SME
University Physics I: Mechanics · Problem type
A block falling from a pulley with mass
Use this for “A 3 kg bucket hangs from a rope wound round a 5 kg solid disk pulley of radius 0.2 m. Find the bucket’s acceleration, the rope’s tension and the pulley’s angular acceleration.”
Diagram
α = τ/I = 0.981/0.02
α
=
49.05
rad/
s
2
This is an example. Type your own values to start a new problem.
a (m/s²)
=
m (kg)
×
9.81
+
÷
(
)
2
M
Pulley mass
example
α
Pulley’s angular acceleration
worked out (not typed) · rad/s²
τ
Torque on the pulley
worked out (not typed) · N·m
T
Rope tension
worked out (not typed)
Formulas
I
=
1
2
×
M
×
r
2
0.02
=
1
2
×
4
×
0.1
2
a
=
m
× 9.81
m
+
I
÷
r
2
4.905
=
2 × 9.81
2 + 0.02 ÷
0.1
2
α
=
a
r
49.05
=
4.905
0.1
τ
=
I
×
α
0.981 = 0.02 × 49.05
τ
=
T
×
r
0.981 = 9.81 × 0.1
Assumptions
•
The pulley is a uniform solid disk on a frictionless axle, so I = ½Mr²; g = 9.81 m/s².
•
The rope is light and doesn’t slip or stretch, so the rim moves with the block: a = rα.
•
The block obeys mg − T = ma and the pulley Tr = Iα; together they give a = mg ÷ (m + I ÷ r²).
Step-by-step
WE KNOW
m = 2 kg, M = 4 kg, r = 0.1 m
FIND
pulley’s moment of inertia (I), block’s acceleration (down) (a), pulley’s angular acceleration (α), torque on the pulley (τ), rope tension (T)
Step 1 · Find pulley’s moment of inertia (I)
Use I = ½ × M × r²
A uniform solid disk about its axle: half the mass times the radius squared.
I
=
1
2
×
4
×
0.1
2
I
=
1
2
×
4
×
0.01
I
=
2
×
0.01
I
=
0.02
kg·
m
2
Step 2 · Find block’s acceleration (down): a
Use a = (m × 9.81) ÷ (m + I ÷ r²)
Put
T
=
Iα
r
=
Ia
r
2
into
mg
−
T
=
ma:
the
pulley
adds
I
r
2
to
the
mass
that
the
weight
speeds
up.
a
=
2 × 9.81
2 + 0.02 ÷
0.1
2
a
=
2
×
9.81
2 + 0.02 ÷ 0.01
a
=
2
×
9.81
2 + 2
a
=
2
×
9.81
4
a
=
19.62
4
a
=
4.905
m/
s
2
Step 3 · Find pulley’s angular acceleration (α)
Use α = a ÷ r
The
rope
doesn’t
slip,
so
the
rim
speeds
up
as
fast
as
the
block:
a
=
rα.
α
=
4.905
0.1
α
=
49.05
rad/
s
2
Step 4 · Find torque on the pulley (τ)
Use τ = I × α
Newton’s
second
law
for
turning:
the
torque
is
I
times
α
.
τ
=
0.02
×
49.05
τ
=
0.981
N·
m
Step 5 · Find rope tension (T)
Use τ = T × r
Only
the
rope
turns
the
pulley,
pulling
at
the
rim:
τ
=
Tr,
so
divide
τ
by
r
.
T
=
τ
r
T
=
0.981
0.1
T
=
9.81
N
Check
Put the numbers back in. Both sides should match.
0.02
=
1
2
×
4
×
0.1
2
✓
4.905
=
2 × 9.81
2 + 0.02 ÷
0.1
2
✓
49.05
=
4.905
0.1
✓
0.981 = 0.02 × 49.05 ✓
0.981 = 9.81 × 0.1 ✓
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