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I am a high school student trying to get an intuitive understanding of centripetal and centrifugal forces. I can calculate the correct solution but I don't understand the meaning of it.

In class today we had the following problem: Given a car weighing 1.5 tons at a speed of 20 m/s compute the centripetal force as multiples of the gravitational force when the car is going through a curve with radius 80 meters and a friction coefficient $\mu = 0.8$. The curve is level (not banked)

If I understand correctly what prevents the car from sliding out of the curve is the frictional force which is the centripetal force in this case. It is computed as: $$F_{friction} = F_{centripetal} = \mu * F_{Gravity}$$ The condition for the frictional and centripetal forces is $$F_{friction} \geq F_{centripetal}$$ So I computed the centripetal force and the friction force using these formulas: $$\mu * g * m \geq m * \frac{v^2}{r}$$ This means that the centripetal force can only be as big as the frictional force correct ? As a result I get $$11.77kN \geq 7.5kN$$ But what does this result tell me ? If the left hand side is greater does that mean the car will not slide ? I can understand that if the centrifugal force is bigger than the frictional force the car will slide off. But what about the centripetal force ? Can someone please explain the meaning of this inequality to me ? The coefficient $F_{gravity} / F_{centripetal}$ is 1,96. What does this tell me ?

Thanks in advance !

IAmTony
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With the car not slipping on a level surface, the maximum frictional force is $\mu N$ were $N = mg$, and $\mu$ is the coefficient of static friction. The frictional force can be less than this maximum; for example, if the car is stationary the frictional force is zero. So the frictional force varies as needed up to the maximum to keep the car from slipping, and for circular motion the frictional force provides the centripetal force and has value $mv^2/r$.

The centripetal force is not a basic force such as friction, gravity, or tension in a string, it is the name given to a basic force when that force maintains circular motion at constant velocity. Here the force of friction is the centripetal force. See answers to Centripetal Force Terminology on this exchange.

If the required centripetal force is greater than the maximum frictional force (velocity too great or radius too small), the car will slip and in this case the coefficient of friction changes to the coefficient of kinetic friction which is less than the coefficient of static friction.

John Darby
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If I understand correctly what prevents the car from sliding out of the curve is the frictional force which is the centripetal force in this case.

Sometimes it can be easier to think about this in the ground frame rather than the car frame. Instead of "sliding out" of the curve, think of the car as normally going straight, but you expect the wheels to force the car onto the curving path.

If you hit a slick patch, the wheels won't grip enough and the car simply doesn't turn as fast as you want.

If the left hand side is greater does that mean the car will not slide ?

Yes. As the wheels are holding the car via static friction, you haven't computed the frictional force, you've computed the maximum frictional force. Friction may actually be less than that. If you turn the wheel slightly, you create small forces. If you turn the wheel excessively at high speed, then at some point the maximum force is exceeded and the wheel skids.

You have computed the maximum possible centripetal force. If you want to take a curve that requires a greater force than that, the car will skid and not follow the intended path.

BowlOfRed
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