Engineering Mechanics

(^548) „„„„„ A Textbook of Engineering Mechanics
∴ Frequency of motion,
2 2
11 1
(^22) G
ggh
n
tLhk
== =
ππ+
Example 26.16. A body of mass 0·5 kg oscillates about an axis at a distance 300 mm from
the centre of gravity. If the mass moment of inertia about the centroidal axis, parallel to the axis of
rotation, be 0·125 kg-m^2 , find the length of the equivalent simple pendulum.
Solution. Given : Mass of the body (m) = 0·5 kg ; Distance of Centre of oscillation from the
centre of gravity (h) = 300 mm = 0·3 m and moment of inertia about centroidal axis (IG) = 0·125 kg-m^2
Let kG = Radius of gyration about the centroidal axis.
We know that mass moment of inertia about the centroidal axis (IG)
0·125 = m kG^2 = 0·5 kG^2
∴
2 0·125 0·25
0·5
kG ==
We also know that the length of equivalent simple pendulum,
(^2) 0·25
0·3 1·133 m
0·3
LhkG
h
=+ = + = Ans.
Example 26.17. The pendulum AB of an Izod impact testing machine makes 40 oscillations
per minute about a knife edge through the hole A. The distance of centre of gravity from the centre of
oscillation 450 mm. Find the radius of gyration of the pendulum about an axis through its centre of
gravity and parallel to the knife edge.
Also find the number of oscillations per minute, which the pendulum will make, if supported
on a knife edge through another hole at B such that AB = 1·5 m. Take g as 9·81 m/s^2.
Solution. Given : No. of oscillations/ min = 40 ; Distance of centre of gravity from the centre
of oscillation (l) = 450 mm = 0·45 m ; Length AB = 1·5 m and centre of gravity (g) = 9·81 m/s^2.
Radius of gyration
Let kC = Radius of gyration about an axis through centre of gravity and
parallel to the knife edge, and
L = Length of the equivalent simple pendulum.
We know that frequency of pendulum
40 2
60 3
n==
∴
3
Time period ( ) 1·5 s
2
T ==
We also know that time period (T).
1·5 2 2
9·81
LL
g
=π =π×
Squaring both sides, 2·25 4 2 4·024
9·81
L
=π× = L
∴
2·25
0·559 m
4·024
L==