# 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?

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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.