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I searched for my question and I found many results but all of them eventually try to solve something specific (like solving for the kinetic energy or finding when does the ball stop).

what I want is why does a ball roll at all?

Pretend I shot a ball and the ball started slipping on the ground, pretend it doesn't leave the ground and that the ground isn't incline.

Now I thought of friction force exerting a torque on the ball, if the force is $F_c$ then the torque should be:
$$\tau=u\times F_c\times r$$(There is no sin because $F_c$ is perpendicular to $r$)

The existence of $u$ is due to the fact that the force doesn't just rotate the ball but also slows it down.

What's missing in my assumption is the speed of the ball, it seems to me that the faster the ball the faster it rotates, I don't have an explanation for this because $F_c$ doesn't relate to velocity, $F_c$ is given by :
$$F_c=u_k\times N$$

where $N$ is the normal force and $u_k$ is the kinetic friction coefficient.

So what's a better explanation to this phenomena?

Edit

Something sparked in my mind, the ball doesn't just slip, in reality it bounces, we just don't see the bounces.

But I don't know how to put this into a good formula or explanation.

PS

what I mean by "the faster the ball" I mean its initial velocity (after shooting it) because the ball slows down.

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Consider two different regimes.

First - when the ball is still slipping, the relative motion of the ground and ball causes a force on the ball which (a) slows down the center of mass and (b) increases the angular speed. Once the ball rolls without slipping this force disappears.

Second - when the ball does not slip, the contact point must be stationary. This means that the angular velocity $\omega=\frac{v}{r}$. Now in that expression $v$ is the velocity of the rolling ball which is less than the velocity you launched it with because of what I wrote above. The time it takes to stop slipping depends on the initial velocity - as you said the torque is independent of the speed but the ball will just keep slipping (and rotating faster) as long as it takes to reach the point of no slip.

I wrote some earlier answers analyzing this motion in more detail - see for example here or there

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  • $\begingroup$ How can you compare and say that w is less than v. Both are directly proportional and radius is constant. How can you compare parameters with different dimensions.. $\endgroup$ – Abhijeet Apr 19 '15 at 17:30
  • $\begingroup$ @Abhijeet I don't compare $\omega$ with $v$ - I compare it with $v/r$ (the "no slip" angular velocity) $\endgroup$ – Floris Apr 19 '15 at 17:53
  • $\begingroup$ thanks, I would really appreciate if you add the links, but I don't understand , does the ball roll after it stops slipping ?(it seems to me that the rotational motion stops with the Withdrawal motion) $\endgroup$ – niceman Apr 19 '15 at 17:55
  • $\begingroup$ @floris V/r is w which itself is the angular velocity while rolling with no slip $\endgroup$ – Abhijeet Apr 19 '15 at 17:59
  • $\begingroup$ @niceman - Once the ball stops slipping, it continues to roll with velocity $v$ and angular velocity $\omega = \frac{v}{r}$. $\endgroup$ – Floris Apr 19 '15 at 18:12
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The ball rolls because: the friction present between the surface pushes the atoms/particles in contact with surface or ground backwards, hence with centre of mass moving in a direction, the ball's bottom surface(ie. The surface in contact) moves in opposite direction or backwards, causing the ball to roll. The ball slips because there is no friction that can push the bottom or atoms/particles in contact backwards. Thanks :)

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  • $\begingroup$ this doesn't explain the effect of speed $\endgroup$ – niceman Apr 19 '15 at 16:22
  • $\begingroup$ As long as the friction is sufficient, the ball won't slip, it'll rotate. But it will not hold if the velocity of the ball is too high, v=r*w. Other way to say it is that the ball will topple over its surface if the initial shooting speed it considerably high. $\endgroup$ – Abhijeet Apr 19 '15 at 16:33
  • $\begingroup$ hmmmm yes , if the speed is high it will topple, that's another question to be asked later "why does the ball topple" but for the mean time your answer still doesn't explain the effect of speed on rolling speed $\endgroup$ – niceman Apr 19 '15 at 16:37
  • $\begingroup$ Effect of speed on rollong speed? What do you wish to ask.. $\endgroup$ – Abhijeet Apr 19 '15 at 16:40
  • $\begingroup$ I want to say "if you shoot stronger, it will rotate faster" why ? sorry bad english , meant rotation speed $\endgroup$ – niceman Apr 19 '15 at 16:43
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Your assumption that the ball will be faster if it rotates faster has a very logical explanation and is completely valid. If your ball has a circumference C it will cover that distance with each rotation. Let us say your C was 1m and the period was 2 rps (revolutions per second) then your ball would cover 2 metres in 1 second. If your period is doubled to be 4 rps then your ball would cover 4 metres in one second. Its linear speed has also doubled.

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  • $\begingroup$ yes but why ? what's the mathematical proof to your sayings ? from where to get "2 rps" ? $\endgroup$ – niceman Apr 19 '15 at 16:21
  • $\begingroup$ Also the ball will never rotate faster than what is given by its linear velocity (if torque is provided by friction). The two rounds per second is a random value (this also works algebraically). $\endgroup$ – Jaywalker Apr 19 '15 at 16:43
  • $\begingroup$ hmmmm your comment needs a proof $\endgroup$ – niceman Apr 19 '15 at 16:44

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