Find the maximum velocity for the overturn of a car moving on a circular track of radius 100 m. The co-efficient of friction between the road and tyre is 0.2

F=μR = mv²/r
R=mg
μmg=mv²/r
μg=v²/r
0.2 × 9.8 = v²/100
v² = 196
v = √196
v = 14ms⁻¹

Centrifugal force F= MV²/r
Friction Fr=UR=UMg
(R=Mg)
F=Fr
MV²/r=UMg
(M is common so it will cancel out)
V²/r=Ug
g=10m/s² or 9.8m/s²
V²/100=0.2×9.8
Cross multiply
V²=100×0.2×9.8
V²=196
V=√196
V=14m/s
V²/100=0.2×9.8

MV²/R=fMg
V²/R=fg
V²=fgR=0.2 × 9.81 × 100 = 196
V²=196
V²=14²
V=14

Bhai how it came? 14m/s? Intuition ?

CF-CENTRIFUGAL FORCE=(MV^2/R)
FF-FRICTIONAL FORCE=(COEFFICENT OF FRICTION(COF) X MASS X ACCELARATION DUE TO GRAVITY)
AS CF=FF
V=SQRT(COF X R X 9.81)

Please show the solution of this question

For a car moving in a circular track, the necessary centripetal force is provided by the friction.
Let the mass of car = m
maximum velocity for over turn = v
radius of track = 100m
coefficient of friction = 0.2

rac{mv^2}{r}=mu mg rac{v^2}{100}=0.2*9.8 v^2=100*0.2*9.8=196 v= sqrt{196}=oxed{14 m/s}

Relavent formula