This preview shows page 1. Sign up to view the full content.
α
ω
==
−
=−
Δ
Δ
t
0 150
22
114
rev / min
h)(60 min /1h)
rev / min
2
(.
..
(b) Using Eq. 1013 with
t
= (2.2) (60) = 132 min, the number of revolutions is
()
2
3
0
11
(150 rev/min)(132 min)
1.14 rev/min
132min
9.9 10 rev.
tt
θω
=+
=
+
−
=×
(c) With
r
= 500 mm, the tangential acceleration is
ar
t
−
F
H
G
I
K
J
F
H
G
I
K
J
114
2
.r
e
v
/
m
i
n
2r
a
d
1rev
1m
in
60 s
(500 mm)
2
ch
π
which yields
a
t
= –0.99 mm/s
2
.
(d) The angular speed of the flywheel is
(75 rev/min)(2 rad/rev)(1 min/ 60 s)
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Acceleration

Click to edit the document details