SME
University Physics I: Mechanics · Problem type
A banked curve with no friction: the design speed
Use this for “A curve of radius 80 m is banked at 12°. At what speed can a 1500 kg car take it with no friction, and how hard does the road push on it?”
Diagram
Up
and
down:
N
cos
θ
=
mg,
so
N
=
mg
cos θ
=
11,770
cos 15
°
=
12,190
N
Level: N sin θ = mv²/r, the centripetal force
Dividing:
tan
θ
=
v
2
rg
,
so
v
=
√
rg tan θ
=
√
50 × 9.81 × tan 15°
=
11.46
m/s
This is an example. Type your own values to start a new problem.
tan(
θ
°)
=
(
v (m/s)
)
2
r (m)
×
9.81
m
Mass of the car
example
N
Normal force
worked out (not typed)
Formulas
tan(
θ
)
=
v
2
r
× 9.81
tan(15°)
=
11.4643
2
50 × 9.81
N
=
m
×
9.81
cos(
θ
)
12,187.2712
=
1,200
×
9.81
cos(15°)
Assumptions
•
No friction: only the normal force N and the weight mg act on the car; g = 9.81 m/s².
•
The car goes round a level circle of radius r at a steady speed, so the net force points level, toward the center.
•
Up and down: N cos θ = mg. Toward the center: N sin θ = mv² ÷ r.
Step-by-step
WE KNOW
r = 50 m, θ = 15°, m = 1,200 kg
FIND
design speed (v), normal force (N)
Step 1 · Find design speed (v)
Use tan(θ) = v² ÷ (r × 9.81)
Divide
N
sin
θ
=
mv
²
r
by
N
cos
θ
=
mg:
N
and
the
mass
cancel,
leaving
tan
θ
=
v
2
rg
.
v
=
√(
r
×
9.81
×
tan(
θ
))
v
=
√(50
×
9.81
×
tan(15°))
v
=
√
50 × 9.81 × 0.26795
v
=
√
490.5 × 0.26795
v
=
√
131.43
v
=
11.4643
m/s
Step 2 · Find normal force (N)
Use N = m × 9.81 ÷ cos(θ)
Up
and
down:
the
upward
part
of
N
,
N
cos
θ
,
holds
the
weight
mg.
N
=
1,200
×
9.81
cos(15°)
N
=
1,200
×
9.81
0.96593
N
=
11,772
0.96593
N
=
12,187.27
N
Check
Put the numbers back in. Both sides should match.
tan(15°)
=
11.4643
2
50 × 9.81
✓
12,187.27
=
1,200
×
9.81
cos(15°)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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