SME
Dynamics
Rigid-body dynamics
Use this for “A 2 kg block hangs from a rope wound round a 4 kg disk pulley of radius 0.2 m. Find a, α and the rope tension.”
Diagram
α = τ/I = 1.962/0.08
α
=
24.53
rad/
s
2
This is an example. Type your own values to start a new problem.
I (kg·m²)
=
1
2
×
M (kg)
×
(
r (m)
)
2
a (m/s²)
=
m (kg)
×
g
÷
(
m (kg)
+
I (kg·m²)
÷
(
r (m)
)
2
)
T (N)
=
m (kg)
(g
−
a (m/s²)
)
α
Pulley’s angular acceleration
calculated · rad/s²
τ
Torque on the pulley
worked out (not typed) · N·m
Formulas
I
=
0.5
×
M
×
r
2
0.08
=
0.5
×
4
×
0.2
2
a
=
m
×
9.81
m
+
I
÷
r
2
4.905
=
2
×
9.81
2 + 0.08 ÷
0.2
2
α
=
a
r
24.525
=
4.905
0.2
T
=
m
×
(9.81
−
a
)
9.81 = 2 × (9.81 − 4.905)
τ
=
I
×
α
1.962 = 0.08 × 24.525
Assumptions
•
The pulley is a uniform disk on a frictionless axle, so I = ½Mr²; g = 9.81 m/s².
•
The rope is light and doesn’t slip, so the rim moves with the block: a = rα.
•
Block: mg − T = ma. Pulley: Tr = Iα. Together they give a = mg ÷ (m + I ÷ r²).
Step-by-step
WE KNOW
m = 2 kg, M = 4 kg, r = 0.2 m
FIND
pulley’s moment of inertia (I), block’s acceleration (down) (a), pulley’s angular acceleration (α), rope tension (T), torque on the pulley (τ)
Step 1 · Find pulley’s moment of inertia (I)
Use I = 0.5 × M × r²
A
uniform
disk
about
its
axle:
I
=
∫
r
2
dm
comes
to
half
the
mass
times
the
radius
squared.
I
=
0.5
×
4
×
0.2
2
I
=
0.5
×
4
×
0.04
I
=
2
×
0.04
I
=
0.08
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
the
weight
speeds
up.
a
=
2
×
9.81
2 + 0.08 ÷
0.2
2
a
=
2
×
9.81
2 + 0.08 ÷ 0.04
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
0.2
α
=
24.525
rad/
s
2
Step 4 · Find rope tension (T)
Use T = m × (9.81 − a)
The
block’s
law
mg
−
T
=
ma,
solved
for
the
rope’s
pull:
less
than
the
weight
while
the
block
speeds
up.
T
=
2
×
(9.81
−
4.905)
T
=
2
×
4.905
T
=
9.81
N
Step 5 · Find torque on the pulley (τ)
Use τ = I × α
Newton’s second law for turning: the rope’s pull at the rim, Tr, is the torque Iα.
τ
=
0.08
×
24.525
τ
=
1.962
N·
m
Check
Put the numbers back in. Both sides should match.
0.08
=
0.5
×
4
×
0.2
2
✓
4.905
=
2
×
9.81
2 + 0.08 ÷
0.2
2
✓
24.525
=
4.905
0.2
✓
9.81 = 2 × (9.81 − 4.905) ✓
1.962 = 0.08 × 24.525 ✓
← Previous lesson
Direct impact and the coefficient of restitution
Next lesson →
Rolling down a slope without slipping
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
Rolling down a slope without slipping
Use this for “A solid ball rolls without slipping down a 30° slope. Find its acceleration.”
A sliding ladder by its instantaneous centre
Use this for “A 5 m ladder slides with its foot moving out at 2 m/s when it makes 60° with the floor. Find ω and the top’s speed.”
Energy stored in a flywheel
Use this for “A flywheel with I = 2 kg·m² spins at 3000 rpm. How much kinetic energy does it store?”