SME
Grade 11 Science · Problem type
Torque and rotational inertia
Use this for “A 3 N·m torque turns a 2 kg disk of radius 0.5 m. What is its angular acceleration? What if it were a hoop?”
Diagram
I
=
cmr²
=
1
2
×
2
×
0.5
2
I
=
0.25
kg·
m
2
α = τ/I = 3/0.25 = 12 rad/s²
The same torque spins the hoop up most slowly: its mass is all at the rim.
This is an example. Type your own values to start a new problem.
τ (N·m)
=
I (kg·m²)
×
α (rad/s²)
c
Shape factor
example
m
Mass
example
r
Radius
example
Formulas
I
=
c
×
m
×
r
2
0.25
=
1
2
×
2
×
0.5
2
τ
=
I
×
α
3 = 0.25 × 12
Assumptions
•
Rotational inertia I is how hard a body is to spin up: mass farther from the axis counts more.
•
I = cmr² about the center: c = 1 for a hoop, ⅔ for a hollow ball, ½ for a solid disk, 0.4 for a solid ball.
•
Newton’s second law for turning: τ = Iα, with α in rad/s².
Step-by-step
WE KNOW
c = 1/2, m = 2 kg, r = 0.5 m, τ = 3 N·m
FIND
rotational inertia (I), angular acceleration (α)
Step 1 · Find rotational inertia (I)
Use I = c × m × r²
The shape factor times the mass times the radius squared.
I
=
1
2
×
2
×
0.5
2
I
=
1
2
×
2
×
0.25
I
=
0.25
kg·
m
2
Step 2 · Find angular acceleration (α)
Use τ = I × α
Divide the torque by the rotational inertia: a bigger I spins up more slowly.
α
=
τ
I
α
=
3/0.25
α
=
12
rad/
s
2
Check
Put the numbers back in. Both sides should match.
0.25
=
1
2
×
2
×
0.5
2
✓
3 = 0.25 × 12 ✓
Topic check
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