Thevalingam, Donald – Homework 18 – Due: Feb 7 2007, midnight – Inst: Eslami
1
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printout
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have
17
questions.
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before answering.
The due time is Central
time.
001
(part 1 of 6) 10 points
A heavy ball swings on a string in a circular
arc of radius 1 m.
Q
1 m
Q
0
P
18
◦
The two highest points of the ball’s trajec
tory are
Q
and
Q
0
; at these points the string is
±
18
◦
from the vertical. Point
P
is the lowest
point of the ball’s trajectory where the string
hangs vertically down.
The acceleration of gravity is 9
.
8 m
/
s
2
.
What is the ball’s speed at the point
P
? Ne
glect air resistance and other frictional forces.
Correct answer: 0
.
979435 m
/
s.
Explanation:
Let :
g
= 9
.
8 m
/
s
2
,
m
= 5 kg
,
‘
= 1 m
,
and
θ
= 18
◦
.
The diagram for the velocity and accelera
tion of the ball at point
P
is as follows,
x
y
v
a
The difference of gravitational potential en
ergy of the ball between point
Q
and point
P
is Δ
U
=
U
Q

U
P
=
m g ‘
(1

cos
θ
).
The
velocity of the ball at point
Q
is zero, so
K
P
+
U
P
=
K
Q
+
U
Q
K
P
+
U
P
= 0 +
U
Q
.
Therefore
1
2
m v
2
P
=
K
P
=
U
Q

U
P
=
m g ‘
(1

cos
θ
)
v
P
=
p
2
g ‘
(1

cos
θ
)
=
q
2 (9
.
8 m
/
s
2
) (1 m) (1

cos 18
◦
)
=
0
.
979435 m
/
s
.
002
(part 2 of 6) 0 points
What is the magnitude of the ball’s accelera
tion at the point
P
?
Correct answer: 0
.
959292 m
/
s
2
.
Explanation:
At point
P
, the centripetal acceleration is
the total acceleration:
a
P
=
v
2
P
‘
=
(0
.
979435 m
/
s)
2
1 m
=
0
.
959292 m
/
s
2
.
003
(part 3 of 6) 10 points
What is the ball’s speed at the point
Q
?
1.
0 m
/
s
<
k
v
Q
k
<
k
v
p
k
2.
k
v
Q
k
= 0 m/s
correct
3.
k
v
Q
k
<
0 m/s
4.
k
v
Q
k
>
k
v
p
k
5.
k
v
Q
k
=
k
v
p
k
Explanation:
Since
Q
is one of the highest points, the
velocity at
Q
is
zero
.
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Thevalingam, Donald – Homework 18 – Due: Feb 7 2007, midnight – Inst: Eslami
2
004
(part 4 of 6) 0 points
What is the magnitude of the ball’s accelera
tion at the point
Q
?
Correct answer: 3
.
02837 m
/
s
2
.
Explanation:
The diagram for the velocity and accelera
tion of the ball at point
Q
is as follows,
x
y
‘
a
v
= 0
θ
Since the velocity at point
Q
is zero, the
centripetal acceleration (which should point
along the string toward the point where the
string is attached) is also zero.
The tangential acceleration is thus the to
tal acceleration. Since it is perpendicular to
the string, the tension of the string does not
contribute to it. It is provided by the tangen
tial component of the gravitational force on
the ball.
a
Q
=
m g
sin
θ
m
=
g
sin
θ
= (9
.
8 m
/
s
2
) sin 18
◦
=
3
.
02837 m
/
s
2
.
005
(part 5 of 6) 10 points
After several swings, the string breaks.
The
mass of the string and air resistance are neg
ligible.
If the break occurs when the ball is at point
P
, describe the motion of the ball after the
break.
1.
It moves horizontally with a constant
speed.
2.
The ball falls vertically straight down.
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 Spring '08
 Turner
 Physics, Acceleration, Energy, Kinetic Energy, Potential Energy, Work, Velocity, Thevalingam

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