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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 ang...

uploaded imageA 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 theta, 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 r (in meters) of the turn if theta = 20.0^\circ (assuming the car continues in uniform circular motion around the turn)?

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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=maF = 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,

FW=mg{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:

ac=v2r{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 θ\theta . A centripetal force is acting on the car towards the center of the circular path. The car is moving with centripetal acceleration ac{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.

N cose
sin
a
mg
Figure 1: Free body diagram

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

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

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

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

F=NsinθF = N\sin \theta

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

Substitute NsinθN\sin \theta for F and v2r\frac{{{v^2}}}{r} for a in equation F=macF = m{a_{\rm{c}}} .

Nsinθ=mv2rN\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.

Ncosθ=mgN\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 θ\theta is the banking angle.

Divide equation Nsinθ=mv2rN\sin \theta = \frac{{m{v^2}}}{r} by equation Ncosθ=mgN\cos \theta = mg and solve for r.

NsinθNcosθ=mv2rmgtanθ=v2rgr=v2gtanθ\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, 9.8m/s29.8{\rm{ m/}}{{\rm{s}}^2} for g, and 20.0o{20.0^{\rm{o}}} for θ\theta in equation r=v2gtanθr = \frac{{{v^2}}}{{g\tan \theta }} and determine the radius of turn.

r=(65km/h)2(9.8m/s2)tan(20o)(103m1km)2(1h3600s)2=91.4m\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.

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