Question
A car of mass kg moves at m/s on a horizontal circular track of radius m. Find the centripetal force needed and the minimum coefficient of friction between tyres and road. Take m/s.
Solution — Step by Step
On a flat horizontal track, only static friction can provide the centripetal force. So N.
The maximum static friction is . For the car not to slip:
Final answers: N, .
Why This Works
Centripetal force is not a new kind of force — it’s the net inward force from whatever real forces (friction, tension, gravity, normal) are acting. On a flat road, friction is the only horizontal force, so it must equal .
Notice that mass cancels out of the friction condition: . Heavier cars need the same coefficient — they have more inertia but also more weight.
Alternative Method
For a banked road at angle , gravity helps — the optimum (no friction needed) banking angle is . With banking, requirement drops sharply. JEE often combines flat and banked versions.
Students sometimes write “centripetal force = friction + something else.” On a flat road there’s no other inward force. Don’t invent forces — only what’s physically there can provide the centripetal pull.
Common Mistake
A persistent error: treating centripetal force as an additional force in the FBD alongside friction, gravity, and normal. It isn’t. Centripetal force is the NAME we give to the net radial force. Drawing it as a separate arrow leads to double counting.