SME
Dynamics · Problem type
A banked curve: the design speed and the top speed
Use this for “A highway curve of radius 50 m is banked at 20°. What speed needs no friction, and how fast can a car take it if μₛ = 0.25?”
Diagram
At the top speed friction μₛN points down the slope
Up and down: N(cos θ − μₛ sin θ) = mg; level: N(sin θ + μₛ cos θ) = mv²/R
vₘₐₓ
=
√(
Rg(tan θ + μₛ)
1 − μₛ tan θ
)
vₘₐₓ
=
√
50 × 9.81 × 0.6754
=
18.2
m/s
This is an example. Type your own values to start a new problem.
v (m/s)
=
g
×
R (m)
×
tan
°
vₘₐₓ (m/s)
=
g
×
R (m)
×
tan
°
+
1
−
tan
°
The top speed’s fraction is (sin θ + μₛ cos θ) ÷ (cos θ − μₛ sin θ), as the Formulas section writes it, with its top and bottom divided by cos θ.
Formulas
v
=
√(9.81
×
R
×
tan(
θ
))
13.3614 = √(9.81 × 50 × tan(20°))
vₘₐₓ
=
√(9.81
×
R
×
sin(
θ
) + μₛ × cos(
θ
)
cos(
θ
) − μₛ × sin(
θ
)
)
18.2016
=
√(9.81
×
50
×
sin(20°) + 0.25 × cos(20°)
cos(20°) − 0.25 × sin(20°)
)
Assumptions
•
The car goes round a level circle at a steady speed, so the net force points level, toward the center; g = 9.81 m/s².
•
With no friction, N cos θ = mg and N sin θ = mv² ÷ R, so v = √(gR tan θ).
•
At the top speed the car is about to slide up the bank, so friction μₛN points down the slope.
•
If μₛ tan θ ≥ 1, friction holds the car at any speed: there is no top speed.
Step-by-step
WE KNOW
R = 50 m, θ = 20°, μₛ = 0.25
FIND
speed with no friction (v), top speed (vₘₐₓ)
Step 1 · Find speed with no friction (v)
Use v = √(9.81 × R × tan(θ))
Divide
N
sin
θ
=
mv²
R
by
N
cos
θ
=
mg:
N
and
the
mass
cancel,
leaving
tan
θ
=
v
2
gR
.
v
=
√(9.81
×
50
×
tan(20°))
v
=
√
9.81 × 50 × 0.364
v
=
√
490.5 × 0.364
v
=
√
178.542
v
=
13.3614
m/s
Step 2 · Find top speed (vₘₐₓ)
Use vₘₐₓ = √(9.81 × R × (sin(θ) + μₛ × cos(θ)) ÷ (cos(θ) − μₛ × sin(θ)))
Toward
the
center
N(sin
θ
+
μₛ
cos
θ
)
=
mv²
R
;
up
and
down
N(cos
θ
−
μₛ
sin
θ
)
=
mg.
Divide
the
two:
N
and
m
cancel.
vₘₐₓ
=
√(9.81
×
50
×
sin(20°) + 0.25 × cos(20°)
cos(20°) − 0.25 × sin(20°)
)
vₘₐₓ
=
√(9.81
×
50
×
0.342 + 0.25 × 0.9397
0.9397 − 0.25 × 0.342
)
vₘₐₓ
=
√(9.81
×
50
×
0.342 + 0.2349
0.9397 − 0.0855
)
vₘₐₓ
=
√(9.81
×
50
×
0.5769
0.8542
)
vₘₐₓ
=
√(490.5
×
0.5769
0.8542
)
vₘₐₓ
=
√(
283
0.8542
)
vₘₐₓ
=
√
331.3
vₘₐₓ = 18.2016 m/s
Check
Put the numbers back in. Both sides should match.
13.3614 = √(9.81 × 50 × tan(20°)) ✓
18.2016
=
√(9.81
×
50
×
sin(20°) + 0.25 × cos(20°)
cos(20°) − 0.25 × sin(20°)
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.