SME
University Physics I: Mechanics
Rotation and torque
Use this for “A 3 kg disk of radius 0.2 m rolls from rest down a ramp that drops 2 m. How fast is it moving at the bottom, and how fast does it spin?”
Diagram
mgh = 2 × 9.81 × 1.5 = 29.43 J at the top.
v
=
√(2
×
9.81
×
1.5
1 + 0.4
)
=
4.58
m/s,
and
ω
=
v
r
=
45.85
rad/s.
Kₜ = ½mv² = 21.02 J and Kᵣ = cKₜ = 8.41 J: together 29.43 J = mgh.
At the bottom: hoop 3.84 m/s, hollow ball 4.2 m/s, disk 4.43 m/s, solid ball 4.58 m/s.
Mass and radius cancel: the smaller c, the less goes into spin, so the solid ball wins.
This is an example. Type your own values to start a new problem.
v (m/s)
=
2
×
9.81
×
h (m)
1
+
m
Mass
example
r
Radius
example
ω
Spin at the bottom
worked out (not typed) · rad/s
Kₜ
Translational kinetic energy
worked out (not typed) · J
Kᵣ
Rotational kinetic energy
worked out (not typed) · J
Formulas
v
=
√(2
×
9.81
×
h
1 +
c
)
4.5849
=
√(2
×
9.81
×
1.5
1 + 0.4
)
ω
=
v
r
45.8491
=
4.5849
0.1
Kₜ
=
1
2
×
m
×
v
2
21.0214
=
1
2
×
2
×
4.5849
2
Kᵣ
=
c
×
Kₜ
8.4086 = 0.4 × 21.0214
Assumptions
•
It rolls without slipping, so v = rω, and static friction at the contact point does no work; g = 9.81 m/s².
•
I = cmr² with the shape factor c: hoop 1, hollow ball ⅔ (type 0.6667), disk ½, solid ball 0.4.
•
It starts from rest, so mgh = ½mv² + ½Iω² = ½(1 + c)mv²: the mass and the radius cancel from v.
Step-by-step
WE KNOW
c = 0.4, m = 2 kg, r = 0.1 m, h = 1.5 m
FIND
speed at the bottom (v), spin at the bottom (ω), translational kinetic energy (Kₜ), rotational kinetic energy (Kᵣ)
Step 1 · Find speed at the bottom (v)
Use v = √(2 × 9.81 × h ÷ (1 + c))
The
drop’s
energy
mgh
becomes
1
2
(1
+
c
)mv².
The
mass
cancels;
solve
for
v
.
v
=
√(2
×
9.81
×
1.5
1 + 0.4
)
v
=
√(2
×
9.81
×
1.5
1.4
)
v
=
√(19.62
×
1.5
1.4
)
v
=
√(
29.43
1.4
)
v
=
√
21.02
v
=
4.58491
m/s
Step 2 · Find spin at the bottom (ω)
Use ω = v ÷ r
Rolling
without
slipping,
the
rim
turns
as
fast
as
the
body
moves:
v
=
rω.
ω
=
4.58491
0.1
ω
=
45.8491
rad/s
Step 3 · Find translational kinetic energy (Kₜ)
Use Kₜ = ½ × m × v²
The energy of moving along is half the mass times the speed squared.
Kₜ
=
1
2
×
2
×
4.58491
2
Kₜ
=
1
2
×
2
×
21.02
Kₜ = 21.0214 J
Step 4 · Find rotational kinetic energy (Kᵣ)
Use Kᵣ = c × Kₜ
The
spin
energy
1
2
Iω²
is
1
2
(cmr²)
(
v
r
)
2
=
c
×
1
2
mv²:
c
times
the
energy
of
moving
along.
Kᵣ = 0.4 × 21.0214
Kᵣ = 8.4086 J
Check
Put the numbers back in. Both sides should match.
4.58491
=
√(2
×
9.81
×
1.5
1 + 0.4
)
✓
45.8491
=
4.58491
0.1
✓
21.0214
=
1
2
×
2
×
4.58491
2
✓
8.4086 = 0.4 × 21.0214 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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