This preview shows page 1. Sign up to view the full content.
()
2
2
sin 2
1 cos
2
1
1 2
3 2
tT
ππ
=−
=
where the trigonometric identity cos
2
θ
+
sin
2
= 1 is used. Thus,
=
22
=
2
0.500
3
2
= 34.2
.
Ω−
F
H
I
K
−
F
H
I
K
F
H
G
I
K
J
−
π
π
T
t
T
m
sin
s
rad
rad / s
a
f
During another portion of the cycle its angular speed is +34.2 rad/s when its angular
displacement is
π
/2 rad.
(c) The angular acceleration is
2
2
cos 2
/
.
m
d
dt
T
T
θπ
π
αθ
§·
==
−
=
−
¨¸
©¹
When
=
π
/4,
2
2
2
= 124 rad/s ,
0.500 s
4
α
§
·
−
¨
¸
©
¹
or
2

 124 rad/s .
=
39. (a) We take the angular displacement of the wheel to be
=
m
cos(2
π
t
/
T
), where
m
is the amplitude and
T
is the period. We differentiate with respect to time to find the
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

Click to edit the document details