Muraj, Hamza – Homework 25 – Due: Mar 31 2006, 4:00 am – Inst: Florin
1
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The due time is Central
time.
001
(part 1 of 2) 10 points
A solid sphere of radius 21 cm is positioned
at the top of an incline that makes 28
◦
angle
with the horizontal.
This initial position of
the sphere is a vertical distance 2
.
5 m above
its position when at the bottom of the incline.
The sphere is released and moves down the
incline.
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
.
21 cm
M
μ
‘
28
◦
2
.
5 m
Calculate the speed of the sphere when it
reaches the bottom of the incline in the case
where it rolls without slipping.
Correct answer: 5
.
91608 m
/
s.
Explanation:
Basic Concepts:
X
E
=
K
trans
+
K
rot
+
U
gravity
=
const
From conservation of energy we have
M g h
=
1
2
M v
2
1
+
1
2
I ω
2
=
1
2
M v
2
1
+
1
2
2
5
M R
2
¶
v
2
1
R
2
¶
=
7
10
M v
2
1
.
Therefore,
v
1
=
r
10
7
g h
=
r
10
7
(9
.
8 m
/
s
2
) (2
.
5 m)
= 5
.
91608 m
/
s
.
002
(part 2 of 2) 10 points
Calculate the speed of the sphere when it
reaches the bottom of the incline in the case
where it slips frictionlessly without rolling.
Correct answer: 7 m
/
s.
Explanation:
From conservation of energy we have
M g h
=
1
2
M v
2
2
.
Therefore,
v
2
=
p
2
g h
=
q
2 (9
.
8 m
/
s
2
) (2
.
5 m)
= 7 m
/
s
.
003
(part 1 of 3) 10 points
Consider a thick cylindrical shell rolling down
along an incline. It has a mass
m
, an outer ra
dius
R
and a moment of inertia
I
cm
=
4
5
mR
2
about its center of mass. The incline is at an
angle
θ
to the horizontal. 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
α
2.
mgR
sin
θ
= (
I
cm
+
mR
2
)
α
correct
3.
mgR
tan
θ
=
I
cm
α
4.
mgR
= (
I
cm
+
mR
2
)
α
5.
mgR
sin
θ
=
I
cm
α
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 Spring '09
 Turner
 Acceleration, Energy, Kinetic Energy, Mass

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