This preview shows page 1. Sign up to view the full content.
awkward. We pursue the calculus method but choose to work with 12
gT
2
/2
π
instead of
T
(it should be clear that 12
gT
2
/2
π
is a minimum whenever
T
is a minimum). The result is
d
dx
dx
dx
L
x
gT
L
x
12
2
2
2
2
2
0
12
12
π
e
j
di
==
+
=−
+
which yields
/ 12
(1.85 m)/ 12
0.53 m
xL
=
as the value of
x
which should produce
the smallest possible value of
T
.
(b) With
L
= 1.85 m and
x
= 0.53 m, we obtain
T
= 2.1 s from the expression derived in
part (a).
49. This is similar to the situation treated in Sample Problem 155, except that
O
is no
longer at the end of the stick. Referring to the center of mass as
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, Center Of Mass, Mass

Click to edit the document details