Question

A flat curve on a highway has a radius of curvature 400 m. A car goes around a curve at a speed $32\dfrac{m}{s}$. What is the minimum value of coefficient of friction that will prevent the car from sliding $\left( {g = 9 \cdot 8\dfrac{m}{{{s^2}}}} \right)$:

Answer 1

Friction is the resistance to the motion of anybody. Friction of the force is only responsible for holding the vehicles on the ground while they take sharp turns but upto a certain level of velocity of the vehicle only, if the speed of the vehicle is increased beyond the upper limit then friction would also not be able to stop the vehicle from skidding.

Formula used: The friction force on the vehicle of mass ‘m’ is given by,
${f_s} = \mu \cdot mg$
Where ${f_s}$ is the friction $\mu$ is the coefficient of friction m is the mass of the vehicle and g is the acceleration due to gravity.
The formula of the centripetal force is given by,
${F_c} = \dfrac{{m{v^2}}}{r}$
Where ${F_c}$ is the centripetal force $v$is the velocity of the vehicle and r is the radius of the turn.

Complete step-by-step answer:
It is given in the problem that a car is taking a flat curve on the road and the speed of the car is $32\dfrac{m}{s}$ and the radius of the turn is 400 ms if the acceleration due to gravity is $g = 9 \cdot 8\dfrac{m}{{{s^2}}}$ then we need to find the value of minimum coefficient of friction.
The friction will provide the necessary centripetal force therefore we get,
$\Rightarrow {f_s} = {F_c}$
Where ${f_s}$ is friction force and ${F_c}$ is the centripetal force.
Now replace the formula of ${f_s}$ and ${F_c}$ in the above relation.
$\Rightarrow {f_s} = {F_c}$
$\Rightarrow \mu \cdot mg = \dfrac{{m{v^2}}}{r}$
$\Rightarrow \mu \cdot g = \dfrac{{{v^2}}}{r}$
$\Rightarrow \mu = \dfrac{{{v^2}}}{{r \cdot g}}$
Replace the value of velocity, radius of the track and acceleration due to gravity.
$\Rightarrow \mu = \dfrac{{{v^2}}}{{r \cdot g}}$
$\Rightarrow \mu = \dfrac{{{{\left( {32} \right)}^2}}}{{\left( {400} \right) \cdot \left( {9 \cdot 8} \right)}}$
$\Rightarrow \mu = 0 \cdot 261$
The minimum coefficient of friction is $\mu = 0 \cdot 261$.

Note: Centripetal force is always acting towards the centre of the circular radius, it is the centripetal force which helps anybody to stay and continue in the circular track .The friction provides the necessary centripetal force to the vehicle which helps the vehicle to continue in the circular turn.