practice 11 – – Due: Apr 10 2007, noon
1
Question 1, chap 13, sect 2.
part 1 of 1
10 points
A constant horizontal force of 120 N is ap
plied to a lawn roller in the form of a uniform
solid cylinder of radius 0
.
35 m and mass 13 kg.
The acceleration of gravity is 9
.
8 m
/
s
2
.
M
R
F
If the roller rolls without slipping, calculate
the acceleration of the center of mass. (
I
cm
=
1
2
M R
2
)
Correct answer: 6
.
15385 m
/
s
2
(tolerance
±
1
%).
Explanation:
Basic Concepts:
summationdisplay
vector
F
=
mvector
a
summationdisplay
vector
τ
=
Ivectorα
If we choose the center of mass as the axis,
summationdisplay
F
=
F
−
f
=
M a
τ
=
fR
=
I
cm
α
=
1
2
M R
2
a
R
then you can solve these two equations to get
a.
The another way to do this is we can
choose the contact point as the axis, using
I
=
1
2
M R
2
+
M R
2
=
3
2
M R
2
, we find
3
2
M R
2
a
R
=
F R .
Therefore
a
=
2
F
3
M
(1)
=
2 (120 N)
3 (13 kg)
= 6
.
15385 m
/
s
2
.
Question 2, chap 13, sect 2.
part 1 of 3
7 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
=
1
2
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.
m g R
=
I
cm
α
2.
m g R
tan
θ
= (
I
cm
+
mR
2
)
α
3.
m g R
cot
θ
= (
I
cm
+
mR
2
)
α
4.
m g R
sin
θ
= (
I
cm
+
mR
2
)
α
correct
5.
m g R
cos
θ
= (
I
cm
+
mR
2
)
α
6.
m g R
sin
θ
=
I
cm
α
7.
m g R
cos
θ
=
I
cm
α
8.
m g R
tan
θ
=
I
cm
α
9.
m g R
cot
θ
=
I
cm
α
10.
m g R
= (
I
cm
+
mR
2
)
α
Explanation:
Basic Concepts:
Energy conservation
U
i
+
K
i
=
U
f
+
K
f
(1)
where the
KE
terms include both rotational
and translational kinetic energy. Equation of
motion
τ
=
Iα
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practice 11 – – Due: Apr 10 2007, noon
2
θ
θ
mg
sin
θ
mg
P
The torque about the contact point P is
given by
τ
=
R m g
sin
θ
=
I
p
α .
Using the law of parallel axes,
I
p
=
I
cm
+
m R
2
.
This leads to the choice of
m g R
sin
θ
= (
I
cm
+
m R
2
)
α.
Question 3, chap 13, sect 2.
part 2 of 3
7 points
The linear acceleration
a
of the shell as it
rolls down the incline is given by
1.
a
=
9
5
g
sin
θ
2.
a
=
5
9
g
sin
θ
3.
a
=
2
3
g
cos
θ
4.
a
=
5
9
g
cos
θ
5.
a
=
9
5
g
cos
θ
6.
a
=
g
cos
θ
7.
a
=
3
2
g
sin
θ
8.
a
=
3
2
g
cos
θ
9.
a
=
2
3
g
sin
θ
correct
10.
a
=
g
sin
θ
Explanation:
Using the law of parallel axes, one finds
that the moment of inertia of the shell about
P
I
=
I
cm
+
mR
2
=
parenleftbigg
1
2
parenrightbigg
m R
2
+
m R
2
=
parenleftbigg
3
2
parenrightbigg
m R
2
Now we apply the equation of motion
τ
=
I α
I α
=
parenleftbigg
3
2
parenrightbigg
m R
2
α
=
parenleftbigg
3
2
parenrightbigg
m R a ,
and
τ
=
R m g
sin
θ .
This leads to
a
=
parenleftbigg
2
3
parenrightbigg
g
sin
θ .
Question 4, chap 13, sect 2.
part 3 of 3
7 points
Release the shell from rest at the top of the
incline.
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 Spring '08
 Turner
 Acceleration, Force, Mass, Correct Answer, kg

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