homework 26 – BAUTISTA, ALDO – Due: Apr 3 2006, 4:00 am
1
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Question 1
Part 1 of 2.
10 points.
Consider the problem of the solid sphere
rolling down an incline without slipping. The
incline has an angle
θ
.
The sphere’s length
up the incline is
‘
, and its height is
h
. At the
beginning, the sphere rests on the very top
of the incline.
The sphere has mass
M
and
radius
R
.
The acceleration of gravity is 9
.
8 m
/
s
2
.
Hint:
The moment of inertia of a sphere
with respect to an axis through its
center
is
2
5
M R
2
.
Choose the instantaneous axis through the
contact point
P
as the axis of origin for the
torque equation.
R
M
μ
‘
θ
h
The acceleration of the center of mass is
1.
a
=
3
5
g
cos
θ .
2.
a
=
5
7
g
cos
θ .
3.
a
=
3
7
g
sin
θ .
4.
a
=
3
5
g
sin
θ .
5.
a
=
2
7
g
cos
θ .
6.
a
=
3
7
g
cos
θ .
7.
a
=
5
7
g
sin
θ .
correct
8.
a
=
2
7
g
sin
θ .
Explanation:
Basic Concepts:
X
~
F
=
m ~a
X
~
τ
=
I ~α .
m g
cos
θ
N
f
m g
sin
θ
With the origin at the point of contact,
X
τ
:
M g R
sin
θ
=
I α ,
(1)
where
I
is obtained using the parallelaxis
theorem
I
=
2
5
M R
2
+
M R
2
=
7
5
M R
2
.
(2)
If the sphere rolls without slipping,
α
=
a
R
.
(3)
Now substituting
I
from Eq. 2 and
α
from Eq.
3 into Eq. 1, we have
M g R
sin
θ
=
7
5
M R
2
a
R
,
so
a
=
5
7
g
sin
θ .
Question 2
Part 2 of 2.
10 points.
The minimum coefficient of friction such
that the sphere rolls without slipping is
1.
μ
=
5
7
cos
θ .
2.
μ
=
3
5
cos
θ .
3.
μ
=
2
7
sin
θ .
4.
μ
=
2
7
tan
θ .
correct
5.
μ
=
3
7
tan
θ .
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homework 26 – BAUTISTA, ALDO – Due: Apr 3 2006, 4:00 am
2
6.
μ
=
3
7
sin
θ .
7.
μ
=
2
7
cos
θ .
8.
μ
=
5
7
tan
θ .
Explanation:
The net force along the direction of the
incline is
X
F
=
M g
sin
θ

f
=
M
5
7
g
sin
θ
,
where
f
≤
μ N
=
μ M g
cos
θ
is the minimum
“no slipping” criterion. Then
M g
sin
θ

μ M g
cos
θ
=
M
5
7
g
sin
θ
μ
cos
θ
=
1

5
7
sin
θ ,
so
μ
=
2
7
tan
θ
.
Question 3
Part 1 of 3.
10 points.
Consider a thick cylindrical shell rolling
down along an incline.
It has a mass
m
,
an outer radius
R
and a moment of iner
tia
I
cm
=
4
5
mR
2
about its center of mass.
The incline is at an angle
θ
to the hori
zontal.
Its length is
s
and its height is
h
.
s
h
θ
v
A
A
The equation of motion
τ
=
Iα
as the shell is
rolling down the incline is given by(using the
contact point as the pivot point):
1.
mgR
= (
I
cm
+
mR
2
)
α
2.
mgR
tan
θ
= (
I
cm
+
mR
2
)
α
3.
mgR
tan
θ
=
I
cm
α
4.
mgR
sin
θ
= (
I
cm
+
mR
2
)
α
correct
5.
mgR
cot
θ
= (
I
cm
+
mR
2
)
α
6.
mgR
=
I
cm
α
7.
mgR
cot
θ
=
I
cm
α
8.
mgR
cos
θ
= (
I
cm
+
mR
2
)
α
9.
mgR
cos
θ
=
I
cm
α
10.
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 Spring '08
 Turner
 Angular Momentum, Energy, Kinetic Energy, Mass, Work

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