Question

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

Answer 1

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

$\frac{mv^2}{r}=\mu mg\\ \\ \frac{v^2}{100}=0.2*9.8\\ \\ v^2=100*0.2*9.8=196\\ \\v= \sqrt{196}=\boxed{14\ m/s}$

Answer 2

Answer:

v = 20 (2) ^ 1/2

Explanation:

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

v = ( 0.2 x 10 x 100 ) ^ 1/2

v = (0.2 x 1000) ^ 1/2

v = (200) ^ 1/2

v = 20 (2) ^ 1/2