A car on a banked curve is governed by two related ideas: mechanical energy controls how fast the car enters
the curve, while centripetal force determines whether the bank angle and friction can support that speed.
1. Mechanical energy before the curve
If the car changes height before entering the banked curve, the energy balance is:
\[
K_1+U_1+W_{\mathrm{nc}}=K_2+U_2
\]
with
\[
K=\frac12mv^2,
\qquad
U=mgh.
\]
Therefore, if the curve-entry height \(h_2\) is known:
\[
\frac12mv_2^2
=
\frac12mv_1^2+mg(h_1-h_2)+W_{\mathrm{nc}}
\]
\[
v_2
=
\sqrt{
v_1^2+2g(h_1-h_2)+\frac{2W_{\mathrm{nc}}}{m}
}.
\]
Positive \(W_{\mathrm{nc}}\) means engines or external work add energy. Negative \(W_{\mathrm{nc}}\) means
friction, braking, or drag remove energy.
2. Frictionless banked curve
On a frictionless banked curve, the horizontal component of the normal force provides the centripetal force.
The vertical component balances weight:
\[
N\cos\theta=mg
\]
\[
N\sin\theta=\frac{mv^2}{r}.
\]
Dividing the second equation by the first gives:
\[
\tan\theta=\frac{v^2}{rg}
\]
\[
v_0=\sqrt{rg\tan\theta}.
\]
This \(v_0\) is the design speed: no static friction is required when the car travels at this speed.
3. If the car is faster or slower than the design speed
If the car travels faster than \(v_0\), it tends to slide up the bank and outward. Static friction acts down the bank.
If the car travels slower than \(v_0\), it tends to slide down the bank and inward. Static friction acts up the bank.
| Condition |
Tendency |
Static friction direction |
| \(v=v_0\) |
No sliding tendency |
No friction needed |
| \(v>v_0\) |
Up the bank / outward |
Down the bank |
\(v
| Down the bank / inward |
Up the bank |
|
4. Required friction coefficient at a given speed
Let
\[
q=\frac{v^2}{rg}.
\]
The minimum coefficient of static friction required to hold the car on the banked curve is:
\[
\mu_{\mathrm{req}}
=
\frac{|q\cos\theta-\sin\theta|}
{q\sin\theta+\cos\theta}.
\]
If \(\mu_{\mathrm{req}}\le\mu_s\), the car can take the curve without slipping. If
\(\mu_{\mathrm{req}}>\mu_s\), the available static friction is not enough.
5. Safe speed range with static friction
If the available coefficient of static friction is \(\mu_s\), the minimum and maximum speeds are:
\[
v_{\min}
=
\sqrt{
rg\,
\frac{\sin\theta-\mu_s\cos\theta}
{\cos\theta+\mu_s\sin\theta}
}
\]
when \(\sin\theta-\mu_s\cos\theta>0\). Otherwise, \(v_{\min}=0\).
\[
v_{\max}
=
\sqrt{
rg\,
\frac{\sin\theta+\mu_s\cos\theta}
{\cos\theta-\mu_s\sin\theta}
}.
\]
If the denominator \(\cos\theta-\mu_s\sin\theta\) is zero or negative, the simplified model gives no finite
upper speed limit from static friction alone.
Worked example
A car travels on a banked curve with \(r=90\ \mathrm{m}\), \(\theta=12^\circ\), and speed
\(v=18\ \mathrm{m/s}\). Let \(g=9.81\ \mathrm{m/s^2}\).
Step 1: design speed.
\[
v_0=\sqrt{rg\tan\theta}
=
\sqrt{(90)(9.81)\tan(12^\circ)}
\approx 13.7\ \mathrm{m/s}.
\]
The car is faster than the no-friction design speed, so static friction must act down the bank.
Step 2: required centripetal acceleration.
\[
a_c=\frac{v^2}{r}
=
\frac{18^2}{90}
=
3.60\ \mathrm{m/s^2}.
\]
Step 3: required friction coefficient.
\[
q=\frac{v^2}{rg}
=
\frac{18^2}{(90)(9.81)}
\approx 0.367.
\]
\[
\mu_{\mathrm{req}}
=
\frac{|q\cos\theta-\sin\theta|}
{q\sin\theta+\cos\theta}
\approx 0.147.
\]
If the road-tire friction coefficient is greater than about \(0.147\), the car can take the curve at
\(18\ \mathrm{m/s}\) without slipping.
Formula summary
| Concept |
Formula |
Meaning |
| Kinetic energy |
\(K=\frac12mv^2\) |
Energy due to car speed. |
| Gravitational potential energy |
\(U=mgh\) |
Energy due to road height. |
| Energy balance |
\(K_1+U_1+W_{\mathrm{nc}}=K_2+U_2\) |
Find curve-entry speed from height and energy changes. |
| Centripetal acceleration |
\(a_c=v^2/r\) |
Inward acceleration needed for circular motion. |
| Frictionless design speed |
\(v_0=\sqrt{rg\tan\theta}\) |
Speed requiring no static friction. |
| Required friction coefficient |
\(\mu_{\mathrm{req}}=\frac{|q\cos\theta-\sin\theta|}{q\sin\theta+\cos\theta}\) |
Minimum static friction needed at speed \(v\). |
| Speed ratio |
\(q=\frac{v^2}{rg}\) |
Compares actual speed with radius and gravity. |
The key idea is that energy finds the car’s speed, while the banked-curve force balance checks whether that speed can be supported.
