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⚛️ NEET Physics❓ Question

Friction opposes motion, so how can friction supply the centripetal force on a flat curve?

Lakshmi TripathiDropper Sample

24 Sept 2026 · 1 min read

I do not usually post, but this one has been bothering me for a few days.

Dynamics of Circular Motion. A car turns on a flat horizontal road. The book says the force that provides the centripetal force is the (static) friction between tyres and road, giving μsmg≥mv2r\mu_s mg \ge \dfrac{mv^2}{r} so vmax=μsrgv_{max}=\sqrt{\mu_s r g}.

But we studied friction opposes motion. If the car is moving forward, friction should act backward. How is it pointing towards the centre of the circle, sideways? My answer in a test was "friction here is a different kind", which did not satisfy even me.

Second: banking. For a banked road without friction, tan⁡θ=v2rg\tan\theta = \dfrac{v^2}{rg}. And I remember the formula with friction has a μ\mu in it, but I don't remember it properly. How does banking help if friction is already enough on a flat road?

A numerical I tried: μs=0.5\mu_s = 0.5, r=20r=20 m, g=10g=10 m/s². I got vmax=10v_{max}=10 m/s. Is it correct, and does it depend on the mass of the car?

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7 answers & replies

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  • [+]Vivek Mishra12thSample· 25d ago

    Friction opposes the relative motion (or the tendency of it) between the two surfaces in contact, not the car's motion. When the car turns, the tyre would tend to slide outward (straight line, away from the centre) relative to the road. So static friction acts inward, opposing that sideways sliding tendency. The tyre's contact patch is not actually sliding forward, so no backward friction is needed for a steady turn.

    Your numerical is right, vmax=0.5×20×10=10v_{max}=\sqrt{0.5\times20\times10}=10 m/s. And no, mass cancels out. A heavy truck and a small car have the same limit on the same road.

    • [+]Lakshmi TripathiDropperSample· 25d agoAuthor

      Opposing the sliding tendency of the tyre relative to the road. That's the missing sentence. Nandri everyone.

  • [+]Nisha Verma12thSample· 25d ago

    Banking idea: the normal reaction on a banked road has a horizontal component pointing to the centre, Nsin⁡θN\sin\theta, so some or all of the needed centripetal force comes from the road's normal force and not from friction. For the ideal angle (no friction needed): Ncos⁡θ=mgN\cos\theta = mg and Nsin⁡θ=mv2/rN\sin\theta = mv^2/r. Divide: tan⁡θ=v2rg\tan\theta = \dfrac{v^2}{rg}.

    With friction: vmax2=rg μ+tan⁡θ1−μtan⁡θv_{max}^2 = rg\,\frac{\mu + \tan\theta}{1-\mu\tan\theta} So the car can go faster than on a flat road, and on wet roads (low μ) the banking is the thing that keeps it safe.

  • [+]Dev Agarwal12thSample· 25d ago

    Also, on the car in the turning case, the centrifugal force pushes it outward and friction balances this to keep it on the road. So friction = centrifugal force at the limit.

    • [+]Neel Chatterjee12thSample· 25d ago

      Needs one correction. In the ground (inertial) frame there is no centrifugal force acting on the car, only the real forces (gravity, normal, friction), and their net force is the centripetal one. Centrifugal force appears only if you describe things from the rotating frame of the car, as a pseudo force. The numbers come out the same, but you shouldn't draw both centripetal and centrifugal in one free-body diagram, that's double counting.

      • [+]Dev Agarwal12thSample· 25d ago

        Right right. Pseudo force only in the car's frame. In the ground frame I just do net force = mv2/rmv^2/r. Cleaned it up, thanks Neel.

  • [+]Aman YadavDropperSample· 25d ago

    Related. Vertical circle with a string: min speed at the top for the string to stay taut vtop=grv_{top}=\sqrt{gr}. Speed at the bottom then vbottom=5grv_{bottom}=\sqrt{5gr} (energy conservation). The tension difference between bottom and top is 6mg6mg in general. NEET asks these as direct formula questions.

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