Question

When the road is dry and coefficient of friction is $\mu$, the maximum speed of a car in a circular path is $10 m{s}^{-1}$. If the road becomes wet and the coefficient of friction becomes $\dfrac {\mu}{2}$, what is the maximum speed permitted?

Answer 1

Hint: To solve this problem, use the formula for maximum velocity in terms of coefficient of friction. Find the maximum speed of the car when the road is dry. Substitute the values in the formula. Then, find the maximum speed of the car when the road becomes wet. Substitute the values in the equation. Divide both these equations of maximum speed and find the maximum speed permitted when the road becomes wet.

Formula used:
$v= \sqrt {\mu gr}$

Complete step by step answer:
Given: Coefficient of friction when road is dry ${v}_{1}= \mu$
            Coefficient of friction when road is wet ${v}_{2}= \dfrac {\mu}{2}$
            Velocity ${v}_{1}=10 m{s}^{-1}$
Let the maximum speed of a car when road becomes wet is ${v}_{2}$
Maximum velocity is given by,
$v= \sqrt {\mu gr}$ …(1)
Where, $\mu$ is the coefficient of friction
             r is the radius of the path
When the road is dry, the velocity is given by,
${v}_{1}= \sqrt {\mu gr}$
Substituting the values in above equation we get,
$10= \sqrt {\mu gr}$ …(2)
When the road becomes wet, the velocity is given by,
${v}_{2}= \sqrt {\dfrac {\mu}{2} gr}$ …(3)
Dividing equation. (3) by equation. (2) we get,
$\dfrac {{v}_{2}}{10}= \dfrac {1}{\sqrt{2}}$
$\Rightarrow {v}_{2}= \dfrac {10}{\sqrt{2}}$
$\therefore {v}_{2}= 7.07$
Hence, when the road is wet, the maximum speed permitted is $7.07 m{s}^{-1}$.

Note:
From the above problem, it can be inferred that the coefficient of friction is directly proportional to the speed of the car. As the coefficient of friction decreases, the speed of the car also decreases. If the coefficient of friction increases then the speed of the car also increases.