roofner (bar784) – Homework 5 – Weathers – (17104)
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001
10.0 points
A ball on the end of a string is whirled around
in a horizontal circle of radius 0
.
317 m. The
plane of the circle is 1
.
82 m above the ground.
The string breaks and the ball lands 2
.
94 m
away from the point on the ground directly
beneath the ball’s location when the string
breaks.
The acceleration of gravity is 9
.
8 m
/
s
2
.
Find the centripetal acceleration of the ball
during its circular motion.
Correct answer: 73
.
4108 m
/
s
2
.
Explanation:
In order to find the centripetal acceleration
of the ball, we need to find the initial velocity
of the ball.
Let
y
be the distance above the
ground. After the string breaks, the ball has
no initial velocity in the vertical direction, so
the time spent in the air may be deduced from
the kinematic equation,
y
=
1
2
g t
2
.
Solving for
t
,
⇒
t
=
radicalbigg
2
y
g
.
Let
d
be the distance traveled by the ball.
Then
v
x
=
d
t
=
d
radicalbigg
2
y
g
.
Hence, the centripetal acceleration of the ball
during its circular motion is
a
c
=
v
2
x
r
=
d
2
g
2
y r
=
73
.
4108 m
/
s
2
.
002
10.0 points
Before throwing a 1
.
03 kg discus, an ath
lete rotates it along a circular path of radius
1
.
27 m. The maximum speed of the discus is
24 m
/
s.
Determine the magnitude of its maximum
radial acceleration.
Correct answer: 453
.
543 m
/
s
2
.
Explanation:
The maximum radial acceleration is
a
r
=
v
2
r
=
(24 m
/
s)
2
1
.
27 m
= 453
.
543 m
/
s
2
.
003
(part 1 of 3) 10.0 points
A train slows down at a constant rate as it
rounds a sharp circular horizontal turn.
Its
initial speed is not known. It takes 18
.
4 s to
slow down from 77 km
/
h to 28 km
/
h.
The
radius of the curve is 154 m.
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 Fall '08
 Weathers
 mechanics, Acceleration, Work, Velocity, Correct Answer, m/s, vw

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