Question

A cyclist is rounding a 20m radius curve at 12 m/s. What is the minimum possible coefficient of static friction between the bike tires and the ground?

Answer 1

Concepts and reason

The concepts required to solve this problem are Newton’s second law and centripetal acceleration.

First use the Newton’s second law to solve for the expression of minimum coefficient of static friction required.

Finally, substitute the values the in the coefficient equation to calculate the minimum possible coefficient of static friction.

Fundamentals

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

$\sum {F = ma}$

Here, $\sum F$ is the sum of all the forces on the object, $m$ is mass of the object, and $a$ is the acceleration of the object.

The expression of static friction over a horizontal surface is,

$f \le {\mu _{\rm{s}}}mg$

Here, ${\mu _{\rm{s}}}$ is the coefficient of static friction, $m$ is mass of the object, and $g$ is the acceleration due to gravity.

The centripetal acceleration is the rate of change of the velocity in tangential direction. The magnitude of acceleration is constant for a uniform circular motion. It is always towards the center along the radius vector of any circular motion. The centripetal acceleration is expressed as,

${a_{\rm{c}}} = \frac{{{v^2}}}{r}$

Here, $v$ is the constant speed, and $r$ is the perpendicular distance from the object to center of rotation.

Use the Newton’s second law for the bike.

Substitute $f$ for $\sum F$ in the equation $\sum {F = ma}$ .

$f = ma$

Substitute ${\mu _{\rm{s}}}mg$ for $f$ in the equation $f = ma$ .

$\begin{array}{c}\\{\mu _{\rm{s}}}mg = m\frac{{{v^2}}}{r}\\\\{\mu _{\rm{s}}} = \frac{{{v^2}}}{{rg}}\\\end{array}$

Use the equation of coefficient of static friction.

Substitute $12{\rm{ m/s}}$ for $v$ , $20{\rm{ m}}$ for $r$ , and $9.81{\rm{ m/}}{{\rm{s}}^2}$ for $g$ in the equation ${\mu _{\rm{s}}} = \frac{{{v^2}}}{{rg}}$ and calculate the ${\mu _{\rm{s}}}$ .

$\begin{array}{c}\\{\mu _{\rm{s}}} = \frac{{{{\left( {{\rm{12 m/s}}} \right)}^2}}}{{\left( {20{\rm{ m}}} \right)\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)}}\\\\ = 0.735\\\end{array}$

Ans:

The minimum coefficient of static friction between the bike tires and the ground is $0.735$ .