This preview shows pages 1–3. Sign up to view the full content.
Oslund (hno56) – Homework 4 – sai – (58015)
1
This printout should have 23 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
(part 1 oF 3) 10.0 points
An athlete swings a 4
.
21 kg ball horizontally
on the end oF a rope. The ball moves in a
circle oF radius 0
.
824 m at an angular speed oF
0
.
663 rev
/
s.
What is the tangential speed oF the ball?
Correct answer: 3
.
43258 m
/
s.
Explanation:
Let :
m
=4
.
21 kg
,
r
=0
.
824 m
,
and
w
=0
.
663 rev
/
s
.
v
t
=
r ω
= (0
.
824 m) (0
.
663 rev
/
s)
·
±
2
π
rad
1 rev
²
=
3
.
43258 m
/
s
.
002
(part 2 oF 3) 10.0 points
What is its centripetal acceleration?
Correct answer: 14
.
2993 m
/
s
2
.
Explanation:
a
c
=
v
2
r
=
r ω
2
= (0
.
824 m) (0
.
663 rev
/
s)
2
·
±
2
π
rad
rev
²
2
=
14
.
2993 m
/
s
2
.
003
(part 3 oF 3) 10.0 points
IF the maximum tension the rope can with
stand beFore breaking is 111 N, what maxi
mum tangential speed can the ball have?
Correct answer: 4
.
66105 m
/
s.
Explanation:
Let :
T
= 111 N
.
The tension supplies the centripetal accel
eration, so
T
=
m
v
t
2
r
v
t
=
³
rT
m
=
´
(0
.
824 m) (111 N)
4
.
21 kg
=
4
.
66105 m
/
s
.
keywords:
004
10.0 points
A space station in the Form oF a large wheel,
324 m in diameter, rotates to provide an “ar
tifcial gravity” oF 7 m
/
s
2
For people located
on the outer rim.
±ind the rotational Frequency oF the wheel
that will produce this e²ect.
Correct answer: 1
.
98501 rpm.
Explanation:
We want the centripetal Force to equal
mg
a
F
r
=
mg
a
=
m
v
2
D
2
.
Solving For
v
,
v
=
³
Dg
a
2
.
The time in seconds to complete one revolu
tion is
T
=
circumFerence
speed
=
π D
³
g
a
D
2
.
Hence, the number oF revolutions per second
is
f
=
1
T
=
³
g
a
D
2
π D
=
³
g
a
2
D π
2
=
´
7m
/
s
2
2(324 m)
π
2
=0
.
0330835 rps
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentOslund (hno56) – Homework 4 – sai – (58015)
2
And the number of revolutions per minute is
f
= (0
.
0330835 rps)
60 s
min
=1
.
98501 rpm
.
005
(part 1 of 2) 10.0 points
At the Six Flags Great America amusement
park in Gurnee, Illinois, there is a roller
coaster that incorporates some of the latest
design technology and some basic physics.
Each vertical loop, instead of being circu
lar, is shaped like a teardrop (shown in the
±gure). The cars ride on the inside of the loop
at the top, and the speeds are high enough
to ensure that the cars remain on the track.
The biggest loop is 35 m high, with a maxi
mum speed of 31 m
/
s (nearly 69
.
44 mph) at
the bottom. Suppose the speed at the top
is 15 m
/
s and the corresponding centripetal
acceleration is 2
.
8
g
(where g is the freefall
acceleration).
What is the radius of the arc of the teardrop
at the top?
Correct answer: 8
.
19971 m.
Explanation:
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 Kaplunovsky
 Physics, Work

Click to edit the document details