Centripetal force

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

Fc=mac=mv2r=4π2mrT2=4π2mrf2\displaystyle F_{\text{c}} = m a_{\text{c}} = \frac{m v^{2}}{r} = \frac{4 \pi^{2} m r}{T^{2}} = 4 \pi^{2} m r f^{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 FT\displaystyle \vec{F}_{\text{T}} or Ff\displaystyle \vec{F}_{\text{f}} will be the centripetal force.

If the specific force isn’t supplying enough (for example, FT<Fc\displaystyle F_{\text{T}} < F_{\text{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 Fc\displaystyle \vec{F}_{\text{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, FT=macmg\displaystyle \left \lvert \vec{F}_{\text{T}} \right \rvert = m a_{\text{c}} - m g, and at the bottom, FT=mac+mg\displaystyle \left \lvert \vec{F}_{\text{T}} \right \rvert = m a_{\text{c}} + m g.


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:

Fnet=Ff=mac\displaystyle \vec{F}_{\text{net}} = \vec{F}_{\text{f}} = m \vec{a}_{\text{c}}.

We can substitute for friction and centripetal acceleration:

μmg=mv2r\displaystyle \mu m g{=} m \frac{v^{2}}{r}.

Now we can solve for speed:

v=rμg=(20.0m)(0.450)(9.80m/s2)=9.391m/s\displaystyle v = \sqrt{r \mu g} = \sqrt{\left ( 20.0 \, \text{m} \right ) \left ( 0.450 \right ) \left ( 9.80 \, \text{m/s}^{2} \right )} = 9.391 \, \text{m/s}.

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