homework 06 – BAUTISTA, ALDO – Due: Feb 3 2006, 4:00 am
1
Version number encoded for clicker entry:
V1:1, V2:4, V3:2, V4:4, V5:3.
Question 1
Part 1 of 3.
0 points.
An athlete swings a 6
.
72 kg ball horizon
tally on the end of a rope. The ball moves in
a circle of radius 0
.
899 m at an angular speed
of 0
.
558 rev
/
s.
What is the tangential speed of the ball?
Correct answer: 3
.
15191 m
/
s (tolerance
±
1
%).
Explanation:
ω
= 0
.
558 rev
/
s = 3
.
50602 rad
/
s .
v
t
=
r ω
= (0
.
899 m)(3
.
50602 rad
/
s)
= 3
.
15191 m
/
s
.
Question 2
Part 2 of 3.
0 points.
What is its centripetal acceleration?
Correct answer: 11
.
0507 m
/
s
2
(tolerance
±
1
%).
Explanation:
a
r
=
r ω
2
= (0
.
899 m)(3
.
50602 rad
/
s)
2
= 11
.
0507 m
/
s
2
.
Question 3
Part 3 of 3.
0 points.
If the maximum tension the rope can with
stand before breaking is 156 N, what is the
maximum tangential speed the ball can have?
Correct answer: 4
.
56833 m
/
s (tolerance
±
1
%).
Explanation:
F
=
m v
2
r
.
If the maximum value of
T
is 156 N, then we
have
156 N =
(6
.
72 kg)
v
2
0
.
899 m
.
From which,
v
max
= 4
.
56833 m
/
s
.
Question 4
Part 1 of 3.
10 points.
The figure below represents, at a given in
stant, the total acceleration of a particle mov
ing clockwise in a circle of radius 3
.
3 m. The
magnitude of the total acceleration of the par
ticle at the given instant is 17
.
7 m
/
s
2
; the an
gle between the vector of total acceleration
and the radiusvector of the particle is 30
◦
.
3
.
3 m
v
17
.
7 m
/
s
2
30
◦
At this instant of time, find the centripetal
acceleration of the particle.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Turner
 Acceleration, Work, Velocity, Correct Answer, Nonuniform circular motion

Click to edit the document details