Centripetal force

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

$F_{\text{c}}=m{a}_{\text{c}}=\frac{m{v}^2}{r}=\frac{4{\pi }^{2}mr}{T^2}=4{\pi }^{2}mr{f}^2\text{.}$

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 ${\overrightarrow{F}}_{\text{T}}$ or ${\overrightarrow{F}}_{\text{f}}$ will be the centripetal force.

If the specific force isn’t supplying enough (for example, $F_{\text{T}}<F_{\text{c}}$), 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 ${\overrightarrow{F}}_{\text{c}}$ if the ball moves in a perfect circle. Since gravity doesn’t change, the tension force must change. At the top of the circle, $\left|{\overrightarrow{F}}_{\text{T}}\right|=m{a}_{\text{c}}-mg\text{,}$ and at the bottom, $\left|{\overrightarrow{F}}_{\text{T}}\right|=m{a}_{\text{c}}+mg\text{.}$

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:

${\overrightarrow{F}}_{\text{net}}={\overrightarrow{F}}_{\text{f}}=m{\overrightarrow{a}}_{\text{c}}\text{.}$

We can substitute for friction and centripetal acceleration:

$\mu mg=m\frac{v^2}{r}\text{.}$

Now we can solve for speed:

$v=\sqrt{r\mu g}=\sqrt{(20.0\text{m})(0.450)(9.80{\text{m/s}}^2)}=9.391\text{m/s}\text{.}$

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