Question

A car of mass M= 1300 kg traveling at 65.0 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. (Intro 1 figure) . Use g= 9.80 m/s^2 throughout this problem.

What is the radius (in meters) of the turn if = 20.0 $^\circ$ (assuming the car continues in uniform circular motion around the turn)?

Answer 1

Concepts and reason

The concepts to solve this question are Newton’s second law of motion, normal force, and centripetal acceleration.

First, draw a free body diagram describing the whole system. Then, find the radius of turn by balancing the forces in the vertical direction and by finding the net force acting in the horizontal direction.

Fundamentals

Newton’s second law of motion states that the net force acting on an object is directly proportional to the product of its mass and acceleration.

According to Newton’s second law, net force acting on an object is,

$F = ma$

Here, m is the mass of the object and a is the acceleration of the object.

Weight of an object is the force on an object due to gravity. Force acting on an object due to its weight is,

${F_{\rm{W}}} = mg$

Here, m is the mass of the object and g is acceleration due to gravity.

Centripetal acceleration is the acceleration that causes an object to move along a circular path and corresponding force is the centripetal force. The expression for centripetal acceleration is:

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

Here, v is the velocity of the object and r is the radius of the circular path.

The figure 1 shows a car travelling on a banked road, banked at an angle $\theta$ . A centripetal force is acting on the car towards the center of the circular path. The car is moving with centripetal acceleration ${a_{\rm{c}}}$ . N is the normal force acting on the car by the road and mg is the force acting on the car due to gravity.

According to Newton’s second law, the net force acting on the car in the horizontal direction is,

$F = m{a_{\rm{c}}}$

Here, m is the mass of the car and ${a_{\rm{c}}}$ is the centripetal acceleration.

The force acting on the car in the horizontal direction is,

$F = N\sin \theta$

Here, N is the normal force acting on the car and $\theta$ is the banking angle.

Substitute $N\sin \theta$ for F and $\frac{{{v^2}}}{r}$ for a in equation $F = m{a_{\rm{c}}}$ .

$N\sin \theta = \frac{{m{v^2}}}{r}$

Here, v is the velocity of the car and r is the radius of turn.

Refer figure 1 and balance the forces acting on the car in the vertical direction.

$N\cos \theta = mg$

Here, N is the normal force acting on the car, m is the mass of the car, g is acceleration due to gravity, and $\theta$ is the banking angle.

Divide equation $N\sin \theta = \frac{{m{v^2}}}{r}$ by equation $N\cos \theta = mg$ and solve for r.

$\begin{array}{c}\\\frac{{N\sin \theta }}{{N\cos \theta }} = \frac{{m{v^2}}}{{rmg}}\\\\\tan \theta = \frac{{{v^2}}}{{rg}}\\\\r = \frac{{{v^2}}}{{g\tan \theta }}\\\end{array}$

Substitute 65.0 km/h for v, $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g, and ${20.0^{\rm{o}}}$ for $\theta$ in equation $r = \frac{{{v^2}}}{{g\tan \theta }}$ and determine the radius of turn.

$\begin{array}{c}\\r = \frac{{{{\left( {65\,{\rm{km/h}}} \right)}^2}}}{{\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\tan \left( {{{20}^{\rm{o}}}} \right)}}{\left( {\frac{{{{10}^3}{\rm{ m}}}}{{1{\rm{ km}}}}} \right)^2}{\left( {\frac{{1{\rm{ h}}}}{{3600{\rm{ s}}}}} \right)^2}\\\\ = 91.4{\rm{ m}}\\\end{array}$

Ans:

The radius of turn is 91.4 m.