Physics · Laws Of Motion · NEET
To move in a circle the car needs a force pointing toward the centre (centripetal force = m*v^2/R). On a flat road the only horizontal force available is friction from the tyres. So static friction points sideways toward the centre and does the whole job of turning the car. There is no banking angle to help, so friction alone must supply m*v^2/R. If the road were perfectly smooth (mu = 0), no centripetal force would exist and the car would skid straight off in a tangent line.
Both sides of the balance contain mass. The required centripetal force is m*v^2/R, and the maximum friction available is mu_s*N = mu_s*m*g (because N = mg on a level road). Set them equal: m*v^2/R = mu_s*m*g. The m divides out from both sides, leaving v^2 = mu_s*R*g. This is why a heavy truck and a light car have the SAME maximum safe speed on the same turn. Heavier vehicles get more friction, but they also need more centripetal force, and the two effects exactly cancel.
Static friction has a ceiling: its maximum value is mu_s*m*g. If the needed centripetal force m*v^2/R rises above this ceiling, friction cannot supply enough inward force. The car then cannot follow the curve — it skids outward (away from the centre) along a tangent. This is exactly the NCERT cyclist result: at v = 5 m/s on R = 3 m with mu_s = 0.1, v^2 = 25 but mu_s*R*g = 2.94, so the condition v^2 <= mu_s*R*g fails and the cyclist slips.
It is STATIC friction, as long as the tyres are rolling and not sliding. The point of the tyre touching the road is momentarily at rest relative to the road, so the friction that grips and turns the car is static, with a maximum of mu_s*m*g. Kinetic friction only takes over once the car is already skidding — and kinetic friction is weaker, which is why a skidding car is very hard to control. For NEET, use mu_s in v_max = sqrt(mu_s*R*g).
On a LEVEL road, only friction turns the car, giving v_max = sqrt(mu_s*R*g). On a BANKED road, the road is tilted, so the horizontal component of the normal reaction ALSO pushes toward the centre and helps friction. This lets a banked road allow a higher safe speed for the same radius. The level-road formula is the simpler special case (banking angle = 0). The banking derivation is the next concept.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
v_max = sqrt(mu_s * R * g), where mu_s is the coefficient of static friction, R is the radius of the turn, and g = 9.8 m/s^2 (or 10 m/s^2 in many NEET problems). It is derived by setting friction mu_s*m*g equal to the centripetal force m*v^2/R.
No. Mass appears on both sides of m*v^2/R = mu_s*m*g and cancels out, so v_max = sqrt(mu_s*R*g) is independent of mass. This is a favourite NEET trap.
Static friction between the tyres and the road provides the entire centripetal force. There is no banking, so friction alone points toward the centre.
Friction can supply only up to mu_s*m*g. If v is large enough that m*v^2/R exceeds this maximum, the inward force is not enough and the car slides outward along a tangent, i.e. it skids off the curve.
Yes. 18 km/h = 5 m/s so v^2 = 25 m^2/s^2, but mu_s*R*g = 0.1*3*9.8 = 2.94 m^2/s^2. Since 25 > 2.94, the safe condition v^2 <= mu_s*R*g fails and the cyclist slips (NCERT Example 4.10).
