Question

What is the maximum speed with which a 1200-kg car can round a turn of radius 93.0 m on a flat road if the coefficient of static friction between tires and road is 0.60

Answer 1

To solve this problem, apply the concepts related to the centripetal force expressed as the product between the mass and the speed squared over the radius of gyration, and the concept of the friction force equivalent to the product between the coefficient of friction, the mass of the body and acceleration due to gravity, for the problem we equate the centripetal force with the frictional force we have

$F_c = F_f$

$\frac{mv^2}{r} = \mu mg$

Rearrange the expression to get the value of speed

$v = \sqrt{r\mu g}$

Replacing with our values we have,

$v = \sqrt{(93m)(0.60)(9.8m/s^2)}$

$v = 23.38m/s$

Therefore the velocity is 23.38m/s and we can conclude also that the maximum speed of the car is independent of the mass of the car.

Answer 2

Answer:

The maximum speed is 23.39m/s

Explanation:

Centripetal force Fc is the force acting on a body moving in a circular path and directed towards the centre of the path.

If F = ma

centripetal acceleration a = v²/r

Centripetal force Fc = mv²/r ... (1)

where;

m is the mass of the object

r is the radius.

Since there is friction between the tyre and the road, then the frictional force Ff acts between the surface and this frictional force is the one that tends to oppose the moving force (centripetal force)

Ff = μsR where

μs is the coefficient of static friction

R is the normal reaction which is also equivalent to the weight of the car i.e R = W = mg

Ff = μsmg ... (2)

For the body to be static, the centripetal force must be equal to the frictional force i.e Fc = Ff

mv²/r = μsmg

Making v the subject of the formula;

v²/r = μsg

v² = μsgr

v = √μsgr

Given the following data;

μs = 0.6

g = 9.81m/s²

r = 93.0m

v = √0.6×9.81×93

v = √547.398

v = 23.39m/s