What is the centripetal force and speed of?A car drives around a circle with a radius of 100 m on a road that is banked (at an angle) to prevent skidding if the road is icy (no static friction.)...

What is the centripetal force and speed of?

A car drives around a circle with a radius of 100 m on a road that is banked (at an angle) to prevent skidding if the road is icy (no static friction.) If the angle is 20 degrees and the car has a mass of 1500 kg, what is the centripetal force on the car if there is an icy road? What speed can the car go and stay on the track?

Asked on by sammy24

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

When the car is negotiating the curve, the car has a centripetal force directed towards the centre of the curvature. The radius of curvature is given to be 100m. The centripetal force if give by Fc = mv^2/r, where m, is the mass of the car (1500kg) and v is the speed of the car.

To balance the centripetal force, the normal reaction at the place acts in an opposing direction which is equal to mgcos x, where x is the angle of banking of the road. .

Since entire system is in equilibrium and no friction is there, the equation of forces is: (mv^2)/r = mgcosx.

Given m =1500 kg,  r = 100meter, x = 20 degrees and angle of slope (or bankig ) x = 20 degrees and g = 9.81m/s^2 we get:

At the maximum velocity 1500*v^2/100 = 1500*9.81 cos20.

v^2 = (9.81cos20 deg)100.

v= sqrt{(9.81cos20 deg)100}.

v = 30.3618 meter/second.

Therefore the maximum velocity the car can go is 30.3618 m/s = 30.3618*60*60.1000 km/hr  = 109.3km/h.

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