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A car starts from rest on a horizontal circular road of $190\ \mathrm m$ and gains speed at a uniform rate of $1.2\ \mathrm{m/s^2}$. The coefficient of static friction between the tyres and the road is $0.37$. Calculate the distance travelled by the car before it begins to skid.

Here are my doubts regarding the solution and the problem statement.

As the solution given in the picture above, we can see that they have considered the centripetal force and the tangential force. But centrifugal force applies if I observe from the car's frame. Then, the direction of the net force becomes different, so is the solution wrong?

Also, this is contradictory because if that happens, there is no resultant force in the inward direction, and so there is no force to balance the centrifugal force; this is a massive dilemma at hand: how will the net force be defined in this system?

Is this free-body diagram even correct? I look forward to the brilliant answers to this question from all the physics lovers.

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  • $\begingroup$ There's effectively 3 forces in play, right? There's centrifugal force, friction, and whatever is causing acceleration around the circle. We are told the acceleration resulting from the net force. Would it suffice if $(\vec{F}-\vec{F}_f)^2>(\vec{F})^2$? $\endgroup$
    – R. Romero
    Commented Jun 7, 2023 at 18:31

2 Answers 2

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if that happens there is no resultant force in the inward direction

In the frame of the car, the car does not accelerate. So we would expect the net forces to be zero.

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  • $\begingroup$ Yes that makes sense but if we think in other way,i mean we have a centrifugal force and a tangential force which are at right angles,so there should be a resultant right?I know that net force should be $0$ since we are at rest but how do we prove the two vectors at right angles logic wrong? $\endgroup$
    – madness
    Commented Sep 27, 2021 at 9:14
  • $\begingroup$ You're used to thinking of "in the frame of the car" as meaning a centrifugal force appears. Well, "in the frame of the car" forces appear to oppose forward and rearward acceleration forces as well, not just centrifugal. Perhaps instead of the cars frame you want to observe from some frame rotating at a fixed speed? $\endgroup$
    – BowlOfRed
    Commented Sep 27, 2021 at 9:30
  • $\begingroup$ There is no radial acceleration at least. There is still tangential acceleration. $\endgroup$ Commented Sep 27, 2021 at 10:01
  • $\begingroup$ That depends on your frame. In the frame of the car, the car has exactly zero acceleration in any direction. Any forces that would create an acceleration would cause an opposite virtual force to appear due to the acceleration in the frame. This is true for all directions. As you specified "the frame of the car", this could make understanding the problem more difficult. $\endgroup$
    – BowlOfRed
    Commented Sep 27, 2021 at 10:05
  • $\begingroup$ @BowlOfRed True; the frame itself would be accelerating to keep up with the car. $\endgroup$ Commented Sep 27, 2021 at 10:10
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But if I observe from the frame of the car,then centrifugal force applies for me.

Yes, this is true!

Then the direction of the net force is becoming different

True; in the frame moving with the car the acceleration of the car is $0$. So the net force is $0$ if we want to bring in pseudo-forces to keep Newton's second law valid.

Also this is kind of contradictory because if that happens there is no resultant force in the inward direction and so there is no force to balance the centrifugal force

Here is the common misconception. The centripetal force (friction here) doesn't go away in the rotating frame! It's a real force, and so it exists in all frames of reference. However, in the frame rotating with the car, the centrifugal force exactly cancels the centripetal force.

If the frame is accelerating with the tangential acceleration of the car as well, then there is a tangential pseudo-force which cancels the tangential force acting on the car as well. This results in no acceleration of the car in the frame moving with the car.

If the frame is rotating at some fixed rate, then we will still have a centripetal and centrifugal force, but they will not always be equal, and so we will still observe circular motion. This makes sense, because a frame not rotating at the rate of the car at all times will see the car moving in a circular path. In this fixed rotation frame, the previously mentioned tangential pseudo-force would no longer be present.

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  • $\begingroup$ Thanks a lot!If you don't mind me asking,how can we be sure that frictional force is providing the centripetal force?*Because the solution in the picture is showing something different meaning it says frictional force acts opposite to the net force*.Please forgive me if i am being dumb $\endgroup$
    – madness
    Commented Sep 27, 2021 at 20:42
  • $\begingroup$ @madness I'm suspicious of that diagram. There should be a component of friction supplying the centripetal force. They're isn't any other force that could do this here. $\endgroup$ Commented Sep 27, 2021 at 23:39
  • $\begingroup$ @madness The problem states the car is accelerating. The frictional force on the tyres needs to provide both the centripetal force, and the force to accelerate the car tangentially. That is, it needs to the provide the total force that is accelerating the car. $\endgroup$
    – fishinear
    Commented Oct 11, 2023 at 13:49

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