This preview shows page 1. Sign up to view the full content.
Engineering Mechanics  Dynamics
Chapter 17
g
9.81
m
s
2
=
L
2m
=
Solution:
Since the deflection of the
spring is unchanged, we have
F
A
Mg
2
=
F
A
L
2
1
12
ML
2
α
=
6
F
A
ML
=
14.7
rad
s
2
=
F
A
Mg
−
M
−
a
Gy
=
a
Gy
g
F
A
M
−
=
a
Gy
4.91
m
s
2
=
a
Gx
0
m
s
2
=
*Problem 1768
In order to experimentally determine the moment of inertia
I
G
of a connecting rod of mass
M
,
the rod is suspended horizontally at
A
by a cord and at
B
by a bearing and piezoelectric sensor,
an instrument used for measuring force. Under these equilibrium conditions, the force at
This is the end of the preview. Sign up
to
access the rest of the document.
Unformatted text preview: B is measured as F 1 . If, at the instant the cord is released, the reaction at B is measured as F 2 , determine the value of I G . The support at B does not move when the measurement is taken. For the calculation, the horizontal location of G must be determined. Given: M 4 kg = F 1 14.6 N = F 2 9.3 N = a 350 mm = g 9.81 m s 2 = Solution: Guesses x 1 mm = I G 1 kg m 2 ⋅ = A y 1 N = 1 rad s 2 = 553...
View
Full
Document
This note was uploaded on 08/16/2011 for the course EGN 3321 taught by Professor Christianfeldt during the Spring '08 term at University of Central Florida.
 Spring '08
 ChristianFeldt
 Dynamics

Click to edit the document details