## Centripetal force

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.

### Example

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.