Question

On an icy winter day, the coefficient of friction between the tires of a car and a roadway is reduced to 1/4 its value on a dry day. As a result, the maximum speed vmax dry at which the car can safely negotiate a curve of radius R is reduced. The new value for this speed is what percentage of its value on a dry day?

Answer 1

Answer:

50%

Explanation:

For the particular case we assume that the centripetal force is equal to the frictional force. Starting from this, we realize that while the car takes a curve, which allows it to not leave the track, it is the friction force that adheres it to the floor, in this way,

$F_c = F_f$

The values of these equations are given by,

$F_c = \frac{mv^2}{R}$

$F_f = \mu N = \mu mg$

Equating the terms

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

$v = \sqrt{\mu Rg}$

Both gravity and radius are values that during the trajectory with constants, so

$v \approx \sqrt{\mu}$

The difference comes in winter and dry, so,

\mu_{dry} = \mu

as in winter it is a quarter, you have to,

\mu_{winter} = \frac{1}{4} \mu_{dry} = \frac{1}{4} \mu

Performing the proportion we have to

$\frac{\mu_{winter}}{\mu_{dry}} = \frac{\sqrt{\mu/4}}{\mu}$

$\frac{\mu_winter}{\mu_{dry}} = \frac{1}{2} = 0.5$

We can conclude that the new value for this speed is 50% of its value on a dry day.