Centripetal force follows directly from our equations for centripetal acceleration that we saw in uniform circular motion:

`F_"c" = ma_"c" = (mv^2)/r = (4pi^2mr)/T^2 = 4pi^2mrf^2`.

Centripetal force causes an object to follow a circular path. That being said, it is not another *type* of force that you can add in the net force equation. Some specific force like `vec F_"T"` or `vec F_"f"` will *be* the centripetal force.

If the specific force isn’t supplying enough (for example, `F_"T" < F_"c"`), then the object will not follow the circular path: it will go off on a tangent. If the original centripetal force is now zero (e.g., the string breaks), then it will travel in a straight line. If it was too small but stays that way (e.g., car turning too fast), then it will follow a circle with a wider radius.

The centripetal force does not have to be provided by a single force. For example, when you spin a ball on a string in a vertical circle, there are two forces on the ball at all times: tension and gravity. They will always sum to `vec F_"c"` if the ball moves in a perfect circle. Since gravity doesn’t change, the tension force must change. At the top of the circle, `|vec F_"T"| = ma_"c" - mg`, and at the bottom, `|vec F_"T"| = ma_"c" + mg`.

A 1105 kg car is entering a level turn with a radius of 20.0 m at 21.5 m/s. If the coefficient of friction is 0.450, what is the maximum safe speed?

There is only one force acting on the car, the force of friction, and it needs to provide a centripetal force for the car to be safe:

`vec F_"net" = vec F_"f" = mvec a_"c"`.

We can substitute for friction and centripetal acceleration:

`mumg = mv^2/r`.

Now we can solve for speed:

`v = sqrt(rmug) = sqrt((20.0\ "m")(0.450)(9.80\ "m/s"^2)) = 9.391\ "m/s"`.

Therefore, the maximum safe speed is 9.39 m/s.