Question

An unbalanced curve has a radius of 60m. The maximum speed at which a car can make a turn if the coefficient of static friction is 0.75 is:
(A) $2.1m{s^{ - 1}}$
(B) $14m{s^{ - 1}}$
(C) $21m{s^{ - 1}}$
(D) $7m{s^{ - 1}}$

Answer 1

Hint:The most important thing to have in mind while solving numerical problems involving friction is that static friction is inversely proportional to radius of the curve and directly proportional to the square of the maximum velocity.

Formula Used:The maximum velocity is related to coefficient of static friction by the following expression:
$\mu = \dfrac{{{V^2}}}{{rg}}$ Here, $\mu $is the coefficient of static friction, $r$ is the radius and $g$ is acceleration due to gravity.

Complete step by step answer:
In the given question, the coefficient of static friction is 0.75 and the radius of the unbanked curve is$60m$. Thus, putting the values in the above equation we get:
\[
  {V^2} = \mu \times rg \\
   \Rightarrow {V^2} = 0.75 \times 60 \times 9.8 \\
 \]
To obtain the value of V we have to find the square root of the terms in the RHS of the equation. Thus,
$V = \sqrt {60 \times 9.8 \times 0.75} = 21m/s$
Hence, the maximum value for velocity that the car can have while taking a turn from the unbanked curve is$21m{s^{ - 1}}$. Thus, option (C) is the correct answer to the above numerical problem.

Additional Information:Coefficient of friction is a dimensionless quantity which is defined as the ratio between normal force and frictional force. Coefficient of friction is the inherent property of a material. A low value of coefficient of friction indicates that the force required for sliding to occur is less. Materials that have coefficient of friction smaller than 0.1 are considered to be lubricous materials.

Note:Often students tend to misinterpret the value of coefficient of friction for the value of $\theta $ as$\mu = \tan \theta $. This blunder can be avoided by seeing that $\theta $has a dimension while $\mu $is a dimensionless quantity.