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If there is a block that is placed at the rim of a turntable, and we start rotating this turntable, I know that while the turntable is rotating, a centripetal force is acting on the block. This force is static friction. If we increase the angular speed of the turntable, the static friction must increase (according to newton's 2nd law with the normal direction $F=mv^2/r$, where $v$ is the speed of the block, $r$ is the radius, and $m$ is the mass of the block). If we continue increasing the angular speed of the turntable, kinetic friction will act as a centripetal force instead of the static friction and the block will slip. My question is, in what direction will the block slip? Will the block slip in the tangential direction? Or will it slip at an angle from the tangential direction?

My expectations(I don't know if this is true) is that according to newton's 2nd law $F=mv^2/r$, since $v$ is very large and $F$ (kinetic friction force) is very small, then $r$ must become very large. When the radius of any circle (in general) is very large, the circle becomes like a straight line (I mean that a straight line is a circle such that the radius goes to infinity, so as $r$ gets larger, the block will kinda move in a straight line), which means that the block will slip in the tangential direction.

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    $\begingroup$ Which reference frame are you in? From the reference frame of the turntable, the block will slip away from the center. From the reference frame of someone standing on the "sidelines" and observing the block, the path of the block will be somewhat more complex, and will depend on the coefficient of kinetic friction and the angular acceleration of the turntable. $\endgroup$ – David White Jun 27 '17 at 15:45
  • $\begingroup$ @Eman.suradi : Thanks for your comments to my answer. Sorry I am confused myself. What I should say is that the static limit applies in the direction of attempted motion. I am deleting my answer temporarily and shall revise it within a couple of days. $\endgroup$ – sammy gerbil Jun 28 '17 at 14:25
  • $\begingroup$ @sammygerbil I am waiting :) $\endgroup$ – Eman.suradi Jun 28 '17 at 15:14
  • $\begingroup$ @sammygerbil .. $\endgroup$ – Eman.suradi Jul 8 '17 at 20:24
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If the block started off stationary and totally massless then yes, there would be no friction force because there would be no normal force and the block would sit in one place while the turntable accelerated under it. This would be a perfect tangential slip condition.

However, if the block is massive and therefore is co-moving with the turntable, then any failure of its static friction will mean that it wants to continue in a straight line, neither staying in one place nor moving along with the rotating reference frame. The turntable will "drag" it a bit in the direction it's turning, but usually the coefficient of kinetic friction is much, much lower than the coefficient of static friction, so to first approximation it simply flies off straight along whatever tangent it was going when it lost traction: in the rotating reference frame this will seem first and foremost like a radial acceleration, but it will seem to be warped by a Coriolis force into the tangential directions too, as it gets underway.

If you are unaware of the Coriolis force, my favorite way to think about it is as the cause behind the global wind patterns on our planet. Look at the "trade winds" pattern on that page, how at the equator the winds move West. The simple explanation is that, since the circumference of a circle scales proportionately to the radius, the outer parts of the circle have to move at higher speeds to complete a revolution in the same time. Therefore what it means to be "at rest" relative to the rotating reference frame -- the co-moving velocity -- involves moving at faster and faster speeds at further and further radii. So, when you move "outward" with your constant momentum, the rotating reference frame thinks you are moving "backwards" relative to it. So the Earth spins towards the east which is why the Sun rises in the east; if the Sun heats up the oceans into steam which starts floating up into the atmosphere, then we would say that this updraft also seems to be drift westward relative to the surface due to the Coriolis effect. This is the dominant effect near the equator. The air that it displaces must come back down somewhere, so it generally comes back down at some other latitude North or South, and as it comes down it is going much faster than the co-moving velocities are, so it appears to be going East relative to the surface. So that's what the Coriolis effect does.

Similarly, as your object begins its radial acceleration it starts to move to further and further radii where it does not have the speed to keep up with the local tangential velocity, so it must appear to accelerate in the backwards tangential direction. The now-lessened static friction force will point opposite to the comoving velocity and therefore it will start pointing towards the center and will begin pointing a little bit forwards of center as we get to higher radii. Finally, there will be another fictitious force if you are not increasing this velocity very slowly, due to your acceleration of the ground relative to the object in question.

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  • $\begingroup$ "However, if the block is massive and therefore is co-moving with the turntable, then any failure of its static friction will mean that it wants to continue in a straight line, neither staying in one place nor moving along with the rotating reference frame. The turntable will "drag" it a bit in the direction it's turning, but usually the coefficient of kinetic friction is much, much lower than the coefficient of static friction, so to first approximation it simply flies off straight along whatever tangent it was going when it lost traction" $\endgroup$ – Eman.suradi Jun 27 '17 at 16:29
  • $\begingroup$ This is an explanation from a fixed reference frame , true. ? $\endgroup$ – Eman.suradi Jun 27 '17 at 16:29
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    $\begingroup$ What do you mean by radial acceleration? Do is it mean that the object is moving in the radial direction ? $\endgroup$ – Eman.suradi Jun 27 '17 at 18:12
  • $\begingroup$ Yes, I mean that using comoving polar coordinates $(r, \theta)$ you see that "straight line" (fixed-reference-frame) trajectory appear to go towards increasing $r$ in a sort of parabolic way, and then, due to the Coriolis force, decreasing $\theta.$ As sammygerbil says in the other answer, this is all based on assuming very low tangential accelerations so that what "gives" is that the centrifugal force exceeds the static friction force limit. $\endgroup$ – CR Drost Jun 28 '17 at 14:13

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