# 2 Problems on Motion on circular Path: Centripetal force

1. A 1000 kg car rounds a turn of radius 30m at a velocity of 9 m/s.

(a) How much centripetal force is required?

(b) Where does this force come from?

2. The maximum force a road can exert on the tires of a 3200 lb car is

2000 lb. What is the maximum velocity at which the car can round a turn of radius 320 ft?

https://brainmass.com/physics/circular-motion/problems-motion-circular-path-centripetal-force-201178

## SOLUTION This solution is **FREE** courtesy of BrainMass!

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1. A 1000 kg car rounds a turn of radius 30m at a velocity of 9 m/s.

(a) How much centripetal force is required?

The centripetal force acting on a body of mass m moving on a circular path of radius R with a constant speed v is given by

Substituting values in this formula the magnitude of required centripetal force on the body is given by

And its direction is always changing and remains towards the center of the circular path.

(b) Where does this force come from?

The force is applied by the friction between the track and the tires of the car.

2. The maximum force a road can exert on the tires of a 3200 lb car is

2000 lb. What is the maximum velocity at which the car can round a turn of radius 320 ft?

The maximum possible frictional force is given by the limiting friction which is acting as centripetal force in case of maximum possible velocity as

F = 2000 lb.

The weight of the body is

W = mg = 3200 lb

'g' is the acceleration due to gravity whose value is 32 f/s2 (FPS system of units)

Hence the maximum possible velocity is given by the formula of centripetal force as

Or

Or v2 = 6400

Or v = 80 f/s

Hence the maximum possible speed of the car is 80 f/s

https://brainmass.com/physics/circular-motion/problems-motion-circular-path-centripetal-force-201178