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I was trying to wrap my head around this solution, and the whole idea of a car moving around a level curve, how the friction supplies the centripetal force. But I still can't wrap my head around the idea. So maybe it would help if I walk you through my logic and you can see where I went wrong in my reasoning.

You can deduce from the centripetal force requirement that while moving on a merry-go-round, there has to be some centripetal force that is keeping you moving in the circular pattern. But let's say for a moment that there's no centripetal force acting on me and I'm standing on a merry-go-round. I'll move at a constant velocity in a direction tangent to the center of the circle. "Static friction acting on an object points opposite to the direction in which the object would slide along the other object if static friction didn't exist." So I'm right now at the moment after I started sliding, sliding with a velocity vector that's orthogonal to the centripetal force. If static friction points in the opposite direction, and I'm sliding tangent to the circle, shouldn't it be pointing in the direction opposite the tangent rather than perpendicular to the tangent line?

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If you are sliding across the surface, then "static friction" is not applicable. Consider first your motion on a merry go round without sliding. At any instant, your tangential velocity is the same as the tangential velocity of the surface under your feet. Since the two velocities are the same, no instantaneous frictional force is required to keep you moving along with the surface in that direction. Let's say that your merry go round is spinning in the x-y plane and you are at the 'top,' such that your velocity has no y component.

An instant later, the point under your feet is pulled toward the middle of the circle by the centripetal molecular forces. These forces are (at this point in the circle) completely in the y direction, but not in the x direction. However, since you are not molecularly bound to the merry go round, those forces do not affect you. So you are moving in the x direction, but not in the y direction. Now there is a small relative motion between you and the surface in the y (i.e. radial) direction. Because that is the direction if the relative motion between your feet and the merry go round, that is the direction of static friction.

Now, if you were to overcome static friction and begin sliding, the direction of kinetic friction would not be purely radial, because in addition to your radial relative motion, your tangential speed would begin to differ from that of the surface so there would be tangential frictional forces as well. But that is dependent on your position with respect to the surface is changing, which does not apply to your original question about static friction.

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  • $\begingroup$ This is a good clear answer. $\endgroup$ – Žarko Tomičić Nov 8 '15 at 18:07
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I believe that firstly we need to get one thing straight and that is, on infinitesimal level, where is the acceleration of a body moving on some circular path pointing? The answer is, it is always pointing away from the center of curvature of your path, road, or whatever. So then, there is always a force in action, force which is trying to move you peprepndicularly off your momentary velocity vector. But this vector is ever changing so the force is too. So the direction of a friction force is always changing too. Now, if we imagine you standing on a circular disc, and a disc starts to rotate, the friction is definitely keeping you in this circular motion. But the nature of this circular motion is such that it always keeps changeing its direction. Always. If you turn this force(friction) that changes its directio off, your velocity would remain unchanged and you would fall off your path. What about the friction related to the direction tangent to the circle? Well, in the case of a moving car, it actually helps it move. Imagine that you are standing on a disc made of ice. Only thing you have to keep you in place is some sort of a removable wall put behind you so the disc accelerates in its angular velocity and the wall is keeping you in place. But you will notice that you are sliding away from the center too. And here this wall that helped you when the disc was starting to rotate now is of no use. But now, with your angular velocity matched with the disc, remove the wall from your back and put it on your side. It would stop your sliding away.Of course, if the disc changes its angular velocity, it would be a different story.Now you would have to have two walls, one on the side and one behind you. In this example, YOU ARE ALWAYS LOOKING IN THE DIRECTION TANGENT TO THAT OF A CIRCLE ON WHICH YOU ARE MOVING.

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  • $\begingroup$ "The answer is, it is always pointing away from the center of curvature of your path, road, or whatever." I presume you mean toward the center here. $\endgroup$ – dmckee Nov 8 '15 at 16:25
  • $\begingroup$ Acceleration is always pointing in the direction of the force. If we are talking about centripetal force, then yes, it is toward. If we are talking about centrifugal, it is away. Friction is suplying centripetal force and in that manner, is pointing towards. But this is only because body is trying to move away. So, yes and no :-) $\endgroup$ – Žarko Tomičić Nov 8 '15 at 17:51
  • $\begingroup$ For introductory questions like this, there are a couple of things that I try to keep in mind: (a) students at this level have a hard time intuiting the frame of reference you're using, so tell them explicitly and (b) if you are going to work in a non-inertial frame you should be clear that this introduces a pseudoforce. I happen to agree that the answer here is clearest in the rotating frame, but that relies on the reader to be able to translate from inertial to non-inertial coordinates with some fluency. $\endgroup$ – dmckee Nov 8 '15 at 17:56
  • $\begingroup$ Might be so.Good point indeed. But, this as much of an effort as I would like to input on this right now. Being the clearest at the moment is ok. I was under the impression that rb612 should be pointed in the direction of constant angular velocity versus angular acceleration because that might be, unknowingly, source of his confusion. $\endgroup$ – Žarko Tomičić Nov 8 '15 at 18:01
  • $\begingroup$ Seen your profile, and quote. I like one myself...Physics is that area of human experience which is best described by harmonic oscilator. $\endgroup$ – Žarko Tomičić Nov 8 '15 at 18:05

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