A car moving around a flat circular path needs an inward net force. This inward force is the centripetal force.
On a flat, unbanked road, the vertical forces are the car’s weight and the normal force from the road. These two
vertical forces balance, so they do not provide the horizontal inward force. The inward force must come from static
friction between the tyres and the road.
Centripetal force requirement
For circular motion at speed \(v\) on a path of radius \(r\), the required inward centripetal force is:
Centripetal force.
\[
\begin{aligned}
F_c &= \frac{mv^2}{r}
\end{aligned}
\]
This force points toward the center of the circular path. If the required inward force is larger than the available static
friction, the tyres cannot keep the car on the circular path and the car skids outward relative to the curve.
Maximum static friction
On a flat road, the normal force is equal in magnitude to the car’s weight:
Vertical balance.
\[
\begin{aligned}
N &= mg
\end{aligned}
\]
The maximum possible static friction before skidding is:
Static friction limit.
\[
\begin{aligned}
f_{s,\max} &= \mu_s N \\
&= \mu_s mg
\end{aligned}
\]
At the maximum safe speed, the required centripetal force exactly equals the maximum available static friction:
No-skid limit.
\[
\begin{aligned}
\frac{mv_{\max}^2}{r} &= \mu_smg
\end{aligned}
\]
The mass appears on both sides, so it cancels:
Maximum speed formula.
\[
\begin{aligned}
v_{\max}^2 &= \mu_sgr \\
v_{\max} &= \sqrt{\mu_sgr}
\end{aligned}
\]
Why mass does not affect the speed limit
In this simplified flat-road model, the maximum speed does not depend on the car’s mass. A heavier car requires a larger
centripetal force, but it also has a larger normal force, which increases the maximum available static friction by the same
factor. Therefore, mass cancels from the speed formula.
Mass is still useful if you want the actual force value. Once \(m\) is known, the maximum static friction force is:
Maximum friction force.
\[
\begin{aligned}
f_{s,\max} &= \mu_smg
\end{aligned}
\]
At the no-skid limit, this is also the required centripetal force:
Force equality at the limit.
\[
\begin{aligned}
F_c &= f_{s,\max}
\end{aligned}
\]
Lateral acceleration
The centripetal acceleration at the friction limit is:
Lateral acceleration limit.
\[
\begin{aligned}
a_c &= \frac{v_{\max}^2}{r} \\
&= \mu_sg
\end{aligned}
\]
This means the maximum lateral acceleration is directly proportional to the friction coefficient. A dry road with large
\(\mu_s\) allows a higher lateral acceleration than a wet or icy road. Since \(v_{\max}\) depends on the square root of
\(\mu_s\), reducing friction by a factor of four reduces the maximum speed by a factor of two.
Assumptions of the flat-curve model
This calculator assumes the road is flat and unbanked. It also assumes static friction is the only horizontal force providing
centripetal force. The model does not include banking, aerodynamic downforce, tyre load sensitivity, suspension effects, road
slope, rolling resistance, or changes in \(\mu_s\) with temperature and surface condition. For introductory physics, however,
the formula \(v_{\max}=\sqrt{\mu_sgr}\) captures the core relationship among grip, gravity, radius, and safe turning speed.
| Quantity |
Formula |
Meaning |
| Centripetal force |
\(F_c=mv^2/r\) |
Required inward net force for circular motion |
| Maximum static friction |
\(f_{s,\max}=\mu_smg\) |
Largest available inward friction force on a flat road |
| Maximum safe speed |
\(v_{\max}=\sqrt{\mu_sgr}\) |
Largest speed before the no-skid limit is exceeded |
| Lateral acceleration limit |
\(a_c=\mu_sg\) |
Maximum centripetal acceleration at the friction limit |
The force called “centripetal force” is not a new kind of force. In this flat-turn case, centripetal force is supplied by
static friction. In other circular-motion problems, it may be supplied by tension, gravity, a normal-force component, or a
combination of forces.
