# Rolling stone on a frictional surface

Consider a spherical rigid stone rotating with angular velocity $\omega$ being dropped vertically onto a horizontal rigid surface with the coefficient of friction $\mu$. Can the stone roll on the surface? If it can, what type of the motion is it? Constant velocity, constant accelaration, or varying acceleration? Ignore air resistance.

In reality if we drop a rotating ball, it will roll on the surface. But I am not sure whether it is because of the deformation of the ball at the surface of contact.

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Presumably you mean dropped from almost zero height, i.e. the vertical kinetic energy can be ignored (it will not bounce). Generally we have two coefficients of friction, static, and dynamic, the later is valid if there is relative motion. The friction will cause a torque between your ball and the surface, so it will start rolling, but intially it is a combination of rolling, and slipping. Eventually when the angular velocity at the contact point matches the horizontal velocity you have a rolling stone. –  Omega Centauri Dec 18 '10 at 4:28
More interestingly, if we assume no deformation and a perfectly elastic collision (so bouncing ball): will the ball bounce up straight or on the side? –  Sklivvz Dec 18 '10 at 19:39

I'll assume that:

1. The stone has horizontal axis of rotation.
2. The objects in question are perfectly rigid.
3. The collision is perfectly inelastic.
4. The static and dynamic coefficients of friction are equal.

In this case, you will get a vertical delta impulse on the stone, and up to $\mu$ of that impulse is converted to an instantaneous jump in horizontal speed. If the angular momentum of the stone is not exhausted by that impulse, it will skid for a while, producing constant acceleration until it grips the surface. If the angular momentum is exhausted, then the stone will just roll.

Relaxing one or more of the assumptions may yield more interesting behavior. For example, under certain conditions, rubber balls will switch the direction of rotation with each bounce.

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I will assume that the collision to the surface is perfectly elastic and $\vec{\omega}$ is parallel to the surface. Let's denote the collision time by $\tau$ and the vertical velocity of the ball at the moment of collision by $v_y$. $\tau$ is supposed to be infinitesimally short. Now we will concentrate in what happens during collision. Let's denote the force exerted by sliding friction while colliding by $F_\mu$.(due to the rotation of the ball) Then $$F_\mu=\mu\frac{\Delta P}{\tau}$$ where $\Delta P=2mv_y$ is the change in vertical momentum of the ball. This force obviously accelerates the ball in horizontal direction. So after the collision the velocity of the ball has a horizontal component $$v_x=\frac{F_\mu\tau}{m}=2\mu v_y$$ and it bounce up at an angle $\alpha$ to the horizontal
$$\alpha=arctan\frac{v_y}{v_x}=arctan\frac{1}{2\mu}$$However this solution is valid only if$\mu<\frac{1}{5}$ because in case of $\mu>\frac{1}{5}$ the rotational motion of the ball ends before the end of the collision.

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It can and will roll (and/or slide) if its inital axis of rotation does not lie in the vertical.

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The sphere is initially going to accelerate in the horizontal direction as it transferes the rotational speed into linear motion, by slipping. This will end when the sphere is going to start rolling with speed

$$\omega_{roll} = \frac{I}{I+m\,R^2}\,\Omega$$

where $\Omega$ is the initial rot. speed, $m$ the mass, $I$ the mass moment of inertia and $R$ the radius of the sphere. Once the sphere is rolling its velocity is going to be constant. This will happen at a time $t_c$ after the impact. You can find the time by plotting the horizontal speed $\dot{x}$ and the rot. speed $\omega$ as a function of time and note when $\dot{x}=\omega\,R$.

If you assume the (vetrical) impact velocity to be $v$, the coefficient of friction $\mu$ and the contact to be fairly stiff (fast), but also critically damped (with no rebound), I get the following equations for the rotational speed as a function of time

$$\omega = \frac{\mu\,m\,R}{I}\left(v\,\left(\exp({-\beta\,t})-1\right)-g\,t\right)+\Omega$$

The linear speed is

$$\dot{x} = \mu\left(v\,\left(1-\exp({-\beta\,t})\right)+g\,t\right)$$

where $\beta = \sqrt{k/m}$ is the contact frequency (sqrt. stiffness over mass), and it is a huge number making the contact last milliseconds.

To solve for all this I start from the vertical penetration (approach) of the part as it makes contact and comes to rest at a value less than zero due to gravity.

$$y = \frac{v}{\beta}\exp({-\beta\,t})-\frac{g}{\beta^2}$$

The above comes from the spring damper contact model

$$\ddot{y} = -\frac{k}{m} y - \frac{d}{m} \dot{y} - g$$

with damping $d=2\sqrt{k\,m}$.

Peace!

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After reading your answer, I feel like the only way I can verify it is to do all the calculations myself because so little detail is provided. –  Mark Eichenlaub Dec 20 '10 at 12:31
I started with these equations $\ddot{x}=\frac{F}{m}$, $\ddot{y}=\frac{N}{m}-g$ and $I\,\dot{\omega}=-R\,F$ with $F=\mathrm{sign}(\omega R-\dot{x})\,\mu\,N$. $N=-\left(k y+d \dot{y}\right)$. Then solve for $y$, then for $\omega$ and then for $x$, by assuming spin is less than horiz. speed. –  ja72 Dec 20 '10 at 22:53