SME
Grade 11 Science · Problem type
A car rounding a flat curve
Use this for “A 1,200 kg car takes a 50 m curve at 15 m/s. How much friction does it need, and what is the least μₛ that holds it?”
Diagram
Centripetal acceleration: v²/r = 15²/50 = 4.5 m/s²
Centripetal force: mv²/r = 1,200 × 4.5 = 5,400 N
Friction
toward
the
center
supplies
it:
the
tires
need
μ
≥
v
2
/(rg)
=
0.459.
Least coefficient of static friction: 0.459
This is an example. Type your own values to start a new problem.
Fc (N)
=
m (kg)
×
(
v (m/s)
)
2
r (m)
ac
Centripetal acceleration
calculated · m/s²
μₛ
Least coefficient of static friction
calculated
Formulas
ac
=
v
2
/
r
4.5
=
15
2
/50
Fc
=
m
×
ac
5,400 = 1,200 × 4.5
μₛ = ac/9.8
0.459 = 4.5/9.8
Assumptions
•
On a flat curve, static friction between the tires and the road pulls the car toward the center.
•
The tires don’t slide sideways, so the friction is static: it can be at most μₛ FN.
•
Too fast, and friction can’t supply mv²/r: the car slides off along the tangent.
Step-by-step
WE KNOW
m = 1,200 kg, r = 50 m, v = 15 m/s
FIND
centripetal acceleration (ac), friction needed (Fc), least coefficient of static friction (μₛ)
Step 1 · Find centripetal acceleration (ac)
Use ac = v²/r
Turning
takes
an
acceleration
toward
the
center:
v
2
over
r
.
ac
=
15
2
/50
ac
=
225
50
ac
=
4.5
m/
s
2
Step 2 · Find friction needed (Fc)
Use Fc = m × ac
Newton’s second law toward the center: the mass times the centripetal acceleration.
Fc = 1,200 × 4.5
Fc = 5,400 N
Step 3 · Find least coefficient of static friction (μₛ)
Use μₛ = ac/9.8
Friction μₛmg must give mac: the mass cancels, leaving ac over g.
μₛ = 4.5/9.8
μₛ = 0.459
Check
Put the numbers back in. Both sides should match.
4.5
=
15
2
/50
✓
5,400 = 1,200 × 4.5 ✓
0.459 ≈ 4.5/9.8 ✓
Topic check
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