diameter. In the end it is the non linear Hertzian contact properties that matter. $\endgroup$ – ja72 Jun 4 '14 at 14:29
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If no other information is actually given, then this sounds to me like a Fermi problem. The idea behind these questions is for you to make reasonable assumptions and justify them.

An exact answer will depend on properties of the ball, and probably how fast the ball is moving. But just assume some reasonable values, or keep them as unknown variables.

Next, assume some model for the ball-ground collision. I myself may assume constant acceleration for its simplicity. Another one might be some type of linear restoring force, where you set the maximum displacement and hence determine the spring constant.

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An estimate can be made; others have suggestions. Here's a simple one.

If you know the height from which it is dropped, then you can calculate its speed when it hits the ground. The ball will be deformed, reaching maximum deformation when it's speed hits zero and the ball starts to go up.

For a crude estimate assume that the ball deforms by, say, 1/3 of its diameter. That is, the ball continues to move downward for and addtional length of 1/3 of its diameter after the ball first makes contact with the ground. That fraction could be 1/10, or 1/2, or 1/10000 for a metal ball, but we have to assume something. Assume also that the ball's acceleration is constant once in contact with the floor. That distance, along with the speed of the ball, gives you the duration between first contact and maximum compression. It has to go back up, so multiply by two.

Without a lot more information about the ball, and potentially very hard math, that's about all you can do. Any assumptions you make about the ball's elasticity are guesses, so it's not worth the effort. I suspect that the estimate you get this way is not a bad estimate at all.

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