SME
Classical Mechanics · Problem type
A pendulum in an accelerating car
Use this for “A car speeds up at 2 m/s². At what angle does a hanging pendulum settle, and what is its period if it is 0.5 m long?”
Diagram
g = 9.81 m/s² at 90° from the horizontal: (0, −9.81) m/s².
−a = 3 m/s² at 0° from the horizontal: (−3, 0) m/s².
ΣFₓ
=
−3
m/
s
2
and
ΣFᵧ
ΣFₓ
=
−9.81
m/
s
2
geff
=
√
−
3
2
+ −
9.81
2
=
10.26
m/
s
2
at
253°
from
+x.
This is an example. Type your own values to start a new problem.
tan
θ
°
=
a (m/s²)
9.81
m/s²
T
=
2π
L (
)
geff (m/s²)
=
T (
)
Formulas
θ
=
tan⁻¹(
a
9.81
)
17.0042
=
tan⁻¹(
3
9.81
)
geff
=
√
9.81
2
+
a
2
10.2585
=
√
9.81
2
+
3
2
T
=
2
×
π
×
√(
L
geff
)
1.9617
=
2
×
π
×
√(
1
10.2585
)
Assumptions
•
The car speeds up at a steady rate a in a straight line; g = 9.81 m/s².
•
In the car’s frame a fictitious force −ma acts backward on the bob, so it feels geff, the sum of g down and a backward.
•
The bob settles along geff, tipped back from the vertical by θ, and swings about there with small swings.
Step-by-step
WE KNOW
a = 3 m/s², L = 1 m
FIND
angle back from vertical (θ), effective gravity (geff), period of small swings (T)
Step 1 · Find angle back from vertical (θ)
Use θ = tan⁻¹(a ÷ 9.81)
The
string
pulls
along
geff:
its
backward
part
is
a
and
its
downward
part
is
g,
so
tan
θ
=
a
g
.
θ
=
tan⁻¹(
3
9.81
)
θ
=
17.0042°
Step 2 · Find effective gravity (geff)
Use geff = √(9.81² + a²)
Gravity g down and the fictitious a backward are at right angles: add them by Pythagoras.
geff
=
√
9.81
2
+
3
2
geff
=
√
96.2361 + 9
geff
=
√
105.2361
geff
=
10.2585
m/
s
2
Step 3 · Find period of small swings (T)
Use T = 2 × π × √(L ÷ geff)
In the car the bob swings as if gravity were geff, so geff takes the place of g in the period.
T
=
2
×
π
×
√(
1
10.2585
)
T
=
2
×
√
0.09748
×
π
T
=
2
×
0.3122
×
π
T
=
0.6244
×
π
T
=
1.9617
s
Check
Put the numbers back in. Both sides should match.
17.0042
=
tan⁻¹(
3
9.81
)
✓
10.2585
=
√
9.81
2
+
3
2
✓
1.9617
=
2
×
π
×
√(
1
10.2585
)
✓
Topic check
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