Pickering, Tracey – Homework 23 – Due: Mar 29 2006, noon – Inst: Drummond
1
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printout
should
have
16
questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
The due time is Central
time.
001
(part 1 oF 3) 10 points
A cylinder oF mass 15 kg rolls without slipping
on a horizontal surFace .
At the instant its center oF mass has a speed
oF 9
.
5 m
/
s determine the translational kinetic
energy oF its center oF mass.
Correct answer: 676
.
875 J.
Explanation:
The translational kinetic energy is given by
T
t
=
1
2
m v
2
= 676
.
875 J
The rotational energy is given by
T
r
=
1
2
I ω
2
= 338
.
438 J
Since the moment oF inertia I For a cylinder is
I
=
1
2
m R
2
and using
v
=
ω R
, we get
T
r
=
1
4
m v
2
= 338
.
438 J
002
(part 2 oF 3) 10 points
At the instant its center oF mass has a speed
oF 9
.
5 m
/
s determine the rotational energy
about its center oF mass
Correct answer: 338
.
438 J.
Explanation:
003
(part 3 oF 3) 10 points
At the instant its center oF mass has a speed
oF 9
.
5 m
/
s determine its total energy.
Correct answer: 1015
.
31 J.
Explanation:
004
(part 1 oF 3) 10 points
A 4
.
65 kg hollow cylinder with inner radius
0
.
26 m and outer radius 0
.
43 m is pulled by
a horizontal string with a Force oF 38
.
9 N, as
shown in the fgure.
38
.
9 N
4
.
65 kg
ω
0
.
26 m
0
.
43 m
The Frictional Force exerted on the wheel by
the ±oor
1.
is to the right.
correct
2.
is to the leFt.
3.
is zero.
Explanation:
IF there were no Friction the bottom oF the
wheel would spin to the leFt; the Frictional
Force will oppose this spinning motion.
ThereFore, the Frictional Force is directed to
the right.
005
(part 2 oF 3) 10 points
What must be the magnitude oF the Force
oF Friction iF the cylinder is to roll without
slipping?
Correct answer: 7
.
33243 .
Explanation:
Let :
R
in
= 0
.
26 m
,
R
out
= 0
.
43 m
,
F
= 38
.
9 N
,
m
= 4
.
65 kg
,
and
f
= Force oF Friction
.
Suppose the cylinder has a length oF
‘
; then
the density oF the cylinder is
ρ
=
m
π
(
R
2
out

R
2
in
)
‘
.
The moment oF inertia about the center is
I
cm
=
Z
ρ r
2
dV
=
m
π
(
R
2
out

R
2
in
)
‘
×
Z
R
out
R
in
r
2
r dr
Z
2
π
0
dθ
Z
‘
0
dz
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View Full DocumentPickering, Tracey – Homework 23 – Due: Mar 29 2006, noon – Inst: Drummond
2
=
m
π
(
R
2
out

R
2
in
)
‘
R
4
out

R
4
in
4
2
π ‘
=
m
2
(
R
2
out
+
R
2
in
)
.
(1)
By the parallel axis theorem, the moment of
inertia about the point of contact with the
ground is
I
=
I
cm
+
m R
2
out
=
1
2
m
(3
R
2
out
+
R
2
in
)
(2)
=
1
2
(4
.
65 kg)
h
3 (0
.
43 m)
2
+ (0
.
26 m)
2
i
= 1
.
44685 kg
·
m
2
.
Using
a
free
body
diagram
(see
part
1
for
the
direction
of
f
),
F
ω
0
.
26 m
0
.
43 m
f
a
we have
F
+
f
=
m a
f
=
m a

F .
(3)
From the torque equation (using the contact
point of the cylinder and the surface as the
axis of rotation), we have
τ
=
I α
F
(2
R
out
) =
I α
(4)
=
I
a
R
out
,
so
a
=
2
R
2
out
I
F
m a
=
2
m R
2
out
I
F ,
using Eq
.
2
=
4
R
2
out
3
R
2
out
+
R
2
in
F .
(5)
Substituting
m a
from Eq. 5 into Eq. 3, we get
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