Question

A car is moving on a horizontal circular track of radius 0.2km with a constant speed.if coefficient of friction between tyres of car and road is 0.45,then speed of car may be (g=10m/s^2)

Answer 1

Lets derive the formula to use :

mv²/r = ωmgωMg... 1

Where :

m = mass

v = maximum velocity

r = radius of the path

g = acceleration due to gravity = 10

ω = coefficient of friction.

Deriving for v we have :

Dividing both sides of 1 by m we have %

v²/r = ωg

v² = rωg

v = √rωg

Doing the substitution :

r = 0.2 × 1000 = 200m

ω = 0.45

g = 10

v = √(10 × 0.45 × 200)

v = √900

v = 30 m/s

= 30 m/s