SME
Classical Mechanics · Problem type
An Atwood machine with a heavy pulley, by Lagrange
Use this for “Blocks of 4 kg and 6 kg hang over a 3 kg disk pulley. Use the Lagrangian to find their acceleration, then find both tensions.”
Diagram
a
=
(m₂
−
m₁)
g
m₁ + m₂ +
1
2
M
a
=
(5
−
3)
×
9.81
3 + 5 +
1
2
× 2
a
=
2.18
m/
s
2
Block 1: m₁a = T₁ − m₁g = 35.97 − 29.43 = 6.54 N
Block 2: m₂a = m₂g − T₂ = 49.05 − 38.15 = 10.9 N
The pulley: T₂ − T₁ = ½Ma = 38.15 − 35.97 = 2.18 N turns it.
This is an example. Type your own values to start a new problem.
a
=
(
−
)g
+
+
½
=
a (m/s²)
meff
Effective mass
worked out (not typed)
T₁
Tension on m₁
worked out (not typed)
T₂
Tension on m₂
worked out (not typed)
Formulas
meff
=
m₁
+
m₂
+
M
2
9
=
3
+
5
+
2
2
a
=
(
m₂
−
m₁
)
×
9.81
meff
+2.18
=
(5
−
3)
×
9.81
9
T₁
=
m₁
×
9.81
×
2 ×
m₂
+
M
÷ 2
meff
35.97
=
3
×
9.81
×
2 × 5 + 2 ÷ 2
9
T₂
=
m₂
×
(9.81
−
a
)
38.15 = 5 × (9.81 − 2.18)
Assumptions
•
The pulley is a uniform disk, I = ½MR², on a frictionless axle; the string is light and doesn’t slip, so the disk turns at ẋ ÷ R.
•
One coordinate x, how far m₂ has dropped: T = ½(m₁ + m₂ + I ÷ R²)ẋ² and V = (m₁ − m₂)gx, with I ÷ R² = ½M.
•
a > 0 means m₂ goes down and m₁ up; g = 9.81 m/s².
•
A light pulley: type a tiny M (0.001 kg); then T₁ and T₂ are all but equal.
Step-by-step
WE KNOW
m₁ = 3 kg, m₂ = 5 kg, M = 2 kg
FIND
effective mass (meff), acceleration (m₂ down) (a), tension on m₁ (T₁), tension on m₂ (T₂)
Step 1 · Find effective mass (meff)
Use meff = m₁ + m₂ + M ÷ 2
Kinetic
energy
stage:
both
blocks
move
at
ẋ
and
the
disk
adds
1
2
Iω²
=
¼Mẋ²,
so
T
=
1
2
meff
ẋ
2
.
meff
=
3
+
5
+
2
2
meff = 3 + 5 + 1
meff = 9 kg
Step 2 · Find acceleration (m₂ down): a
Use a = (m₂ − m₁) × 9.81 ÷ meff
Euler–Lagrange
stage:
d/dt(∂L/∂ẋ)
=
meff
ẍ
and
∂L/∂x
=
(
m₂
−
m₁
)g,
so
ẍ
=
a
is
their
ratio.
a
=
(5
−
3)
×
9.81
9
a
=
2
×
9.81
9
a
=
19.62
9
a
=
+2.18
m/
s
2
Step 3 · Find tension on m₁ (T₁)
Use T₁ = m₁ × 9.81 × (2 × m₂ + M ÷ 2) ÷ meff
Tensions
never
enter
L,
so
use
Newton
on
m₁
alone:
T₁
−
m₁g
=
m₁
a
,
so
T₁
=
m₁
(g
+
a
).
Putting
a
in
gives
this
form,
with
nothing
to
cancel.
T₁
=
3
×
9.81
×
2 × 5 + 2 ÷ 2
9
T₁
=
3
×
9.81
×
10 + 1
9
T₁
=
3
×
9.81
×
11
9
T₁
=
29.43
×
11
9
T₁
=
323.73
9
T₁
=
35.97
N
Step 4 · Find tension on m₂ (T₂)
Use T₂ = m₂ × (9.81 − a)
Newton
on
m₂
alone:
it
falls
with
a
,
so
m₂g
−
T₂
=
m₂
a
.
T₂
−
T₁
=
1
2
Ma
is
what
turns
the
disk.
T₂
=
5
×
(9.81
−
2.18)
T₂
=
5
×
7.63
T₂
=
38.15
N
Check
Put the numbers back in. Both sides should match.
9
=
3
+
5
+
2
2
✓
+2.18
=
(5
−
3)
×
9.81
9
✓
35.97
=
3
×
9.81
×
2 × 5 + 2 ÷ 2
9
✓
38.15 = 5 × (9.81 − 2.18) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.