Question

A car of mass M = 800 kg traveling at 55.0 km/hour enters a banked turn covered with ice. The road is banked at an angle ?, and there is no friction between the road and the car's tires as shown in(Figure 1) . Use g = 9.80 m/s2 throughout this problem.

Now, suppose that the curve is level (?=0) and that the ice has melted, so that there is a coefficient of static friction ? between the road and the car's tires as shown in (Figure 2) . What is ?min, the minimum value of the coefficient of static friction between the tires and the road required to prevent the car from slipping? Assume that the car's speed is still 55.0 km/hour and that the radius of the curve is 65.4 m .

Answer 1

Concepts and reason

The concept behind this question is a frictional force and centripetal force.

Initially, when an object is rotating in a circular motion then the two types of force acts on the object first one is a centripetal force and the second one is centrifugal force. Centripetal force is proportional to the velocity and the radius of the circular field.

Then, find the coefficient of friction by using the friction formula.

Fundamentals

The expression of the Centripetal force is express as follows,

${\vec F_c} = {\rm{ }}\frac{{m{v^2}}}{r}$

Here,${F_c}$is the centripetal force,$m$is the mass of the object,$v$is the circular velocity and $r$is the radius of the circle.

The expression of the frictional force is equal to,

${\vec F_{fr}} = {\rm{ }}{\vec F_N}\mu$

Here,${\vec F_{fr}}$is the frictional force,${\vec F_N}$is the normal force and $\mu$is the friction coefficient.

The expression of the normal force is equal to,

${\vec F_N} = mg$

Here,$g$is the acceleration due to gravity.

Calculate the centripetal acceleration.

The expression for the centripetal acceleration is equal to,

${a_c} = \frac{{{v^2}}}{r}$

Here,${a_c}$is the centripetal acceleration.

Substitute $55{\rm{ km/h}}$for$v$ and$65.4{\rm{ m}}$for$r$in the above expression of the acceleration,

$\begin{array}{c}\\{a_c} = \frac{{{{\left( {55{\rm{ km/h}}} \right)}^2}}}{{\left( {65.4{\rm{ m}}} \right)}}\\\\ = \frac{{{{\left( {\left( {55{\rm{ km/h}}} \right)\left( {\frac{{1000{\rm{ m}}}}{{3600{\rm{ s}}}}} \right)\left( {\frac{{\rm{1}}}{{{\rm{km/h}}}}} \right)} \right)}^2}}}{{\left( {65.4{\rm{ m}}} \right)}}\\\\ = 3.568{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

Calculate the coefficient of friction.

The expression of the friction force is equal to,

${F_{fr}} = mg\mu$

The expression of the centripetal force is equal to,

${F_c} = \frac{{m{v^2}}}{r}$

Compare the above force equation,

${F_{fr}} = {F_c}$

Substitute$mg\mu$for${F_{fr}}$and$\frac{{m{v^2}}}{r}$for${F_c}$ in the above expression,

$\begin{array}{c}\\mg\mu = \frac{{m{v^2}}}{r}\\\\\mu = \frac{{{v^2}}}{{gr}}\\\end{array}$

Substitute ${a_c}$for$\frac{{{v^2}}}{r}$in the above expression of the friction coefficient.

$\mu = \frac{{{a_c}}}{g}$

Substitute $3.568{\rm{ m/}}{{\rm{s}}^2}$for${a_c}$and $9.8{\rm{ m/}}{{\rm{s}}^2}$for$g$in the above expression of$\mu$,

$\begin{array}{c}\\\mu = \frac{{3.568{\rm{ m/}}{{\rm{s}}^2}}}{{9.8{\rm{ m/}}{{\rm{s}}^2}}}\\\\ = 0.364\\\end{array}$

Ans:

The coefficient of kinematic friction is equal to$0.364$.