SME
Dynamics · Problem type
Rolling down a slope without slipping
Use this for “A solid ball rolls without slipping down a 30° slope. Find its acceleration.”
Diagram
Rolling
without
slipping:
the
contact
is
the
IC,
so
v
=
0
there,
v
at
the
centre
and
2v
at
the
top.
a
=
g
sin θ
1 + c
=
3.5
m/
s
2
for
a
solid
ball.
This is an example. Type your own values to start a new problem.
a (m/s²)
=
g
sin
°
1
+
m
Mass
example · kg
r
Radius
example · m
I
Moment of inertia
calculated · kg·m²
α
Angular acceleration
worked out (not typed) · rad/s²
Formulas
a
=
9.81
×
sin(
θ
)
1 +
c
3.5036
=
9.81
×
sin(30°)
1 + 0.4
I
=
c
×
m
×
r
2
0.008
=
0.4
×
2
×
0.1
2
α
=
a
r
35.0357
=
3.5036
0.1
Assumptions
•
It rolls without slipping: the contact point is the instantaneous centre, so a = rα.
•
I = cmr²: hoop c = 1, hollow ball 2/3, disk 1/2, solid ball 0.4; g = 9.81 m/s².
•
Friction at the contact does no work; it only turns the body.
•
It rolls only if friction can supply cma: μₛ ≥ c tan θ ÷ (1 + c).
Step-by-step
WE KNOW
θ = 30°, c = 0.4, m = 2 kg, r = 0.1 m
FIND
acceleration down the slope (a), moment of inertia (I), angular acceleration (α)
Step 1 · Find acceleration down the slope (a)
Use a = 9.81 × sin(θ) ÷ (1 + c)
Along
the
slope
mg
sin
θ
−
f
=
ma;
about
the
centre
fr
=
cmr²
α
=
cmra,
so
f
=
cma.
Add
them:
a
(1
+
c
)
=
g
sin
θ
.
a
=
9.81
×
sin(30°)
1 + 0.4
a
=
9.81
×
sin(30°)
1.4
a
=
9.81
×
0.5
1.4
a
=
4.905
1.4
a
=
3.50357
m/
s
2
Step 2 · Find moment of inertia (I)
Use I = c × m × r²
The shape factor times mr²: how far the mass sits from the axle.
I
=
0.4
×
2
×
0.1
2
I
=
0.4
×
2
×
0.01
I
=
0.8
×
0.01
I
=
0.008
kg·
m
2
Step 3 · Find angular acceleration (α)
Use α = a ÷ r
Rolling
without
slipping,
the
centre
speeds
up
as
fast
as
the
rim:
a
=
rα.
α
=
3.50357
0.1
α
=
35.0357
rad/
s
2
Check
Put the numbers back in. Both sides should match.
3.50357
=
9.81
×
sin(30°)
1 + 0.4
✓
0.008
=
0.4
×
2
×
0.1
2
✓
35.0357
=
3.50357
0.1
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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